For example, getting "5" in "256". The closest I've gotten is Math.floor(256/10)), but that'll still return the numbers in front. Is there any simple way to get what I want or would I have to make a big function for it? Also, for clarity: "n digit" would be defined. Example, getDigit(2,256) would return 5 (second digit)
Math.floor((256 / 10) % 10)
or more generally:
Math.floor(N / (Math.pow(10, n)) % 10)
where N is the number to be extracted, and n is the position of the digit. Note that this counts from 0 starting from the right (i.e., the least significant digit = 0), and doesn't account for invalid values of n.
how about
(12345 + "")[3]
or
(12345 + "").charAt(3)
to count from the other end
[length of string - digit you want] so if you want the 2 it's:
5 - 4 = 1
(12345 + "")[1] = "2"
function getNumber (var num, var pos){
var sNum = num + "";
if(pos > sNum.length || pos <= 0){return "";}
return sNum[sNum.length - pos];
}
First, you need to cast the number to a string, then you can access the character as normal:
var num = 256;
var char = num.toString()[1]; // get the 2nd (0-based index) character from the stringified version of num
Edit: Note also that, if you want to access it without setting the number as a variable first, you need a double dot .. to access the function:
var char = 256..toString()[1];
The first dot tells the interpreter "this is a number"; the second accesses the function.
Convert to string and substring(2,2)?
This should do it:
function getDigit ( position, number ) {
number = number + ""; // convert number to string
return number.substr ( position + 1, 1 ); // I'm adding 1 to position, since 0 is the position of the first character and so on
}
Try this, last line is key:
var number = 12345;
var n = 2;
var nDigit = parseInt((number + '').substr(1,1));
If you want to try to do everything mathematically:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,digitNum-1))%10;
This code counts the digit from the right starting with 1, not 0. If you wish to change it to start at 0, delete the -1 portion in the call.
If you wish to count from the left, it gets more complicated and similar to other solutions:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,number.tostring().length-digitNum))%10;
edit:
Also, this assumes you want base 10 for your number system, but both of those will work with other bases. All you need to do is change instances of 10 in the final line of code to the number representing the base for the number system you'd like to use. (ie. hexadecimal =16, binary = 2)
// You do not say if you allow decimal fractions or negative numbers-
// the strings of those need adjusting.
Number.prototype.nthDigit= function(n){
var s= String(this).replace(/\D+/g,'');
if(s.length<=n) return null;
return Number(s.charAt(n))
}
use variable "count" to control loop
var count = 1; //starting 1
for(i=0; i<100; i++){
console.log(count);
if(i%10 == 0) count++;
}
output will fill
1
2
3
4
5
6
7
8
9
Related
Given any number between 0 and 1, such as 0.84729347293923, is there a simple way to make it into 84729347293923 without string or regex manipulation? I can think of using a loop, which probably is no worse than using a string because it is O(n) with n being the number of digits. But is there a better way?
function getRandom() {
let r = Math.random();
while (Math.floor(r) !== r) r *= 10;
return r;
}
for (let i = 0; i < 10; i++)
console.log(getRandom());
Integers mod 1 = 0, non integers mod 1 != 0.
while ((r*=10) % 1);
Ok, just want to refactor my code (i realized that was bad so this is what i discovered to correctly get the value as you requested).
NOTE: As the question says that "given any number between 0 and 1", this solution only works for values between 0 and 1:
window.onload = ()=>{
function getLen(num){
let currentNumb = num;
let integratedArray = [];
let realLen = 0;
/*While the number is not an integer, we will multiply the copy of the original
*value by ten, and when the loop detects that the number is already an integer
*the while simply breaks, in this process we are storing each transformations
*of the number in an array called integratedArray*/
while(!(Number.isInteger(currentNumb))){
currentNumb *= 10;
integratedArray.push(currentNumb);
}
/*We iterate over the array and compare each value of the array with an operation
*in which the resultant value should be exactly the same as the actual item of the
*array, in the case that both are equal we assign the var realLen to i, and
*in case that the values were not the same, we simply breaks the loop, if the
*values are not the same, this indicates that we found the "trash numbers", so
*we simply skip them.*/
for(let i = 0; i < integratedArray.length; i++){
if(Math.floor(integratedArray[i]) === Math.floor(num * Math.pow(10, i + 1))){
realLen = i;
}else{
break;
}
}
return realLen;
}
//Get the float value of a number between 0 and 1 as an integer.
function getShiftedNumber(num){
//First we need the length to get the float part of the number as an integer
const len = getLen(num);
/*Once we have the length of the number we simply multiply the number by
*(10) ^ numberLength, this eliminates the comma (,), or point (.), and
*automatically transforms the number to an integer in this case a large integer*/
return num * (Math.pow(10, len));
}
console.log(getShiftedNumber(0.84729347293923));
}
So the explanation is the next:
Because we want to convert this number without using any string, regex or any another thing, first we need to get the length of the number, this is a bit hard to do without using string conversions... so i did the function getLen for this purpose.
In the function getLen, we have 3 variables:
currentNumb: This var is a copy of the original value (the original number), this value help us to found the length of the number and we can do some transforms to this value whitout changing the original reference of the number.
We need to multiply this value any times is needed to transform the number to an integer and then multiplyng this value by ten to ten.
with the help of a while (this method makes the number a false integer).
NOTE: I saw "False integer" because when i was making the tests i realized that in the number is being adding more digits than normal... (Very very strange), so this stupid but important thing makes neccesary the filter of these "trash numbers", so later we proccess them.
integratedArray: This array stores the values of the result of the first while operations, so the last number stored in this array is an integer, but this number is one of the "fake integers", so with this array we need to iterate later to compare what of those stored values are different to the original value multiplied by (10 * i + 1), so here is the hint:
In this case the first 12 values of this array are exactly the same with the operation of Math.floor(num * Math.pow(10, i + 1))), but in the 13th value of the array these values are not the same so... yes!, there are those "trash numbers" that we were searching for.
realLen: This is the variable where we will store the real length of the number converting the float part of this number in an integer.
Some binary search approach:
Its useless if avarage length < 8;
It contains floating point issues.
But hey it is O(log n) with tons of wasted side computations - i guess if one counts them its event worse than just plain multiplication.
I prefer #chiliNUT answer. One line stamp.
function floatToIntBinarySearch(number){
const max_safe_int_length = 16;
const powers = [
1,
10,
100,
1000,
10000,
100000,
1000000,
10000000,
100000000,
1000000000,
10000000000,
100000000000,
1000000000000,
10000000000000,
100000000000000,
1000000000000000,
10000000000000000
]
let currentLength = 16
let step = 16
let _number = number * powers[currentLength]
while(_number % 1 != 0 || (_number % 10 | 0) == 0){
step /= 2
if( (_number % 10 | 0) == 0 && !(_number % 1 != 0)){
currentLength = currentLength - step;
} else {
currentLength = step + currentLength;
}
if(currentLength < 1 || currentLength > max_safe_int_length * 2) throw Error("length is weird: " + currentLength)
_number = number * powers[currentLength]
console.log(currentLength, _number)
if(Number.isNaN(_number)) throw Error("isNaN: " + ((number + "").length - 2) + " maybe greater than 16?")
}
return number * powers[currentLength]
}
let randomPower = 10 ** (Math.random() * 10 | 0)
let test = (Math.random() * randomPower | 0) / randomPower
console.log(test)
console.log(floatToIntBinarySearch(test))
var ShortURL = new function() {
var _alphabet = '23456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ-_',
_base = _alphabet.length;
this.encode = function(num) {
var str = '';
while (num > 0) {
str = _alphabet.charAt(num % _base) + str;
num = Math.floor(num / _base);
}
return str;
};
this.decode = function(str) {
var num = 0;
for (var i = 0; i < str.length; i++) {
num = num * _base + _alphabet.indexOf(str.charAt(i));
}
return num;
};
};
I understand encode works by converting from decimal to custom base (custom alphabet/numbers in this case)
I am not quite sure how decode works.
Why do we multiply base by a current number and then add the position number of the alphabet? I know that to convert 010 base 2 to decimal, we would do
(2 * 0^2) + (2 * 1^1) + (2 * 0 ^ 0) = 2
Not sure how it is represented in that decode algorithm
EDIT:
My own decode version
this.decode2 = function (str) {
var result = 0;
var position = str.length - 1;
var value;
for (var i = 0; i < str.length; i++) {
value = _alphabet.indexOf(str[i]);
result += value * Math.pow(_base, position--);
}
return result;
}
This is how I wrote my own decode version (Just like I want convert this on paper. I would like someone to explain more in detail how the first version of decode works. Still don't get why we multiply num * base and start num with 0.
OK, so what does 376 mean as a base-10 output of your encode() function? It means:
1 * 100 +
5 * 10 +
4 * 1
Why? Because in encode(), you divide by the base on every iteration. That means that, implicitly, the characters pushed onto the string on the earlier iterations gain in significance by a factor of the base each time through the loop.
The decode() function, therefore, multiplies by the base each time it sees a new character. That way, the first digit is multiplied by the base once for every digit position past the first that it represents, and so on for the rest of the digits.
Note that in the explanation above, the 1, 5, and 4 come from the positions of the characters 3, 7, and 6 in the "alphabet" list. That's how your encoding/decoding mechanism works. If you feed your decode() function a numeric string encoded by something trying to produce normal base-10 numbers, then of course you'll get a weird result; that's probably obvious.
edit To further elaborate on the decode() function: forget (for now) about the special base and encoding alphabet. The process is basically the same regardless of the base involved. So, let's look at a function that interprets a base-10 string of numeric digits as a number:
function decode10(str) {
var num = 0, zero = '0'.charCodeAt(0);
for (var i = 0; i < str.length; ++i) {
num = (num * 10) + (str[i] - zero);
}
return num;
}
The accumulator variable num is initialized to 0 first, because before examining any characters of the input numeric string the only value that makes sense to start with is 0.
The function then iterates through each character of the input string from left to right. On each iteration, the accumulator is multiplied by the base, and the digit value at the current string position is added.
If the input string is "214", then, the iteration will proceed as follows:
num is set to 0
First iteration: str[i] is 2, so (num * 10) + 2 is 2
Second iteration: str[i] is 1, so (num * 10) + 1 is 21
Third iteration: str[i] is 4, so (num * 10) + 4 is 214
The successive multiplications by 10 achieve what the call to Math.pow() does in your code. Note that 2 is multiplied by 10 twice, which effectively multiplies it by 100.
The decode() routine in your original code does the same thing, only instead of a simple character code computation to get the numeric value of a digit, it performs a lookup in the alphabet string.
Both the original and your own version of the decode function achieve the same thing, but the original version does it more efficiently.
In the following assignment:
num = num * _base + _alphabet.indexOf(str.charAt(i));
... there are two parts:
_alphabet.indexOf(str.charAt(i))
The indexOf returns the value of a digit in base _base. You have this part in your own algorithm, so that should be clear.
num * _base
This multiplies the so-far accumulated result. The rest of my answer is about that part:
In the first iteration this has no effect, as num is still 0 at that point. But at the end of the first iteration, num contains the value as if the str only had its left most character. It is the base-51 digit value of the left most digit.
From the next iteration onwards, the result is multiplied by the base, which makes room for the next value to be added to it. It functions like a digit shift.
Take this example input to decode:
bd35
The individual characters represent value 8, 10, 1 and 3. As there are 51 characters in the alphabet, we're in base 51. So bd35 this represents value:
8*51³ + 10*51² + 1*51 + 3
Here is a table with the value of num after each iteration:
8
8*51 + 10
8*51² + 10*51 + 1
8*51³ + 10*51² + 1*51 + 3
Just to make the visualisation cleaner, let's put the power of 51 in a column header, and remove that from the rows:
3 2 1 0
----------------------------
8
8 10
8 10 1
8 10 1 3
Note how the 8 shifts to the left at each iteration and gets multiplied with the base (51). The same happens with 10, as soon as it is shifted in from the right, and the same with the 1, and 3, although that is the last one and doesn't shift any more.
The multiplication num * _base represents thus a shift of base-digits to the left, making room for a new digit to shift in from the right (through simple addition).
At the last iteration all digits have shifted in their correct position, i.e. they have been multiplied by the base just enough times.
Putting your own algorithm in the same scheme, you'd have this table:
3 2 1 0
----------------------------
8
8 10
8 10 1
8 10 1 3
Here, there is no shifting: the digits are immediately put in the right position, i.e. they are multiplied with the correct power of 51 immediately.
You ask
I would like to understand how the decode function works from logical perspective. Why are we using num * base and starting with num = 0.
and write that
I am not quite sure how decode works. Why do we multiply base by a
current number and then add the position number of the alphabet? I
know that to convert 010 base 2 to decimal, we would do
(2 * 0^2) + (2 * 1^1) + (2 * 0 ^ 0) = 2
The decode function uses an approach to base conversion known as Horner's rule, used because it is computationally efficient:
start with a variable set to 0, num = 0
multiply the variable num by the base
take the value of the most significant digit (the leftmost digit) and add it to num,
repeat step 2 and 3 for as long as there are digits left to convert,
the variable num now contains the converted value (in base 10)
Using an example of a hexadecimal number A5D:
start with a variable set to 0, num = 0
multiply by the base (16), num is now still 0
take the value of the most significant digit (the A has a digit value of 10) and add it to num, num is now 10
repeat step 2, multiply the variable num by the base (16), num is now 160
repeat step 3, add the hexadecimal digit 5 to num, num is now 165
repeat step 2, multiply the variable num by the base (16), num is now 2640
repeat step 3, add the hexadecimal digit D to num (add 13)
there are no digits left to convert, the variable num now contains the converted value (in base 10), which is 2653
Compare the expression of the standard approach:
(10 × 162) + (5 × 161) + (13 × 160) = 2653
to the use of Horner's rule:
(((10 × 16) + 5) × 16) + 13 = 2653
which is exactly the same computation, but rearranged in a form making it easier to compute. This is how the decode function works.
Why are we using num * base and starting with num = 0.
The conversion algorithm needs a start value, therefore num is set to 0. For each repetition (each loop iteration), num is multiplied by base. This only has any effect on the second iteration, but is written like this to make it easier to write the conversion as a for loop.
I need to get the first 2 non zero digits from a decimal number. How can this be achieved?
Suppose I have number like 0.000235 then I need 0.00023, if the number is 0.000000025666 then my function should return 0.000000025.
Can any one have an idea of how this can be achieved in javascript?
The result should be a float number not a string.
Here are two faster solutions (see jsperf) :
Solution 1 :
var n = 0.00000020666;
var r = n.toFixed(1-Math.floor(Math.log(n)/Math.log(10)));
Note that this one doesn't round to the smallest value but to the nearest : 0.0256 gives 0.026, not 0.025. If you really want to round to the smallest, use this one :
Solution 2 :
var r = n.toFixed(20).match(/^-?\d*\.?0*\d{0,2}/)[0];
It works with negative numbers too.
var myNum = 0.000256
var i = 1
while(myNum < 10){
myNum *= 10
i *= 10
}
var result = parseInt(myNum) / i
With numbers that has that many decimals, you'd have to return a string, as any parsing as number would return scientific notation, as in 0.000000025666 would be 2.5666e-8
function round(n, what) {
var i = 0;
if (n < 1) {
while(n < 1) {
n = n*10;
i++;
}
}
return '0.' + (new Array(i)).join('0') + n.toFixed(what).replace('.','').slice(0,-1);
}
FIDDLE
This question already has answers here:
Truncate (not round off) decimal numbers in javascript
(32 answers)
Closed 8 years ago.
Im trying to get a number with precision to 2 decimals, for example this is what I want, if I have the numbers:
3.456 it must returns me 3.45
3.467 = 3.46
3.435 = 3.43
3.422 = 3.42
I don't want to round up or down or whatever just to get the numbers I see 2 places after .
Thanks
Okay, here is the answer
var a = 5.469923;
var truncated = Math.floor(a * 100) / 100; // = 5.46
Thanks everyone for helping.
Assuming Positive Numbers:
The code:
function roundDown(num,dec) {
return Math.floor(num*Math.pow(10,dec))/Math.pow(10,dec);
}
The test:
function test(num, expected) {
var val = roundDown(num,2);
var pass = val === expected;
var result = pass ? "PASS" : "FAIL";
var color = pass ? "GREEN" : "RED";
console.log("%c" + result + " : " + num + " : " + val, "background-color:" + color);
}
test(3.456, 3.45);
test(3.467, 3.46);
test(3.435, 3.43);
test(3.422, 3.42);
Basic idea:
Take number
Multiply the number to move decimal place to number of significant figures you want
Floor the number to remove the trailing numbers
Divide number back to get the correct value
If you want to have a trailing zero, you need to use toFixed(2) which will turn the number to a string.
function roundDown(num,dec) {
return Math.floor(num*Math.pow(10,dec))/Math.pow(10,dec).toFixed(2);
}
and the test cases would need to change to
test(3.456, "3.45");
test(3.467, "3.46");
test(3.435, "3.43");
test(3.422, "3.42");
Another option is a regular expression.
function roundDown(num,dec) {
var x = num.toString().match(/(\d*(\.\d{2}))?/);
return x ? parseFloat(x[0]) : "";
//return x ? parseFloat(x[0]).toFixed(2) : "";
}
Use String operation to achieve it.
var n = 4.56789;
var numbers = n.toString().split('.');
result = Number(numbers[0]+"."+numbers[1].substr(0,2));
alert(result);
Fiddle
You are looking at the number as if it were a string of digits, rather than a single value, so treat it like a string.-
function cutoff(n, cut){
var parts= String(n).split('.'), dec= parts[1];
if(!cut) return parts[0];
if(dec && dec.length>cut) parts[1]= dec.substring(0, cut);
return parts.join('.');
}
var n= 36.938;
cutoff(n,2)
/* returned value: (String)
36.93
*/
If you want a number, +cutoff(n,2) will do.
function truncateDec(num, decplaces) {
return (num*Math.pow(10,decplaces) - num*Math.pow(10,decplaces) % 1)/Math.pow(10,decplaces);
}
alert(truncateDec(105.678, 2)); // Returns 105.67
alert(truncateDec(105.678, 1)); // Returns 105.6
This could be simplified further if you do not require a dynamic number of decimal places
function truncateDec(num) {
return (num*100 - num*100 % 1)/100;
}
alert(truncateDec(105.678)); // Returns 105.67
How does it work?
The concept is that the main truncation works by getting the remainder from dividing the original decimal by 1. The remainder will be whatever is in the decimals places. The remainder operator is %
105.678 % 1 = 0.678
By subtracting this remainder from the original number, we will be left with only the integer.
105.678 - 0.678 = 105
To include x number of decimal places, we need to first multiply the original number by 10 to the power of that number of decimal places, thereby shifting the decimal backward by x positions. In this example, we will take x = 2.
105.678 * 10^2
= 105.678 * 100
= 10567.8
Now, we repeat the same procedure by subtracting the remainder again.
10567.8 % 1 = 0.8
10567.8 - 0.8 = 10567
And to return back to the number of places as requested, we divide it back by 10^x
10567 / 10^2
= 10567 / 100
= 105.67
Hope it helps!
So, I have successfully written the Fibonacci sequence to create an array with the sequence of numbers, but I need to know the length (how many digits) the 500th number has.
I've tried the below code, but its finding the length of the scientific notation (22 digits), not the proper 105 it should be returning.
Any ideas how to convert a scientific notation number into an actual integer?
var fiblength = function fiblength(nth) {
var temparr = [0,1];
for(var i = 2; i<=nth; i++){
var prev = temparr[temparr.length-2],
cur = temparr[temparr.length-1],
next = prev + cur;
temparr.push(next);
}
var final = temparr[temparr.length-1].toString().length;
console.log(temparr[temparr.length-1]);
return final;
};
a = fiblength(500);
console.log(a);
Why not use the simple procedure of dividing the number by 10 until the number is less than 1.
Something as simple as this should work (a recursive def obv works as well)
function getDigits(n) {
var digits = 0;
while(n >= 1) {
n/=10;
digits += 1;
}
return digits;
}
getDigits(200);//3
getDigits(3.2 * 10e20);//=>22
Here's a solution in constant time:
function fiblength(n) {
return Math.floor((n>1)?n*.2089+.65051:1);
}
Let's explain how I arrived to it.
All previous solutions will probably not work for N>300 unless you have a BigNumber library in place. Also they're pretty inneficient.
There is a formula to get any Fibonacci number, which uses PHI (golden ratio number), it's very simple:
F(n) = ABS((PHI^n)/sqrt(5))
Where PHI=1.61803399 (golden ratio, found all over the fibonacci sequence)
If you want to know how many digits a number has, you calculate the log base 10 and add 1 to that. Let's call that function D(n) = log10(n) + 1
So what you want fiblength to be is in just the following function
fiblength(n) = D(F(n)) // number of digits of a fibonacci number...
Let's work it out, so you see what the one liner code will be like once you use math.
Substitute F(n)
fiblength(n) = D(ABS((PHI^n)/sqrt(5)))
Now apply D(n) on that:
fiblength(n) = log10(ABS((PHI^n)/sqrt(5))) + 1
So, since log(a/b) = log(a) - log(b)
fiblength(n) = log10(ABS((PHI^n))) - log10(sqrt(5))) + 1
and since log(a^n) = n * log(a)
fiblength(n) = n*log10(PHI) - log10(sqrt(5))) + 1
Then we evaluate those logarithms since they're all on constants
and add the special cases of n=0 and n=1 to return 1
function fiblength(n) {
return Math.floor((n>1)?n*.2089+.65051:1);
}
Enjoy :)
fiblength(500) => 105 //no iterations necessary.
Most of the javascript implementations, internally use 64 bit numbers. So, if the number we are trying to represent is very big, it uses scientific notation to represent those numbers. So, there is no pure "javascript numbers" based solution for this. You may have to look for other BigNum libraries.
As far as your code is concerned, you want only the 500th number, so you don't have to store the entire array of numbers in memory, just previous and current numbers are enough.
function fiblength(nth) {
var previous = 0, current = 1, temp;
for(var i = 2; i<=nth; i++){
temp = current;
current = previous + current;
previous = temp;
}
return current;
};
My Final Solution
function fiblength(nth) {
var a = 0, b = 1, c;
for(var i=2;i<=nth;i++){
c=b;
b=a+b;
a=c;
}
return Math.floor(Math.log(b)/Math.log(10))+1;
}
console.log(fiblength(500));
Thanks for the help!!!
The problem is because the resulting number was converted into a string before any meaningful calculations could be made. Here's how it could have been solved in the original code:
var fiblength = function fiblength(nth) {
var temparr = [0,1];
for(var i = 2; i<=nth; i++){
var prev = temparr[temparr.length-2],
cur = temparr[temparr.length-1],
next = prev + cur;
temparr.push(next);
}
var x = temparr[temparr.length-1];
console.log(x);
var length = 1;
while (x > 1) {
length = length + 1;
x = x/10;
}
return length;
};
console.log ( fiblength(500) );