I want to match a number between 2-16, spanning 1 digit to 2 digits.
Regular-Expressions.info has examples for 1 or 2 digit ranges, but not something that spans both:
The regex [0-9] matches single-digit numbers 0 to 9. [1-9][0-9] matches double-digit numbers 10 to 99.
Something like ^[2-9][1-6]$ matches 21 or even 96! Any help would be appreciated.
^([2-9]|1[0-6])$
will match either a single digit between 2 and 9 inclusive, or a 1 followed by a digit between 0 and 6, inclusive.
With delimiters (out of habit): /^([2-9]|1[0-6])$/
The regex itself is just: ^([2-9]|1[0-6])$
Use the python package regex_engine for generating regular expressions for numerical ranges
You can install this package with pip.
pip install regex-engine
from regex_engine import generator
generate = generator()
regex = generate.numerical_range(2, 16)
print(regex)
^([2-9]|1[0-6])$
You can also generate regexes for floating point and negative ranges.
from regex_engine import generator
generate = generator()
regex1 = generate.numerical_range(5, 89)
regex2 = generate.numerical_range(81.78, 250.23)
regex3 = generate.numerical_range(-65, 12)
^([2-9]|1[0-6])$
(Edit: Removed quotes for clarification.)
Just replace your input formatters with
inputFormatters: [
FilteringTextInputFormatter(
RegExp(
r'^([0-9]|[2-8][0-9]|1[0-9]|9[0-9]|[2-8][0-9][0-9]|1[1-9][0-9]|10[0-9]|9[0-8][0-9]|99[0-9]|[2-4][0-9][0-9][0-9]|1[1-9][0-9][0-9]|10[1-9][0-9]|100[0-9]|500[0-0])$'),
allow: true,
)
],
(^[2-9]$|^1[0-6]$)
By specifying start and stop for each set of numbers you are looking for your regex won't also return 36, 46, ... and so on. I tried the above solution and found that this works best for staying within the range of 2-16.
Related
I have a list of strings each telling me after how many iterations an algorithm converged.
string_list = [
"Converged after 1 iteration",
"Converged after 20 iterations",
"Converged after 7 iterations"
]
How can I extract the number of iterations? The result woudl be [1, 20, 7]. I tried with regex. Apparently (?<=after )(.*)(?= iteration*) will give me anything in between after and iteration but then this doesn't work:
occursin(string_list[1], r"(?<=after )(.*)(?= iteration*)")
There's a great little Julia package that makes creating regexes easier called ReadableRegex, and as luck would have it the first example in the readme is an example of finding every integer in a string:
julia> using ReadableRegex
julia> reg = #compile look_for(
maybe(char_in("+-")) * one_or_more(DIGIT),
not_after = ".",
not_before = NON_SEPARATOR)
r"(?:(?<!\.)(?:(?:[+\-])?(?:\d)+))(?!\P{Z})"
That regex can now be broadcast over your list of strings:
julia> collect.(eachmatch.(reg, string_list))
3-element Vector{Vector{RegexMatch}}:
[RegexMatch("1")]
[RegexMatch("20")]
[RegexMatch("7")]
To extract information out of a regex, you want to use match and captures:
julia> convergeregex = r"Converged after (\d+) iteration"
r"Converged after (\d+) iteration"
julia> match(convergeregex, string_list[2]).captures[1]
"20"
julia> parse.(Int, [match(convergeregex, s).captures[1] for s in string_list])
3-element Vector{Int64}:
1
20
7
\d+ matches a series of digits (so, the number of iterations here), and the parantheses around it indicates that you want the part of the string matched by that to be placed in the results captures array.
You don't need the lookbehind and lookahead operators (?<=, ?=) here.
How do I match negative numbers as well by this regular expression? This regex works fine with positive values, but I want it to also allow negative values e.g. -10, -125.5 etc.
^[0-9]\d*(\.\d+)?$
Thanks
You should add an optional hyphen at the beginning by adding -? (? is a quantifier meaning one or zero occurrences):
^-?[0-9]\d*(\.\d+)?$
I verified it in Rubular with these values:
10.00
-10.00
and both matched as expected.
let r = new RegExp(/^-?[0-9]\d*(\.\d+)?$/);
//true
console.log(r.test('10'));
console.log(r.test('10.0'));
console.log(r.test('-10'));
console.log(r.test('-10.0'));
//false
console.log(r.test('--10'));
console.log(r.test('10-'));
console.log(r.test('1-0'));
console.log(r.test('10.-'));
console.log(r.test('10..0'));
console.log(r.test('10.0.1'));
Some Regular expression examples:
Positive Integers:
^\d+$
Negative Integers:
^-\d+$
Integer:
^-?\d+$
Positive Number:
^\d*\.?\d+$
Negative Number:
^-\d*\.?\d+$
Positive Number or Negative Number:
^-?\d*\.{0,1}\d+$
Phone number:
^\+?[\d\s]{3,}$
Phone with code:
^\+?[\d\s]+\(?[\d\s]{10,}$
Year 1900-2099:
^(19|20)[\d]{2,2}$
Date (dd mm yyyy, d/m/yyyy, etc.):
^([1-9]|0[1-9]|[12][0-9]|3[01])\D([1-9]|0[1-9]|1[012])\D(19[0-9][0-9]|20[0-9][0-9])$
IP v4:
^(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]){3}$
I don't know why you need that first [0-9].
Try:
^-?\d*(\.\d+)?$
Update
If you want to be sure that you'll have a digit on the ones place, then use
^-?\d+(\.\d+)?$
Adding "-" minus followed by ? quantifier in front of [0-9] expression should do the work.
-?[0-9]
For reference
^b - ^ quantifier matches for any string begin with "b"
c? - ? quantifier matches for 0 or 1 occurrence of "c"
[0-9] - find any character between the [] brackets
\d - find a digit from 0-9
d* - * quantifier matches for zero or more occurrence of "d"
\. - matches a "." character
z+ - + quantifier matches for one or more occurrence of "z"
e$ - $ quantifier matches for any string end with "e"
Hope, it'll help to understand posted regex in the question.
UPDATED(13/08/2014): This is the best code for positive and negative numbers =)
(^-?0\.[0-9]*[1-9]+[0-9]*$)|(^-?[1-9]+[0-9]*((\.[0-9]*[1-9]+[0-9]*$)|(\.[0-9]+)))|(^-?[1-9]+[0-9]*$)|(^0$){1}
I tried with this numbers and works fine:
-1234454.3435
-98.99
-12.9
-12.34
-10.001
-3
-0.001
-000
-0.00
0
0.00
00000001.1
0.01
1201.0000001
1234454.3435
7638.98701
This will allow a - or + character only when followed by a number:
^([+-](?=\.?\d))?(\d+)?(\.\d+)?$
I have some experiments about regex in django url, which required from negative to positive numbers
^(?P<pid>(\-\d+|\d+))$
Let's we focused on this (\-\d+|\d+) part and ignoring others, this semicolon | means OR in regex, then the negative value will match with this \-\d+ part, and positive value into this \d+
This will allow both positive and negative integers
ValidationExpression="^-?[0-9]\d*(\d+)?$"
^[+-]?\d{1,18}(\.\d{1,2})?$
accepts positive or negative decimal values.
This worked for me, allowing both negative and positive numbers:
\-*\d+
If using C#:
Regex.Match(someString, #"\-*\d+").Value;
If you have this val="-12XXX.0abc23" and you want to extract only the decimal number, in this case this regex (^-?[0-9]\d*(\.\d+)?$) will not help you to achieve it. this is the proper code with the correct detection regex:
var val="-12XXX.0abc23";
val = val.replace(/^\.|[^-?\d\.]|\.(?=.*\.)|^0+(?=\d)/g, '');
console.log(val);
I had a case recently where students were entering only the accepted characters in a numeric response field, yet still managed to break things. Thus, I ended up using the following catch-all.
^[+-]?((\d*\.?\d+)|(\d+\.?\d*))$
This ensures everything that should work will work, including:
0
.0
0.0
-.11
+.2
-0.2
+01.
-123.
+123.4567890
-012.0
+1
-1.
The expression also rejects things that mischievous kids might enter which, while still being valid character input, would not be a valid number, such as:
+.
-
.
(nul or newline)
I found that the expression as most have it written here (ending with \d+$) will reject numbers if they include a decimal point without any numbers after it. And making that expression instead end with \d* would make the entire expression optional, thus causing it to match the entries in the second list above. But by using the capturing group with the boolean OR operator (|) to require at least one digit either after or before a decimal point, all bases are covered.
Just add a 0 or 1 token:
^-?[0-9]\d*(.\d+)?$
For negative number only, this is perfect.
^-\d*\.?\d+$
Regular expression for number, optional decimal point, optional negative:
^-?(\d*\.)?\d+$;
works for negative integer, decimal, negative with decimal
^(-?\d+\.)?-?\d+$
allow:
23425.23425
10.10
100
0
0.00
-100
-10.10
10.-10
-10.-10
-23425.23425
-23425.-23425
0.234
Simply /\d/ works as expected for all cases I can think of:
let ns = {
regex: {
num: /\d/
}
}
for (let i of [-2, -1, 0, 1, 2, 'one', 'negative one', Math.PI, Math.E, (42 * -1), (42 / -1), '-1', '0', '1', 1.01, -1.01, .1, -.1, Math.sqrt(42)]) {
console.log(ns.regex.num.test(i) + ': ' + i);
}
For more Math fun, check out https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math
i have had a go at matching a regular expression that is given in hh:mm:ss:nnn
with the range being :
00:00:00:000 for lowest
23:59:59:999 as the highest
a valid example is 06:07:22:575
invalid example being 6:7:22:475
so far i have given it a go and hasnt worked with this :
^(([0-1][1-9])|(2[1-3])):((0[1-9])|(1-5[0-9]:((0[1-9])|(1-5[0-9]:([1-9][1-9][1-9])$
but it hasnt worked.. can someone help me out
Try this regex:
^(?:[01][0-9]|2[0-3]):[0-5][0-9]:[0-5][0-9]:[0-9][0-9][0-9]$
The only possibly tricky part here which might require explanation is the portion of the regex which covers the first two digits. In this case, we only want to support numbers from 0 to 23 inclusive. So, we use this pattern:
(?:[01][0-9]|2[0-3])
This says to match (and not capture) either a 0 or 1 followed by 0 to 9, i.e. 00-19, or a 2 followed by 0 to 3, i.e. 20-23.
Here is a working demo:
Demo
I think you just missed some square and round brackets?
^(([0-1][1-9])|(2[1-3])):((0[1-9])|([1-5][0-9])):((0[1-9])|([1-5][0-9])):([1-9][1-9][1-9])$
When you are trying to use check number between 00 and 59, it should be
( (0[1-9])|([1-5][0-9]) )
The one you tried (0[1-9])|(1-5[0-9] will only fetch 01~09
I would recommend:
/^[0-1][0-9]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]|^2[1-3]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]$/
Comprising two parts:
^[0-1][0-9]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]
^2[1-3]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]
This will match any time under 24 hours in the format listed above.
This can be seen working on Regex101 here.
const regex = /^[0-1][0-9]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]|^2[1-3]:[0-6][0-9]:[0-6][0-9]:[0-9][0-9][0-9]$/gm;
const strings = [
"00:00:00:000",
"23:59:59:999",
"06:07:22:575",
"6:7:22:475"
]
for (let i = 0; i < strings.length; i++) {
console.log(regex.test(strings[i]));
}
I need a little help with Regex.
I want the regex to validate the following sentences:
fdsufgdsugfugh PCL 6
dfdagf PCL 11
fdsfds PCL6
fsfs PCL13
kl;klkPCL6
fdsgfdsPCL13
some chars, than PCL and than 6 or a greater number.
How this can be done?
I'd go with something like this:
^(.*)(PCL *)([6-9][0-9]*|[1-5][0-9]+)$
Meaning:
(.*) = some chars
(PCL *) = then PCL with optional whitespaces afterwards
([6-9][0-9]*|[1-5][0-9]+) then 6 or a greater number
This one should suit your needs:
^.*PCL\s*(?:[6-9]|\d{2,})$
Visualization by Debuggex
In bash:
EXPR=^[a-zA-Z]\+ *PCL *\([6-9]\|[0-9]\{2,\}\)
Translated:
Line begins with at least 1 occurence of a character (ignore caps)
Any amount of spaces, PCL, any amount of spaces
Either a number between 6 or 9, or a number with at least 2 digits
This expression used with something like grep "$EXPR" file.txt will output in stdout the lines that are valid.
This worked well for me. Reads logically too according to the way you described the matching
/[^PCL]+PCL\s?*[6-9]\d*/
I have a need to search all numbers with 4 digits between 2000 and 3000.
It can be that letters are before and after.
I thought I can use [2000-3000]{4}, but doesnt work, why?
thank you.
How about
^2\d{3}|3000$
Or as Amarghosh & Bart K. & jleedev pointed out, to match multiple instances
\b(?:2[0-9]{3}|3000)\b
If you need to match a3000 or 3000a but not 13000, you would need lookahead and lookbefore like
(?<![0-9])(?:2[0-9]{3}|3000)(?![0-9])
Regular expressions are rarely suitable for checking ranges since for ranges like 27 through 9076 inclusive, they become incredibly ugly. It can be done but you're really better off just doing a regex to check for numerics, something like:
^[0-9]+$
which should work on just about every regex engine, and then check the range manually.
In toto:
def isBetween2kAnd3k(s):
if not s.match ("^[0-9]+$"):
return false
i = s.toInt()
if i < 2000 or i > 3000:
return false
return true
What your particular regex [2000-3000]{4} is checking for is exactly four occurrences of any of the following character: 2,0,0,0-3,0,0,0 - in other words, exactly four digits drawn from 0-3.
With letters before an after, you will need to modify the regex and check the correct substring, something like:
def isBetween2kAnd3kWithLetters(s):
if not s.match ("^[A-Za-z]*[0-9]{4}[A-Za-z]*$"):
return false
idx = s.locate ("[0-9]")
i = s.substring(idx,4).toInt()
if i < 2000 or i > 3000:
return false
return true
As an aside, a regex for checking the range 27 through 9076 inclusive would be something like this hideous monstrosity:
^2[7-9]|[3-9][9-9]|[1-9][0-9]{2}|[1-8][0-9]{3}|90[0-6][0-9]|907[0-6]$
I think that's substantially less readable than using ^[1-9][0-9]+$ then checking if it's between 27 and 9076 with an if statement?
Hum tricky one. The dash - only applies to the character immediately before and after so what your regex is actually matching is exactly 4 characters between 0 and 3 inclusive (ie, 0, 1, 2 and 3). eg, 3210, 1230, 3333, etc... Try the expression below.
(2[0-9]{3})|(3000)
Here's explanation why and ways to detect ranges: http://www.regular-expressions.info/numericranges.html
Correct regex will be \b(2\d{3}|3000)\b. That means: match character '2' then exactly three digits (this will match any from 2000 to 2999) or just match '3000'. There are some good tutorials on regular expressions:
http://gnosis.cx/publish/programming/regular_expressions.html
http://immike.net/blog/2007/04/06/the-absolute-bare-minimum-every-programmer-should-know-about-regular-expressions/
http://www.regular-expressions.info/
why don't you check for greater or less than? its simpler than a regex
num >= 2000 and num <=3000