I have a very long conditional statement like the following:
if(test.type == 'itema' || test.type == 'itemb' || test.type == 'itemc' || test.type == 'itemd'){
// do something.
}
I was wondering if I could refactor this expression/statement into a more concise form.
Any idea on how to achieve this?
Put your values into an array, and check if your item is in the array:
if ([1, 2, 3, 4].includes(test.type)) {
// Do something
}
If a browser you support doesn't have the Array#includes method, you can use this polyfill.
Short explanation of the ~ tilde shortcut:
Update: Since we now have the includes method, there's no point in using the ~ hack anymore. Just keeping this here for people that are interested in knowing how it works and/or have encountered it in other's code.
Instead of checking if the result of indexOf is >= 0, there is a nice little shortcut:
if ( ~[1, 2, 3, 4].indexOf(test.type) ) {
// Do something
}
Here is the fiddle: http://jsfiddle.net/HYJvK/
How does this work? If an item is found in the array, indexOf returns its index. If the item was not found, it'll return -1. Without getting into too much detail, the ~ is a bitwise NOT operator, which will return 0 only for -1.
I like using the ~ shortcut, since it's more succinct than doing a comparison on the return value. I wish JavaScript would have an in_array function that returns a Boolean directly (similar to PHP), but that's just wishful thinking (Update: it now does. It's called includes. See above). Note that jQuery's inArray, while sharing PHP's method signature, actually mimics the native indexOf functionality (which is useful in different cases, if the index is what you're truly after).
Important note: Using the tilde shortcut seems to be swathed in controversy, as some vehemently believe that the code is not clear enough and should be avoided at all costs (see the comments on this answer). If you share their sentiment, you should stick to the .indexOf(...) >= 0 solution.
A little longer explanation:
Integers in JavaScript are signed, which means that the left-most bit is reserved as the sign bit; a flag to indicate whether the number is positive or negative, with a 1 being negative.
Here are some sample positive numbers in 32-bit binary format:
1 : 00000000000000000000000000000001
2 : 00000000000000000000000000000010
3 : 00000000000000000000000000000011
15: 00000000000000000000000000001111
Now here are those same numbers, but negative:
-1 : 11111111111111111111111111111111
-2 : 11111111111111111111111111111110
-3 : 11111111111111111111111111111101
-15: 11111111111111111111111111110001
Why such weird combinations for the negative numbers? Simple. A negative number is simply the inverse of the positive number + 1; adding the negative number to the positive number should always yield 0.
To understand this, let's do some simple binary arithmetic.
Here is how we would add -1 to +1:
00000000000000000000000000000001 +1
+ 11111111111111111111111111111111 -1
-------------------------------------------
= 00000000000000000000000000000000 0
And here is how we would add -15 to +15:
00000000000000000000000000001111 +15
+ 11111111111111111111111111110001 -15
--------------------------------------------
= 00000000000000000000000000000000 0
How do we get those results? By doing regular addition, the way we were taught in school: you start at the right-most column, and you add up all the rows. If the sum is greater than the greatest single-digit number (which in decimal is 9, but in binary is 1) we carry the remainder over to the next column.
Now, as you'll notice, when adding a negative number to its positive number, the right-most column that is not all 0s will always have two 1s, which when added together will result in 2. The binary representation of two being 10, we carry the 1 to the next column, and put a 0 for the result in the first column. All other columns to the left have only one row with a 1, so the 1 carried over from the previous column will again add up to 2, which will then carry over... This process repeats itself till we get to the left-most column, where the 1 to be carried over has nowhere to go, so it overflows and gets lost, and we're left with 0s all across.
This system is called 2's Complement. You can read more about this here:
2's Complement Representation for Signed Integers.
Now that the crash course in 2's complement is over, you'll notice that -1 is the only number whose binary representation is 1's all across.
Using the ~ bitwise NOT operator, all the bits in a given number are inverted. The only way to get 0 back from inverting all the bits is if we started out with 1's all across.
So, all this was a long-winded way of saying that ~n will only return 0 if n is -1.
You can use switch statement with fall thru:
switch (test.type) {
case "itema":
case "itemb":
case "itemc":
case "itemd":
// do something
}
Using Science: you should do what idfah said and this for fastest speed while keep code short:
THIS IS FASTER THAN ~ Method
var x = test.type;
if (x == 'itema' ||
x == 'itemb' ||
x == 'itemc' ||
x == 'itemd') {
//do something
}
http://jsperf.com/if-statements-test-techsin
(Top set: Chrome, bottom set: Firefox)
Conclusion :
If possibilities are few and you know that certain ones are more likely to occur than you get maximum performance out if || ,switch fall through , and if(obj[keyval]).
If possibilities are many, and anyone of them could be the most occurring one, in other words, you can't know that which one is most likely to occur than you get most performance out of object lookup if(obj[keyval]) and regex if that fits.
http://jsperf.com/if-statements-test-techsin/12
i'll update if something new comes up.
If you are comparing to strings and there is a pattern, consider using regular expressions.
Otherwise, I suspect attempting to shorten it will just obfuscate your code. Consider simply wrapping the lines to make it pretty.
if (test.type == 'itema' ||
test.type == 'itemb' ||
test.type == 'itemc' ||
test.type == 'itemd') {
do something.
}
var possibilities = {
"itema": 1,
"itemb": 1,
"itemc": 1,
…};
if (test.type in possibilities) { … }
Using an object as an associative array is a pretty common thing, but since JavaScript doesn't have a native set you can use objects as cheap sets as well.
if( /^item[a-d]$/.test(test.type) ) { /* do something */ }
or if the items are not that uniform, then:
if( /^(itema|itemb|itemc|itemd)$/.test(test.type) ) { /* do something */ }
Excellent answers, but you could make the code far more readable by wrapping one of them in a function.
This is complex if statement, when you (or someone else) read the code in a years time, you will be scanning through to find the section to understand what is happening. A statement with this level of business logic will cause you to stumble for a few seconds at while you work out what you are testing. Where as code like this, will allow you to continue scanning.
if(CheckIfBusinessRuleIsTrue())
{
//Do Something
}
function CheckIfBusinessRuleIsTrue()
{
return (the best solution from previous posts here);
}
Name your function explicitly so it immediately obvious what you are testing and your code will be much easier to scan and understand.
You could put all the answers into a Javascript Set and then just call .contains() on the set.
You still have to declare all the contents, but the inline call will be shorter.
Something like:
var itemSet = new Set(["itema","itemb","itemc","itemd"]);
if( itemSet.contains( test.type ){}
One of my favorite ways of accomplishing this is with a library such as underscore.js...
var isItem = _.some(['itema','itemb','itemc','itemd'], function(item) {
return test.type === item;
});
if(isItem) {
// One of them was true
}
http://underscorejs.org/#some
another way or another awesome way i found is this...
if ('a' in oc(['a','b','c'])) { //dosomething }
function oc(a)
{
var o = {};
for(var i=0;i<a.length;i++) o[a[i]]='';
return o;
}
of course as you can see this takes things one step further and make them easy follow logic.
http://snook.ca/archives/javascript/testing_for_a_v
using operators such as ~ && || ((),()) ~~ is fine only if your code breaks later on. You won't know where to start. So readability is BIG.
if you must you could make it shorter.
('a' in oc(['a','b','c'])) && statement;
('a' in oc(['a','b','c'])) && (statements,statements);
('a' in oc(['a','b','c']))?statement:elseStatement;
('a' in oc(['a','b','c']))?(statements,statements):(elseStatements,elseStatements);
and if you want to do inverse
('a' in oc(['a','b','c'])) || statement;
Just use a switch statement instead of if statement:
switch (test.type) {
case "itema":case "itemb":case "itemc":case "itemd":
// do your process
case "other cases":...:
// do other processes
default:
// do processes when test.type does not meet your predictions.
}
Switch also works faster than comparing lots of conditionals within an if
For very long lists of strings, this idea would save a few characters (not saying I'd recommend it in real life, but it should work).
Choose a character that you know won't occur in your test.type, use it as a delimiter, stick them all into one long string and search that:
if ("/itema/itemb/itemc/itemd/".indexOf("/"+test.type+"/")>=0) {
// doSomething
}
If your strings happen to be further constrained, you could even omit the delimiters...
if ("itemaitembitemcitemd".indexOf(test.type)>=0) {
// doSomething
}
...but you'd have to be careful of false positives in that case (e.g. "embite" would match in that version)
For readability create a function for the test (yes, a one line function):
function isTypeDefined(test) {
return test.type == 'itema' ||
test.type == 'itemb' ||
test.type == 'itemc' ||
test.type == 'itemd';
}
then call it:
…
if (isTypeDefined(test)) {
…
}
...
I think there are 2 objectives when writing this kind of if condition.
brevity
readability
As such sometimes #1 might be the fastest, but I'll take #2 for easy maintenance later on. Depending on the scenario I will often opt for a variation of Walter's answer.
To start I have a globally available function as part of my existing library.
function isDefined(obj){
return (typeof(obj) != 'undefined');
}
and then when I actually want to run an if condition similar to yours I'd create an object with a list of the valid values:
var validOptions = {
"itema":1,
"itemb":1,
"itemc":1,
"itemd":1
};
if(isDefined(validOptions[test.type])){
//do something...
}
It isn't as quick as a switch/case statement and a bit more verbose than some of the other examples but I often get re-use of the object elsewhere in the code which can be quite handy.
Piggybacking on one of the jsperf samples made above I added this test and a variation to compare speeds. http://jsperf.com/if-statements-test-techsin/6 The most interesting thing I noted is that certain test combos in Firefox are much quicker than even Chrome.
This can be solved with a simple for loop:
test = {};
test.type = 'itema';
for(var i=['itema','itemb','itemc']; i[0]==test.type && [
(function() {
// do something
console.log('matched!');
})()
]; i.shift());
We use the first section of the for loop to initialize the arguments you wish to match, the second section to stop the for loop from running, and the third section to cause the loop to eventually exit.
This question already has answers here:
Repeat a string in JavaScript a number of times
(24 answers)
Closed last year.
What is the best or most concise method for returning a string repeated an arbitrary amount of times?
The following is my best shot so far:
function repeat(s, n){
var a = [];
while(a.length < n){
a.push(s);
}
return a.join('');
}
Note to new readers: This answer is old and and not terribly practical - it's just "clever" because it uses Array stuff to get
String things done. When I wrote "less process" I definitely meant
"less code" because, as others have noted in subsequent answers, it
performs like a pig. So don't use it if speed matters to you.
I'd put this function onto the String object directly. Instead of creating an array, filling it, and joining it with an empty char, just create an array of the proper length, and join it with your desired string. Same result, less process!
String.prototype.repeat = function( num )
{
return new Array( num + 1 ).join( this );
}
alert( "string to repeat\n".repeat( 4 ) );
I've tested the performance of all the proposed approaches.
Here is the fastest variant I've got.
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
return result + pattern;
};
Or as stand-alone function:
function repeat(pattern, count) {
if (count < 1) return '';
var result = '';
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
return result + pattern;
}
It is based on wnrph's algorithm.
It is really fast. And the bigger the count, the faster it goes compared with the traditional new Array(count + 1).join(string) approach.
I've only changed 2 things:
replaced pattern = this with pattern = this.valueOf() (clears one obvious type conversion);
added if (count < 1) check from prototypejs to the top of function to exclude unnecessary actions in that case.
applied optimisation from Dennis answer (5-7% speed up)
UPD
Created a little performance-testing playground here for those who interested.
variable count ~ 0 .. 100:
constant count = 1024:
Use it and make it even faster if you can :)
This problem is a well-known / "classic" optimization issue for JavaScript, caused by the fact that JavaScript strings are "immutable" and addition by concatenation of even a single character to a string requires creation of, including memory allocation for and copying to, an entire new string.
Unfortunately, the accepted answer on this page is wrong, where "wrong" means by a performance factor of 3x for simple one-character strings, and 8x-97x for short strings repeated more times, to 300x for repeating sentences, and infinitely wrong when taking the limit of the ratios of complexity of the algorithms as n goes to infinity. Also, there is another answer on this page which is almost right (based on one of the many generations and variations of the correct solution circulating throughout the Internet in the past 13 years). However, this "almost right" solution misses a key point of the correct algorithm causing a 50% performance degradation.
JS Performance Results for the accepted answer, the top-performing other answer (based on a degraded version of the original algorithm in this answer), and this answer using my algorithm created 13 years ago
~ October 2000 I published an algorithm for this exact problem which was widely adapted, modified, then eventually poorly understood and forgotten. To remedy this issue, in August, 2008 I published an article http://www.webreference.com/programming/javascript/jkm3/3.html explaining the algorithm and using it as an example of simple of general-purpose JavaScript optimizations. By now, Web Reference has scrubbed my contact information and even my name from this article. And once again, the algorithm has been widely adapted, modified, then poorly understood and largely forgotten.
Original string repetition/multiplication JavaScript algorithm by
Joseph Myers, circa Y2K as a text multiplying function within Text.js;
published August, 2008 in this form by Web Reference:
http://www.webreference.com/programming/javascript/jkm3/3.html (The
article used the function as an example of JavaScript optimizations,
which is the only for the strange name "stringFill3.")
/*
* Usage: stringFill3("abc", 2) == "abcabc"
*/
function stringFill3(x, n) {
var s = '';
for (;;) {
if (n & 1) s += x;
n >>= 1;
if (n) x += x;
else break;
}
return s;
}
Within two months after publication of that article, this same question was posted to Stack Overflow and flew under my radar until now, when apparently the original algorithm for this problem has once again been forgotten. The best solution available on this Stack Overflow page is a modified version of my solution, possibly separated by several generations. Unfortunately, the modifications ruined the solution's optimality. In fact, by changing the structure of the loop from my original, the modified solution performs a completely unneeded extra step of exponential duplicating (thus joining the largest string used in the proper answer with itself an extra time and then discarding it).
Below ensues a discussion of some JavaScript optimizations related to all of the answers to this problem and for the benefit of all.
Technique: Avoid references to objects or object properties
To illustrate how this technique works, we use a real-life JavaScript function which creates strings of whatever length is needed. And as we'll see, more optimizations can be added!
A function like the one used here is to create padding to align columns of text, for formatting money, or for filling block data up to the boundary. A text generation function also allows variable length input for testing any other function that operates on text. This function is one of the important components of the JavaScript text processing module.
As we proceed, we will be covering two more of the most important optimization techniques while developing the original code into an optimized algorithm for creating strings. The final result is an industrial-strength, high-performance function that I've used everywhere--aligning item prices and totals in JavaScript order forms, data formatting and email / text message formatting and many other uses.
Original code for creating strings stringFill1()
function stringFill1(x, n) {
var s = '';
while (s.length < n) s += x;
return s;
}
/* Example of output: stringFill1('x', 3) == 'xxx' */
The syntax is here is clear. As you can see, we've used local function variables already, before going on to more optimizations.
Be aware that there's one innocent reference to an object property s.length in the code that hurts its performance. Even worse, the use of this object property reduces the simplicity of the program by making the assumption that the reader knows about the properties of JavaScript string objects.
The use of this object property destroys the generality of the computer program. The program assumes that x must be a string of length one. This limits the application of the stringFill1() function to anything except repetition of single characters. Even single characters cannot be used if they contain multiple bytes like the HTML entity .
The worst problem caused by this unnecessary use of an object property is that the function creates an infinite loop if tested on an empty input string x. To check generality, apply a program to the smallest possible amount of input. A program which crashes when asked to exceed the amount of available memory has an excuse. A program like this one which crashes when asked to produce nothing is unacceptable. Sometimes pretty code is poisonous code.
Simplicity may be an ambiguous goal of computer programming, but generally it's not. When a program lacks any reasonable level of generality, it's not valid to say, "The program is good enough as far as it goes." As you can see, using the string.length property prevents this program from working in a general setting, and in fact, the incorrect program is ready to cause a browser or system crash.
Is there a way to improve the performance of this JavaScript as well as take care of these two serious problems?
Of course. Just use integers.
Optimized code for creating strings stringFill2()
function stringFill2(x, n) {
var s = '';
while (n-- > 0) s += x;
return s;
}
Timing code to compare stringFill1() and stringFill2()
function testFill(functionToBeTested, outputSize) {
var i = 0, t0 = new Date();
do {
functionToBeTested('x', outputSize);
t = new Date() - t0;
i++;
} while (t < 2000);
return t/i/1000;
}
seconds1 = testFill(stringFill1, 100);
seconds2 = testFill(stringFill2, 100);
The success so far of stringFill2()
stringFill1() takes 47.297 microseconds (millionths of a second) to fill a 100-byte string, and stringFill2() takes 27.68 microseconds to do the same thing. That's almost a doubling in performance by avoiding a reference to an object property.
Technique: Avoid adding short strings to long strings
Our previous result looked good--very good, in fact. The improved function stringFill2() is much faster due to the use of our first two optimizations. Would you believe it if I told you that it can be improved to be many times faster than it is now?
Yes, we can accomplish that goal. Right now we need to explain how we avoid appending short strings to long strings.
The short-term behavior appears to be quite good, in comparison to our original function. Computer scientists like to analyze the "asymptotic behavior" of a function or computer program algorithm, which means to study its long-term behavior by testing it with larger inputs. Sometimes without doing further tests, one never becomes aware of ways that a computer program could be improved. To see what will happen, we're going to create a 200-byte string.
The problem that shows up with stringFill2()
Using our timing function, we find that the time increases to 62.54 microseconds for a 200-byte string, compared to 27.68 for a 100-byte string. It seems like the time should be doubled for doing twice as much work, but instead it's tripled or quadrupled. From programming experience, this result seems strange, because if anything, the function should be slightly faster since work is being done more efficiently (200 bytes per function call rather than 100 bytes per function call). This issue has to do with an insidious property of JavaScript strings: JavaScript strings are "immutable."
Immutable means that you cannot change a string once it's created. By adding on one byte at a time, we're not using up one more byte of effort. We're actually recreating the entire string plus one more byte.
In effect, to add one more byte to a 100-byte string, it takes 101 bytes worth of work. Let's briefly analyze the computational cost for creating a string of N bytes. The cost of adding the first byte is 1 unit of computational effort. The cost of adding the second byte isn't one unit but 2 units (copying the first byte to a new string object as well as adding the second byte). The third byte requires a cost of 3 units, etc.
C(N) = 1 + 2 + 3 + ... + N = N(N+1)/2 = O(N^2). The symbol O(N^2) is pronounced Big O of N squared, and it means that the computational cost in the long run is proportional to the square of the string length. To create 100 characters takes 10,000 units of work, and to create 200 characters takes 40,000 units of work.
This is why it took more than twice as long to create 200 characters than 100 characters. In fact, it should have taken four times as long. Our programming experience was correct in that the work is being done slightly more efficiently for longer strings, and hence it took only about three times as long. Once the overhead of the function call becomes negligible as to how long of a string we're creating, it will actually take four times as much time to create a string twice as long.
(Historical note: This analysis doesn't necessarily apply to strings in source code, such as html = 'abcd\n' + 'efgh\n' + ... + 'xyz.\n', since the JavaScript source code compiler can join the strings together before making them into a JavaScript string object. Just a few years ago, the KJS implementation of JavaScript would freeze or crash when loading long strings of source code joined by plus signs. Since the computational time was O(N^2) it wasn't difficult to make Web pages which overloaded the Konqueror Web browser or Safari, which used the KJS JavaScript engine core. I first came across this issue when I was developing a markup language and JavaScript markup language parser, and then I discovered what was causing the problem when I wrote my script for JavaScript Includes.)
Clearly this rapid degradation of performance is a huge problem. How can we deal with it, given that we cannot change JavaScript's way of handling strings as immutable objects? The solution is to use an algorithm which recreates the string as few times as possible.
To clarify, our goal is to avoid adding short strings to long strings, since in order to add the short string, the entire long string also must be duplicated.
How the algorithm works to avoid adding short strings to long strings
Here's a good way to reduce the number of times new string objects are created. Concatenate longer lengths of string together so that more than one byte at a time is added to the output.
For instance, to make a string of length N = 9:
x = 'x';
s = '';
s += x; /* Now s = 'x' */
x += x; /* Now x = 'xx' */
x += x; /* Now x = 'xxxx' */
x += x; /* Now x = 'xxxxxxxx' */
s += x; /* Now s = 'xxxxxxxxx' as desired */
Doing this required creating a string of length 1, creating a string of length 2, creating a string of length 4, creating a string of length 8, and finally, creating a string of length 9. How much cost have we saved?
Old cost C(9) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 = 45.
New cost C(9) = 1 + 2 + 4 + 8 + 9 = 24.
Note that we had to add a string of length 1 to a string of length 0, then a string of length 1 to a string of length 1, then a string of length 2 to a string of length 2, then a string of length 4 to a string of length 4, then a string of length 8 to a string of length 1, in order to obtain a string of length 9. What we're doing can be summarized as avoiding adding short strings to long strings, or in other words, trying to concatenate strings together that are of equal or nearly equal length.
For the old computational cost we found a formula N(N+1)/2. Is there a formula for the new cost? Yes, but it's complicated. The important thing is that it is O(N), and so doubling the string length will approximately double the amount of work rather than quadrupling it.
The code that implements this new idea is nearly as complicated as the formula for the computational cost. When you read it, remember that >>= 1 means to shift right by 1 byte. So if n = 10011 is a binary number, then n >>= 1 results in the value n = 1001.
The other part of the code you might not recognize is the bitwise and operator, written &. The expression n & 1 evaluates true if the last binary digit of n is 1, and false if the last binary digit of n is 0.
New highly-efficient stringFill3() function
function stringFill3(x, n) {
var s = '';
for (;;) {
if (n & 1) s += x;
n >>= 1;
if (n) x += x;
else break;
}
return s;
}
It looks ugly to the untrained eye, but it's performance is nothing less than lovely.
Let's see just how well this function performs. After seeing the results, it's likely that you'll never forget the difference between an O(N^2) algorithm and an O(N) algorithm.
stringFill1() takes 88.7 microseconds (millionths of a second) to create a 200-byte string, stringFill2() takes 62.54, and stringFill3() takes only 4.608. What made this algorithm so much better? All of the functions took advantage of using local function variables, but taking advantage of the second and third optimization techniques added a twenty-fold improvement to performance of stringFill3().
Deeper analysis
What makes this particular function blow the competition out of the water?
As I've mentioned, the reason that both of these functions, stringFill1() and stringFill2(), run so slowly is that JavaScript strings are immutable. Memory cannot be reallocated to allow one more byte at a time to be appended to the string data stored by JavaScript. Every time one more byte is added to the end of the string, the entire string is regenerated from beginning to end.
Thus, in order to improve the script's performance, one must precompute longer length strings by concatenating two strings together ahead of time, and then recursively building up the desired string length.
For instance, to create a 16-letter byte string, first a two byte string would be precomputed. Then the two byte string would be reused to precompute a four-byte string. Then the four-byte string would be reused to precompute an eight byte string. Finally, two eight-byte strings would be reused to create the desired new string of 16 bytes. Altogether four new strings had to be created, one of length 2, one of length 4, one of length 8 and one of length 16. The total cost is 2 + 4 + 8 + 16 = 30.
In the long run this efficiency can be computed by adding in reverse order and using a geometric series starting with a first term a1 = N and having a common ratio of r = 1/2. The sum of a geometric series is given by a_1 / (1-r) = 2N.
This is more efficient than adding one character to create a new string of length 2, creating a new string of length 3, 4, 5, and so on, until 16. The previous algorithm used that process of adding a single byte at a time, and the total cost of it would be n (n + 1) / 2 = 16 (17) / 2 = 8 (17) = 136.
Obviously, 136 is a much greater number than 30, and so the previous algorithm takes much, much more time to build up a string.
To compare the two methods you can see how much faster the recursive algorithm (also called "divide and conquer") is on a string of length 123,457. On my FreeBSD computer this algorithm, implemented in the stringFill3() function, creates the string in 0.001058 seconds, while the original stringFill1() function creates the string in 0.0808 seconds. The new function is 76 times faster.
The difference in performance grows as the length of the string becomes larger. In the limit as larger and larger strings are created, the original function behaves roughly like C1 (constant) times N^2, and the new function behaves like C2 (constant) times N.
From our experiment we can determine the value of C1 to be C1 = 0.0808 / (123457)2 = .00000000000530126997, and the value of C2 to be C2 = 0.001058 / 123457 = .00000000856978543136. In 10 seconds, the new function could create a string containing 1,166,890,359 characters. In order to create this same string, the old function would need 7,218,384 seconds of time.
This is almost three months compared to ten seconds!
I'm only answering (several years late) because my original solution to this problem has been floating around the Internet for more than 10 years, and apparently is still poorly-understood by the few who do remember it. I thought that by writing an article about it here I would help:
Performance Optimizations for High Speed JavaScript / Page 3
Unfortunately, some of the other solutions presented here are still some of those that would take three months to produce the same amount of output that a proper solution creates in 10 seconds.
I want to take the time to reproduce part of the article here as a canonical answer on Stack Overflow.
Note that the best-performing algorithm here is clearly based on my algorithm and was probably inherited from someone else's 3rd or 4th generation adaptation. Unfortunately, the modifications resulted in reducing its performance. The variation of my solution presented here perhaps did not understand my confusing for (;;) expression which looks like the main infinite loop of a server written in C, and which was simply designed to allow a carefully-positioned break statement for loop control, the most compact way to avoid exponentially replicating the string one extra unnecessary time.
Good news! String.prototype.repeat is now a part of JavaScript.
"yo".repeat(2);
// returns: "yoyo"
The method is supported by all major browsers, except Internet Explorer. For an up to date list, see MDN: String.prototype.repeat > Browser compatibility.
MDN has a polyfill for browsers without support.
This one is pretty efficient
String.prototype.repeat = function(times){
var result="";
var pattern=this;
while (times > 0) {
if (times&1)
result+=pattern;
times>>=1;
pattern+=pattern;
}
return result;
};
String.prototype.repeat is now ES6 Standard.
'abc'.repeat(3); //abcabcabc
Expanding P.Bailey's solution:
String.prototype.repeat = function(num) {
return new Array(isNaN(num)? 1 : ++num).join(this);
}
This way you should be safe from unexpected argument types:
var foo = 'bar';
alert(foo.repeat(3)); // Will work, "barbarbar"
alert(foo.repeat('3')); // Same as above
alert(foo.repeat(true)); // Same as foo.repeat(1)
alert(foo.repeat(0)); // This and all the following return an empty
alert(foo.repeat(false)); // string while not causing an exception
alert(foo.repeat(null));
alert(foo.repeat(undefined));
alert(foo.repeat({})); // Object
alert(foo.repeat(function () {})); // Function
EDIT: Credits to jerone for his elegant ++num idea!
Use Array(N+1).join("string_to_repeat")
Here's a 5-7% improvement on disfated's answer.
Unroll the loop by stopping at count > 1 and perform an additional result += pattnern concat after the loop. This will avoid the loops final previously unused pattern += pattern without having to use an expensive if-check.
The final result would look like this:
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
result += pattern;
return result;
};
And here's disfated's fiddle forked for the unrolled version: http://jsfiddle.net/wsdfg/
/**
#desc: repeat string
#param: n - times
#param: d - delimiter
*/
String.prototype.repeat = function (n, d) {
return --n ? this + (d || '') + this.repeat(n, d) : '' + this
};
this is how to repeat string several times using delimeter.
Tests of the various methods:
var repeatMethods = {
control: function (n,s) {
/* all of these lines are common to all methods */
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return '';
},
divideAndConquer: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
with(Math) { return arguments.callee(floor(n/2), s)+arguments.callee(ceil(n/2), s); }
},
linearRecurse: function (n,s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return s+arguments.callee(--n, s);
},
newArray: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return (new Array(isNaN(n) ? 1 : ++n)).join(s);
},
fillAndJoin: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
var ret = [];
for (var i=0; i<n; i++)
ret.push(s);
return ret.join('');
},
concat: function (n,s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
var ret = '';
for (var i=0; i<n; i++)
ret+=s;
return ret;
},
artistoex: function (n,s) {
var result = '';
while (n>0) {
if (n&1) result+=s;
n>>=1, s+=s;
};
return result;
}
};
function testNum(len, dev) {
with(Math) { return round(len+1+dev*(random()-0.5)); }
}
function testString(len, dev) {
return (new Array(testNum(len, dev))).join(' ');
}
var testTime = 1000,
tests = {
biggie: { str: { len: 25, dev: 12 }, rep: {len: 200, dev: 50 } },
smalls: { str: { len: 5, dev: 5}, rep: { len: 5, dev: 5 } }
};
var testCount = 0;
var winnar = null;
var inflight = 0;
for (var methodName in repeatMethods) {
var method = repeatMethods[methodName];
for (var testName in tests) {
testCount++;
var test = tests[testName];
var testId = methodName+':'+testName;
var result = {
id: testId,
testParams: test
}
result.count=0;
(function (result) {
inflight++;
setTimeout(function () {
result.start = +new Date();
while ((new Date() - result.start) < testTime) {
method(testNum(test.rep.len, test.rep.dev), testString(test.str.len, test.str.dev));
result.count++;
}
result.end = +new Date();
result.rate = 1000*result.count/(result.end-result.start)
console.log(result);
if (winnar === null || winnar.rate < result.rate) winnar = result;
inflight--;
if (inflight==0) {
console.log('The winner: ');
console.log(winnar);
}
}, (100+testTime)*testCount);
}(result));
}
}
Here's the JSLint safe version
String.prototype.repeat = function (num) {
var a = [];
a.length = num << 0 + 1;
return a.join(this);
};
For all browsers
This is about as concise as it gets :
function repeat(s, n) { return new Array(n+1).join(s); }
If you also care about performance, this is a much better approach :
function repeat(s, n) { var a=[],i=0;for(;i<n;)a[i++]=s;return a.join(''); }
If you want to compare the performance of both options, see this Fiddle and this Fiddle for benchmark tests. During my own tests, the second option was about 2 times faster in Firefox and about 4 times faster in Chrome!
For moderns browsers only :
In modern browsers, you can now also do this :
function repeat(s,n) { return s.repeat(n) };
This option is not only shorter than both other options, but it's even faster than the second option.
Unfortunately, it doesn't work in any version of Internet explorer. The numbers in the table specify the first browser version that fully supports the method :
function repeat(pattern, count) {
for (var result = '';;) {
if (count & 1) {
result += pattern;
}
if (count >>= 1) {
pattern += pattern;
} else {
return result;
}
}
}
You can test it at JSFiddle. Benchmarked against the hacky Array.join and mine is, roughly speaking, 10 (Chrome) to 100 (Safari) to 200 (Firefox) times faster (depending on the browser).
Just another repeat function:
function repeat(s, n) {
var str = '';
for (var i = 0; i < n; i++) {
str += s;
}
return str;
}
There are many ways in the ES-Next ways
1. ES2015/ES6 has been realized this repeat() method!
/**
* str: String
* count: Number
*/
const str = `hello repeat!\n`, count = 3;
let resultString = str.repeat(count);
console.log(`resultString = \n${resultString}`);
/*
resultString =
hello repeat!
hello repeat!
hello repeat!
*/
({ toString: () => 'abc', repeat: String.prototype.repeat }).repeat(2);
// 'abcabc' (repeat() is a generic method)
// Examples
'abc'.repeat(0); // ''
'abc'.repeat(1); // 'abc'
'abc'.repeat(2); // 'abcabc'
'abc'.repeat(3.5); // 'abcabcabc' (count will be converted to integer)
// 'abc'.repeat(1/0); // RangeError
// 'abc'.repeat(-1); // RangeError
2. ES2017/ES8 new add String.prototype.padStart()
const str = 'abc ';
const times = 3;
const newStr = str.padStart(str.length * times, str.toUpperCase());
console.log(`newStr =`, newStr);
// "newStr =" "ABC ABC abc "
3. ES2017/ES8 new add String.prototype.padEnd()
const str = 'abc ';
const times = 3;
const newStr = str.padEnd(str.length * times, str.toUpperCase());
console.log(`newStr =`, newStr);
// "newStr =" "abc ABC ABC "
refs
http://www.ecma-international.org/ecma-262/6.0/#sec-string.prototype.repeat
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd
function repeat(s, n) { var r=""; for (var a=0;a<n;a++) r+=s; return r;}
This may be the smallest recursive one:-
String.prototype.repeat = function(n,s) {
s = s || ""
if(n>0) {
s += this
s = this.repeat(--n,s)
}
return s}
Fiddle: http://jsfiddle.net/3Y9v2/
function repeat(s, n){
return ((new Array(n+1)).join(s));
}
alert(repeat('R', 10));
Simple recursive concatenation
I just wanted to give it a bash, and made this:
function ditto( s, r, c ) {
return c-- ? ditto( s, r += s, c ) : r;
}
ditto( "foo", "", 128 );
I can't say I gave it much thought, and it probably shows :-)
This is arguably better
String.prototype.ditto = function( c ) {
return --c ? this + this.ditto( c ) : this;
};
"foo".ditto( 128 );
And it's a lot like an answer already posted - I know this.
But why be recursive at all?
And how about a little default behaviour too?
String.prototype.ditto = function() {
var c = Number( arguments[ 0 ] ) || 2,
r = this.valueOf();
while ( --c ) {
r += this;
}
return r;
}
"foo".ditto();
Because, although the non recursive method will handle arbitrarily large repeats without hitting call stack limits, it's a lot slower.
Why did I bother adding more methods that aren't half as clever as those already posted?
Partly for my own amusement, and partly to point out in the simplest way I know that there are many ways to skin a cat, and depending on the situation, it's quite possible that the apparently best method isn't ideal.
A relatively fast and sophisticated method may effectively crash and burn under certain circumstances, whilst a slower, simpler method may get the job done - eventually.
Some methods may be little more than exploits, and as such prone to being fixed out of existence, and other methods may work beautifully in all conditions, but are so constructed that one simply has no idea how it works.
"So what if I dunno how it works?!"
Seriously?
JavaScript suffers from one of its greatest strengths; it's highly tolerant of bad behaviour, and so flexible it'll bend over backwards to return results, when it might have been better for everyone if it'd snapped!
"With great power, comes great responsibility" ;-)
But more seriously and importantly, although general questions like this do lead to awesomeness in the form of clever answers that if nothing else, expand one's knowledge and horizons, in the end, the task at hand - the practical script that uses the resulting method - may require a little less, or a little more clever than is suggested.
These "perfect" algorithms are fun and all, but "one size fits all" will rarely if ever be better than tailor made.
This sermon was brought to you courtesy of a lack of sleep and a passing interest.
Go forth and code!
Firstly, the OP's questions seems to be about conciseness - which I understand to mean "simple and easy to read", while most answers seem to be about efficiency - which is obviously not the same thing and also I think that unless you implement some very specific large data manipulating algorithms, shouldn't worry you when you come to implement basic data manipulation Javascript functions. Conciseness is much more important.
Secondly, as André Laszlo noted, String.repeat is part of ECMAScript 6 and already available in several popular implementations - so the most concise implementation of String.repeat is not to implement it ;-)
Lastly, if you need to support hosts that don't offer the ECMAScript 6 implementation, MDN's polyfill mentioned by André Laszlo is anything but concise.
So, without further ado - here is my concise polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : this.concat(this.repeat(n-1));
}
Yes, this is a recursion. I like recursions - they are simple and if done correctly are easy to understand. Regarding efficiency, if the language supports it they can be very efficient if written correctly.
From my tests, this method is ~60% faster than the Array.join approach. Although it obviously comes nowhere close disfated's implementation, it is much simpler than both.
My test setup is node v0.10, using "Strict mode" (I think it enables some sort of TCO), calling repeat(1000) on a 10 character string a million times.
If you think all those prototype definitions, array creations, and join operations are overkill, just use a single line code where you need it. String S repeating N times:
for (var i = 0, result = ''; i < N; i++) result += S;
Use Lodash for Javascript utility functionality, like repeating strings.
Lodash provides nice performance and ECMAScript compatibility.
I highly recommend it for UI development and it works well server side, too.
Here's how to repeat the string "yo" 2 times using Lodash:
> _.repeat('yo', 2)
"yoyo"
Recursive solution using divide and conquer:
function repeat(n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
with(Math) { return repeat(floor(n/2), s)+repeat(ceil(n/2), s); }
}
I came here randomly and never had a reason to repeat a char in javascript before.
I was impressed by artistoex's way of doing it and disfated's results. I noticed that the last string concat was unnecessary, as Dennis also pointed out.
I noticed a few more things when playing with the sampling disfated put together.
The results varied a fair amount often favoring the last run and similar algorithms would often jockey for position. One of the things I changed was instead of using the JSLitmus generated count as the seed for the calls; as count was generated different for the various methods, I put in an index. This made the thing much more reliable. I then looked at ensuring that varying sized strings were passed to the functions. This prevented some of the variations I saw, where some algorithms did better at the single chars or smaller strings. However the top 3 methods all did well regardless of the string size.
Forked test set
http://jsfiddle.net/schmide/fCqp3/134/
// repeated string
var string = '0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789';
// count paremeter is changed on every test iteration, limit it's maximum value here
var maxCount = 200;
var n = 0;
$.each(tests, function (name) {
var fn = tests[name];
JSLitmus.test(++n + '. ' + name, function (count) {
var index = 0;
while (count--) {
fn.call(string.slice(0, index % string.length), index % maxCount);
index++;
}
});
if (fn.call('>', 10).length !== 10) $('body').prepend('<h1>Error in "' + name + '"</h1>');
});
JSLitmus.runAll();
I then included Dennis' fix and decided to see if I could find a way to eek out a bit more.
Since javascript can't really optimize things, the best way to improve performance is to manually avoid things. If I took the first 4 trivial results out of the loop, I could avoid 2-4 string stores and write the final store directly to the result.
// final: growing pattern + prototypejs check (count < 1)
'final avoid': function (count) {
if (!count) return '';
if (count == 1) return this.valueOf();
var pattern = this.valueOf();
if (count == 2) return pattern + pattern;
if (count == 3) return pattern + pattern + pattern;
var result;
if (count & 1) result = pattern;
else result = '';
count >>= 1;
do {
pattern += pattern;
if (count & 1) result += pattern;
count >>= 1;
} while (count > 1);
return result + pattern + pattern;
}
This resulted in a 1-2% improvement on average over Dennis' fix. However, different runs and different browsers would show a fair enough variance that this extra code probably isn't worth the effort over the 2 previous algorithms.
A chart
Edit: I did this mostly under chrome. Firefox and IE will often favor Dennis by a couple %.
Simple method:
String.prototype.repeat = function(num) {
num = parseInt(num);
if (num < 0) return '';
return new Array(num + 1).join(this);
}
People overcomplicate this to a ridiculous extent or waste performance. Arrays? Recursion? You've got to be kidding me.
function repeat (string, times) {
var result = ''
while (times-- > 0) result += string
return result
}
Edit. I ran some simple tests to compare with the bitwise version posted by artistoex / disfated and a bunch of other people. The latter was only marginally faster, but orders of magnitude more memory-efficient. For 1000000 repeats of the word 'blah', the Node process went up to 46 megabytes with the simple concatenation algorithm (above), but only 5.5 megabytes with the logarithmic algorithm. The latter is definitely the way to go. Reposting it for the sake of clarity:
function repeat (string, times) {
var result = ''
while (times > 0) {
if (times & 1) result += string
times >>= 1
string += string
}
return result
}
Concatenating strings based on an number.
function concatStr(str, num) {
var arr = [];
//Construct an array
for (var i = 0; i < num; i++)
arr[i] = str;
//Join all elements
str = arr.join('');
return str;
}
console.log(concatStr("abc", 3));
Hope that helps!
With ES8 you could also use padStart or padEnd for this. eg.
var str = 'cat';
var num = 23;
var size = str.length * num;
"".padStart(size, str) // outputs: 'catcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcat'
To repeat a string in a specified number of times, we can use the built-in repeat() method in JavaScript.
Here is an example that repeats the following string for 4 times:
const name = "king";
const repeat = name.repeat(4);
console.log(repeat);
Output:
"kingkingkingking"
or we can create our own verison of repeat() function like this:
function repeat(str, n) {
if (!str || !n) {
return;
}
let final = "";
while (n) {
final += s;
n--;
}
return final;
}
console.log(repeat("king", 3))
(originally posted at https://reactgo.com/javascript-repeat-string/)