The setup: I have a div with a bunch of radio buttons, each of which has been associated with a custom attribute and value using $(element).data(attr_name,attr_value);. When an underlying data structure is changed, I iterate over the fields and set the appropriate buttons to checked:true by using the ':data' selector found here: https://stackoverflow.com/a/2895933/1214731
$($('#style-options').find(':radio').filter(':data('+key+'=='+value+')'))
.prop('checked',true).button('refresh');
This works great: it finds the appropriate elements, even with floating-point values.
Performance depends on value:
I noticed that when I clicked on certain buttons, the page took fractionally longer to respond (for most buttons there was no noticeable delay). Digging a little deeper, this seems to be occurring when certain floating point values are being searched for.
Using chrome dev tools, I logged the following:
> key='fill-opacity';
"fill-opacity"
> value=.2*2;
0.4
> console.time('find data'); for(var i=0;i<100;++i){$('#style-options').find(':radio').filter(':data('+key+'=='+value+')')} console.timeEnd('find data');
find data: 43.352ms undefined
> value=.2*3;
0.6000000000000001
> console.time('find data'); for(var i=0;i<100;++i){$('#style-options').find(':radio').filter(':data('+key+'=='+value+')')} console.timeEnd('find data');
find data: 10322.866ms undefined
The difference in speed is a factor of >200!
Next, I tried typing the number in manually (e.g. decimal place, six, 14x zeros, one) - same speed. All numbers with the same number of digits were the same speed. I then reduced the number of digits progressively:
# of digits time (ms)
16 10300
15 5185
14 2665
13 1314
12 673
11 359
10 202
9 116
8 77
7 60
6 50
5 41
4 39
I quickly ruled out the equality check between numeric and string - no dependence on string length there.
The regex execution is strongly dependent on string length
In the linked answer above, the regex that parses the data string is this:
var matcher = /\s*(?:((?:(?:\\\.|[^.,])+\.?)+)\s*([!~><=]=|[><])\s*("|')?((?:\\\3|.)*?)\3|(.+?))\s*(?:,|$)/g;
The string passed in is of the form [name operator value]. The length of name doesn't seem to make much difference; the length of value has a big impact on speed however.
Specific questions:
1) Why does the length of name have minimal effect on performance, while the length of value has a large effect?
2) Doubling the execution time with each additional character in name seems excessive - is this just a characteristic of the particular regex the linked solution uses, or is it a more general feature?
3) How can I improve performance without sacrificing a lot of flexibility? I'd like to still be able to pass arguments as a single string to a jQuery selector so type checking up front seems difficult, though I'm open to suggestions.
Basic test code for regex matching speeds:
matcher = /\s*(?:((?:(?:\\\.|[^.,])+\.?)+)\s*([!~><=]=|[><])\s*("|')?((?:\\\3|.)*?)\3|(.+?))\s*(?:,|$)/g;
console.time('regex'); for(var i=0;i<1000;++i){matcher.lastIndex=0; matcher.exec('x=='+.1111111111111)}; console.timeEnd('regex')
regex: 538.018ms
//add an extra digit - doubles duration of test
console.time('regex'); for(var i=0;i<1000;++i){matcher.lastIndex=0; matcher.exec('x=='+.11111111111111)}; console.timeEnd('regex')
regex: 1078.742ms
//add a bunch to the length of 'name' - minimal effect
console.time('regex'); for(var i=0;i<1000;++i){matcher.lastIndex=0; matcher.exec('xxxxxxxxxxxxxxxxxxxx=='+.11111111111111)}; console.timeEnd('regex')
regex: 1084.367ms
A characteristic of regexp matching is that they are greedy. If you try to match the expression a.*b to the string abcd, it will happen in these steps:
the first "a" will match
the .* will match the second char, then the third, till the end of the string
reaching the end of the string, there is still a "b" to matched, the matching will fail
the regexp processing starts to backtrack
the last char will be "unmatched" and it will try to match "b" to "d". Fails again. More backtracking
tries to match "b" to "c". Fail. Backtrack.
match "b" to "b". Success. Matching ends.
Although you matched just a few chars, you iterated all the string. If you have more than one greedy operator, you can easily get an input string that will match with exponential complexity.
Understanding backtracking will prevent a lot of errors and performance problems. For example, 'a.*b' will match all the string 'abbbbbbb', instead of just the first 'ab'.
The easiest way to prevent these kind of errors in modern regexp engines, is to use the non-greedy version of the operators * and +. They are usually represented by the same operators followed by a question mark: *? and +?.
I confess that I really didn't stop to debug the complicate regexp that you posted, but I believe that the problem is before matching the '=' symbol. The greedy operator is in this subexpression:
(?:\\\.|[^.,])+\.?)+
I'd try to change it to a non-greedy version:
(?:\\\.|[^.,])+\.?)+?
but this is really just a wild guess. I'm using pattern recognition to solve the problem :-) It makes sense because it is backtracking for each character of the "value" till matching the operator. The name is matched linearly.
This regular expression is just too complex for my taste. I love regular expressions, but it looks like this one matches this famous quotation:
Some people, when confronted with a problem, think "I know, I'll use
regular expressions." Now they have two problems.
Related
I have a regEx for checking a number is less than 15 significant figures, Borrowed from this SO answer
/^-?(?=\d{1,15}(?:[.,]0+)?0*$|(?:(?=.{1,16}0*$)(?:\d+[.,]\d+))).+$/
The the other is used to check that same number is upto 2 decimal places(truncate)
/^-?(\d*\.?\d{0,2}).*/
I have almost 0 regex skill.
Question: How do I combine the 2 regexes to do the work of both, AND not just either OR( accomplished by | character - i am not sure if it achieves same function as combining both)
something like:
/^-?(?=\d{1,15}(?:[.,]0+)?0*$|(?:(?=.{1,16}0*$)(?:\d+[.,]\d+))).+$ <AND&&NOTOR>(\d*\.?\d{0,2}).*/
Thanks in advance
EDIT: edit moved to a seperate SO question
If you add only one condition of maximum 2 decimal places to first regex, try this..
^-?(?=\d{1,15}(?:[.,]0+)?0*$|(?:(?=[,.\d]{1,16}0*$)(?:\d+[.,]\d{1,2}$))).+$
Demo,,, in which I only changed original \d+ to d{1,2}$
Edited for the reguest to extract 15 significant figures and capture group 1 ($1). Try this which is wrapped to capture group 1 ($1) and limited 15 significant figures to be extracted easily.
^(-?(?=\d{1,15}(?:[.,]0+)?0*$|(?:(?=[,.\d]{1,16}0*$)(?:\d+[.,]\d{1,2}$))).{1,16}).*$
Demo,,, in which changed to .{1,16} from .+$.
If the number matches, then able to be replaced $1, but if not so, replaced nothing, thus remains original unmatched number.
Therefore, if you want to extract 15 significant figures by replacing with $1 only when your condition is satisfied, try this regex to your function.
^(-?(?=\d{1,15}(?:[.,]0+)?0*$|(?:(?=[,.\d]{1,16}0*$)(?:\d+[.,]\d{1,2}$))).{1,16}).*$|^.*$
Demo,,, in which all numbers are matched, but only the numbers satisfying your condition are captured to $1 in format of 15 significant figures.
I am parsing a string of multiple numbers between 1 and 10 with the eventual goal of adding them to a set.
There will be multiple concatenated numbers after a text identifier such as {text}12345678910.
I am currently using match(/\d/g) to grab the numbers but it separates 1 and 0 in 10. I then look for 0 in my String Array, see if there's a 1 in the element before it, turn it into a 10 and delete the other entry. Not very elegant.
How can I clean up my matching code? I definitely don't need to use regex for this, but it makes grabbing the numbers fairly easy.
You could just match with this regex:
/10|\d/g
(instead of the one you use currently, not additionally)
Regex is executed left-to-right, so first it finds any occurrences of 10, and then of other digits (so using, for example /\d|10/g or even /\d|(10)/g won't work either).
I need to match a number range:
-9223372036854775808 to 9223372036854775807
^(?:922337203685477580[0-7]|9223372036854775[0-7]\d{2}|922337203685477[0-4]\d{3}|92233720368547[0-6]\d{4}|9223372036854[0-6]\d{5}|922337203685[0-3]\d{6}|92233720368[0-4]\d{7}|9223372036[0-7]\d{8}|922337203[0-5]\d{9}|92233720[0-2]\d{10}|922337[0-1]\d{12}|92233[0-6]\d{13}|9223[0-2]\d{14}|922[0-2]\d{15}|92[0-1]\d{16}|9[01]\d{17}|[1-8]\d{18}|\d{0,18}|-(?:922337203685477580[0-8]|9223372036854775[0-7]\d{2}|922337203685477[0-4]\d{3}|92233720368547[0-6]\d{4}|9223372036854[0-6]\d{5}|922337203685[0-3]\d{6}|92233720368[0-4]\d{7}|9223372036[0-7]\d{8}|922337203[0-5]\d{9}|92233720[0-2]\d{10}|922337[0-1]\d{12}|92233[0-6]\d{13}|9223[0-2]\d{14}|922[0-2]\d{15}|92[0-1]\d{16}|9[01]\d{17}|[1-8]\d{18}|\d{0,18}))?$
// space for easier copy and paste
Yes, I know it sounds crazy, but there's a long story behind this. I can't figure out how to do this in JavaScript by just checking a range, because of the size of the number, and this must be accurate.
Here's the thought process in breaking this thing down. I just started with the max number and worked my way down, then worked on the negative by just adding the - in the regex. You'll obviously have to copy and paste this thing somewhere to see it all. Also, could be mistakes. Made my head nearly explode.
9,223,372,036,854,775,807
922337203685477580[0-7]
9223372036854775[0-7][0-9]{2}
922337203685477[0-4][0-9]{3}
92233720368547[0-6][0-9]{4}
9223372036854[0-6][0-9]{5}
922337203685[0-3][0-9]{6}
92233720368[0-4][0-9]{7}
9223372036[0-7][0-9]{8}
922337203[0-5][0-9]{9}
92233720[0-2][0-9]{10}
922337[0-1][0-9]{12}
92233[0-6][0-9]{13}
9223[0-2][0-9]{14}
922[0-2][0-9]{15}
92[0-1][0-9]{16}
9[01][0-9]{17}
[1-8][0-9]{18}
[0-9]{0,18}
There's a single digit different in the negative vs. positive, so you'll see where I had to basically duplicate most of this.
So a few question:
Did I do this right?
If not, what's a better way?
Can this be done without regular expressions considering the size of the number? I need to validate client-side.
Can it be refactored and still retain strict rules?
Suggestions appreciated :)
Can this be done without regular expressions considering the size of the number?
It can be done in a series of if statements using only string operations (no need to convert to numbers).
all strings that don't match [0-9]{1,19} are out
all candidates that are of length 18 or less are good
for length 19 you can work with string comparison to see if they are numerically less than your upper limit
tweak the above to take care of negative numbers
Your regex is correct.
This is a shorter version
^(?:-9223372036854775808|-?(?:\d{0,18}|(?!922337203685477580[8-9]|92233720368547758[1-9]|92233720368547759|922337203685477[6-9]|92233720368547[8-9]|9223372036854[8-9]|922337203685[5-9]|92233720368[6-9]|92233720369|922337203[7-9]|92233720[4-9]|9223372[1-9]|922337[3-9]|92233[8-9]|9223[4-9]|922[4-9]|92[3-9]|9[3-9])\d{19}))$
Regex demo
How to generate that regex without mistake:
Input max number:
9223372036854775807
Output:
9223372036854775807
922337203685477580
92233720368547758
9223372036854775
922337203685477
92233720368547
9223372036854
922337203685
92233720368
9223372036
922337203
92233720
9223372
922337
92233
9223
922
92
9
Replace last number letter
9->remove all line
8->9
7->[8-9]
6->[7-9]
5->[6-9]
4->[5-9]
3->[4-9]
2->[3-9]
1->[2-9]
0->[1-9]
Output:
922337203685477580[8-9]
92233720368547758[1-9]
92233720368547759
922337203685477[6-9]
92233720368547[8-9]
9223372036854[8-9]
922337203685[5-9]
92233720368[6-9]
92233720369
922337203[7-9]
92233720[4-9]
9223372[1-9]
922337[3-9]
92233[8-9]
9223[4-9]
922[4-9]
92[3-9]
9[3-9]
Regex [Output]
922337203685477580[8-9]|92233720368547758[1-9]|92233720368547759|922337203685477[6-9]|92233720368547[8-9]|9223372036854[8-9]|922337203685[5-9]|92233720368[6-9]|92233720369|922337203[7-9]|92233720[4-9]|9223372[1-9]|922337[3-9]|92233[8-9]|9223[4-9]|922[4-9]|92[3-9]|9[3-9]
Add these[output] to regex
(?!output)\d{19}
Will become [output2]
(?!922337203685477580[8-9]|92233720368547758[1-9]|92233720368547759|922337203685477[6-9]|92233720368547[8-9]|9223372036854[8-9]|922337203685[5-9]|92233720368[6-9]|92233720369|922337203[7-9]|92233720[4-9]|9223372[1-9]|922337[3-9]|92233[8-9]|9223[4-9]|922[4-9]|92[3-9]|9[3-9])\d{19}
Matches \d{19} <= 9223372036854775807
Add
^(?:-9223372036854775808|-?(?:\d{0,18}|[output2]))$
^(?:-9223372036854775808|-?(?:\d{0,18}|(?!922337203685477580[8-9]|92233720368547758[1-9]|92233720368547759|922337203685477[6-9]|92233720368547[8-9]|9223372036854[8-9]|922337203685[5-9]|92233720368[6-9]|92233720369|922337203[7-9]|92233720[4-9]|9223372[1-9]|922337[3-9]|92233[8-9]|9223[4-9]|922[4-9]|92[3-9]|9[3-9])\d{19}))$
Will match
-9223372036854775808 or
+/- \d{0,18} or
+/- \d{19} <= 9223372036854775807
Demo
i need a regular expression for decimal/float numbers like 12 12.2 1236.32 123.333 and +12.00 or -12.00 or ...123.123... for using in javascript and jQuery.
Thank you.
Optionally match a + or - at the beginning, followed by one or more decimal digits, optional followed by a decimal point and one or more decimal digits util the end of the string:
/^[+-]?\d+(\.\d+)?$/
RegexPal
The right expression should be as followed:
[+-]?([0-9]*[.])?[0-9]+
this apply for:
+1
+1.
+.1
+0.1
1
1.
.1
0.1
Here is Python example:
import re
#print if found
print(bool(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0')))
#print result
print(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0').group(0))
Output:
True
1.0
If you are using mac, you can test on command line:
python -c "import re; print(bool(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0')))"
python -c "import re; print(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0').group(0))"
You can check for text validation and also only one decimal point validation using isNaN
var val = $('#textbox').val();
var floatValues = /[+-]?([0-9]*[.])?[0-9]+/;
if (val.match(floatValues) && !isNaN(val)) {
// your function
}
This is an old post but it was the top search result for "regular expression for floating point" or something like that and doesn't quite answer _my_ question. Since I worked it out I will share my result so the next person who comes across this thread doesn't have to work it out for themselves.
All of the answers thus far accept a leading 0 on numbers with two (or more) digits on the left of the decimal point (e.g. 0123 instead of just 123) This isn't really valid and in some contexts is used to indicate the number is in octal (base-8) rather than the regular decimal (base-10) format.
Also these expressions accept a decimal with no leading zero (.14 instead of 0.14) or without a trailing fractional part (3. instead of 3.0). That is valid in some programing contexts (including JavaScript) but I want to disallow them (because for my purposes those are more likely to be an error than intentional).
Ignoring "scientific notation" like 1.234E7, here is an expression that meets my criteria:
/^((-)?(0|([1-9][0-9]*))(\.[0-9]+)?)$/
or if you really want to accept a leading +, then:
/^((\+|-)?(0|([1-9][0-9]*))(\.[0-9]+)?)$/
I believe that regular expression will perform a strict test for the typical integer or decimal-style floating point number.
When matched:
$1 contains the full number that matched
$2 contains the (possibly empty) leading sign (+/-)
$3 contains the value to the left of the decimal point
$5 contains the value to the right of the decimal point, including the leading .
By "strict" I mean that the number must be the only thing in the string you are testing.
If you want to extract just the float value out of a string that contains other content use this expression:
/((\b|\+|-)(0|([1-9][0-9]*))(\.[0-9]+)?)\b/
Which will find -3.14 in "negative pi is approximately -3.14." or in "(-3.14)" etc.
The numbered groups have the same meaning as above (except that $2 is now an empty string ("") when there is no leading sign, rather than null).
But be aware that it will also try to extract whatever numbers it can find. E.g., it will extract 127.0 from 127.0.0.1.
If you want something more sophisticated than that then I think you might want to look at lexical analysis instead of regular expressions. I'm guessing one could create a look-ahead-based expression that would recognize that "Pi is 3.14." contains a floating point number but Home is 127.0.0.1. does not, but it would be complex at best. If your pattern depends on the characters that come after it in non-trivial ways you're starting to venture outside of regular expressions' sweet-spot.
Paulpro and lbsweek answers led me to this:
re=/^[+-]?(?:\d*\.)?\d+$/;
>> /^[+-]?(?:\d*\.)?\d+$/
re.exec("1")
>> Array [ "1" ]
re.exec("1.5")
>> Array [ "1.5" ]
re.exec("-1")
>> Array [ "-1" ]
re.exec("-1.5")
>> Array [ "-1.5" ]
re.exec(".5")
>> Array [ ".5" ]
re.exec("")
>> null
re.exec("qsdq")
>> null
For anyone new:
I made a RegExp for the E scientific notation (without spaces).
const floatR = /^([+-]?(?:[0-9]+(?:\.[0-9]+)?|\.[0-9]+)(?:[eE][+-]?[0-9]+)?)$/;
let str = "-2.3E23";
let m = floatR.exec(str);
parseFloat(m[1]); //=> -2.3e+23
If you prefer to use Unicode numbers, you could replace all [0-9] by \d in the RegExp.
And possibly add the Unicode flag u at the end of the RegExp.
For a better understanding of the pattern see https://regexper.com/.
And for making RegExp, I can suggest https://regex101.com/.
EDIT: found another site for viewing RegExp in color: https://jex.im/regulex/.
EDIT 2: although op asks for RegExp specifically you can check a string in JS directly:
const isNum = (num)=>!Number.isNaN(Number(num));
isNum("123.12345678E+3");//=> true
isNum("80F");//=> false
converting the string to a number (or NaN) with Number()
then checking if it is NOT NaN with !Number.isNaN()
If you want it to work with e, use this expression:
[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?
Here is a JavaScript example:
var re = /^[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?$/;
console.log(re.test('1'));
console.log(re.test('1.5'));
console.log(re.test('-1'));
console.log(re.test('-1.5'));
console.log(re.test('1E-100'));
console.log(re.test('1E+100'));
console.log(re.test('.5'));
console.log(re.test('foo'));
Here is my js method , handling 0s at the head of string
1- ^0[0-9]+\.?[0-9]*$ : will find numbers starting with 0 and followed by numbers bigger than zero before the decimal seperator , mainly ".". I put this to distinguish strings containing numbers , for example, "0.111" from "01.111".
2- ([1-9]{1}[0-9]\.?[0-9]) : if there is string starting with 0 then the part which is bigger than 0 will be taken into account. parentheses are used here because I wanted to capture only parts conforming to regex.
3- ([0-9]\.?[0-9]): to capture only the decimal part of the string.
In Javascript , st.match(regex), will return array in which first element contains conformed part. I used this method in the input element's onChange event , by this if the user enters something that violates the regex than violating part is not shown in element's value at all but if there is a part that conforms to regex , then it stays in the element's value.
const floatRegexCheck = (st) => {
const regx1 = new RegExp("^0[0-9]+\\.?[0-9]*$"); // for finding numbers starting with 0
let regx2 = new RegExp("([1-9]{1}[0-9]*\\.?[0-9]*)"); //if regx1 matches then this will remove 0s at the head.
if (!st.match(regx1)) {
regx2 = new RegExp("([0-9]*\\.?[0-9]*)"); //if number does not contain 0 at the head of string then standard decimal formatting takes place
}
st = st.match(regx2);
if (st?.length > 0) {
st = st[0];
}
return st;
}
Here is a more rigorous answer
^[+-]?0(?![0-9]).[0-9]*(?![.])$|^[+-]?[1-9]{1}[0-9]*.[0-9]*$|^[+-]?.[0-9]+$
The following values will match (+- sign are also work)
.11234
0.1143424
11.21
1.
The following values will not match
00.1
1.0.00
12.2350.0.0.0.0.
.
....
How it works
The (?! regex) means NOT operation
let's break down the regex by | operator which is same as logical OR operator
^[+-]?0(?![0-9]).[0-9]*(?![.])$
This regex is to check the value starts from 0
First Check + and - sign with 0 or 1 time ^[+-]
Then check if it has leading zero 0
If it has,then the value next to it must not be zero because we don't want to see 00.123 (?![0-9])
Then check the dot exactly one time and check the fraction part with unlimited times of digits .[0-9]*
Last, if it has a dot follow by fraction part, we discard it.(?![.])$
Now see the second part
^[+-]?[1-9]{1}[0-9]*.[0-9]*$
^[+-]? same as above
If it starts from non zero, match the first digit exactly one time and unlimited time follow by it [1-9]{1}[0-9]* e.g. 12.3 , 1.2, 105.6
Match the dot one time and unlimited digit follow it .[0-9]*$
Now see the third part
^[+-]?.{1}[0-9]+$
This will check the value starts from . e.g. .12, .34565
^[+-]? same as above
Match dot one time and one or more digits follow by it .[0-9]+$
i have a word from A to Z. all word should in small latter (Capital not include) and 1 to 9 (included all special word who can be used in email address (just for a test)).
how i can generate unique 1 lacs text who never repeat itself. can anyone solve this puzzle.
i want a another thing that all words should not more then 10 char and not should minimum 6 char long
Put the characters in an array. Copy the array as the source of a new line. Randomly slice words from the array and put them in the line (use Math.random() * array.length | 0). Keep going for the required number of words.
You can also just use a string and charAt(index) if you only want single characters, but you have to keep cutting out the character that you select which is likely less efficient than using array.slice.
Whatever suits though, since performance is likely irrelevant.