I'm trying to use regular expressions to find three or more of the same character in a string. So for example:
'hello' would not match
'ohhh' would.
I've tried doing things like:
re.compile('(?!.*(.)\1{3,})^[a-zA-Z]*$')
re.compile('(\w)\1{5,}')
but neither seem to work.
(\w)\1{2,} is the regex you are looking for.
In Python it could be quoted like r"(\w)\1{2,}"
if you're looking for the same character three times consecutively, you can do this:
(\w)\1\1
if you want to find the same character three times anywhere in the string, you need to put a dot and an asterisk between the parts of the expression above, like so:
(\w).*\1.*\1
The .* matches any number of any character, so this expression should match any string which has any single word character that appears three or more times, with any number of any characters in between them.
Hope that helps.
Related
I have an input that is valid if it has this parts:
starts with letters(upper and lower), numbers and some of the following characters (!,#,#,$,?)
begins with = and contains only of numbers
begins with "<<" and may contain anything
example: !!Hel##lo!#=7<<vbnfhfg
what is the right regex expression in python to identify if the input is valid?
I am trying with
pattern= r"([a-zA-Z0-9|!|#|#|$|?]{2,})([=]{1})([0-9]{1})([<]{2})([a-zA-Z0-9]{1,})/+"
but apparently am wrong.
For testing regex I can really recommend regex101. Makes it much easier to understand what your regex is doing and what strings it matches.
Now, for your regex pattern and the example you provided you need to remove the /+ in the end. Then it matches your example string. However, it splits it into four capture groups and not into three as I understand you want to have from your list. To split it into four caputre groups you could use this:
"([a-zA-Z0-9!##$?]{2,})([=]{1}[0-9]+)(<<.*)"
This returns the capture groups:
!!Hel##lo!#
=7
<<vbnfhfg
Notice I simplified your last group a little bit, using a dot instead of the list of characters. A dot matches anything, so change that back to your approach in case you don't want to match special characters.
Here is a link to your regex in regex101: link.
I am attempting to match paragraph numbers inside my block of text. Given the following sentence:
Refer to paragraph C.2.1a.5 for examples.
I would like to match the word C.2.1a.5.
My current code like so:
([0-9a-zA-Z]{1,2}\.)
Only matches C.2.1a. and es., which is not what I want. Is there a way to match the full C.2.1a.5 and not match es.?
https://regex101.com/r/cO8lqs/13723
I have attempted to use ^ and $, but doing so returns no matches.
You should use following regex to match the paragraph numbers in your text.
\b(?:[0-9a-zA-Z]{1,2}\.)+[0-9a-zA-Z]\b
Try this demo
Here is the explanation,
\b - Matches a word boundary hence avoiding matching partially in a large word like examples.
(?:[0-9a-zA-Z]{1,2}\.)+ - This matches an alphanumeric text with length one or two as you tried to match in your own regex.
[0-9a-zA-Z] - Finally the match ends with one alphanumeric character at the end. In case you want it to match one or two alphanumeric characters at the end too, just add {1,2} after it
\b - Matches a word boundary again to ensure it doesn't match partially in a large word.
EDIT:
As someone pointed out, in case your text has strings like A.A.A.A.A.A. or A.A.A or even 1.2 and you don't want to match these strings and only want to match strings that has exactly three dots within it, you should use following regex which is more specific in matching your paragraph numbers.
(?<!\.)\b(?:[0-9a-zA-Z]{1,2}\.){3}[0-9a-zA-Z]\b(?!\.)
This new regex matches only paragraph numbers having exactly three dots and those negative look ahead/behind ensures it doesn't match partially in large string like A.A.A.A.A.A
Updated regex demo
Check these python sample codes,
import re
s = 'Refer to paragraph C.2.1a.5 for examples. Refer to paragraph A.A.A.A.A.A.A for examples. Some more A.A.A or like 1.22'
print(re.findall(r'(?<!\.)\b(?:[0-9a-zA-Z]{1,2}\.){3}[0-9a-zA-Z]\b(?!\.)', s))
Output,
['C.2.1a.5']
Also for trying to use ^ and $, they are called start and end anchors respectively, and if you use them in your regex, then they will expect matching start of line and end of line which is not what you really intend to do hence you shouldn't be using them and like you already saw, using them won't work in this case.
If simple version is required, you can use this easy to understand and modify regex ([A-Z]{1}\.[0-9]{1,3}\.[0-9]{1,3}[a-z]{1}\.[0-9]{1,3})
I think we should keep the regex expression simple and readable.
You can use the regex
**(?:[a-zA-Z]+\.){3}[a-zA-Z]+**
Explanation -
The expression (?:[a-zA-Z]+.){3} ensures that the group (?:[a-zA-Z]+.) is to be repeated 3 times within the word. The group contains an alphabetic character followed a dot.
The word would end with an alphabetic character.
Output:
['C.2.1a.5']
I wanted to search a string for a substring beginning with ">"
Does this syntax say what I want it to say: this character followed by anything.
regex_firstline = re.compile("[>]{1}.*")
As a pythonic way for such tasks you can use str.startswith() method, and don't need to use regex.
But about your regex "[>]{1}.*" you don't need {1} after your character class and you can specify the start of your regex with anchor ^.So it can be "^>.*"
Using http://regex101.com:
[>]{1} matches the single character > literally exactly one time (but it denotes {1} is a meaningless quantifier), and
.* then matches any character as many times as possible.
If a list was provided inside square brackets (as opposed to a single character), regex would attempt to match a single character within the list exactly one time. http://regex101.com has a good listing of tokens and what they mean.
An ideal regex expression would be ^[>].*, meaning at the beginning of a string find exactly one > character followed by anything else (and with only one character in the square brackets, you can remove those to simplify it even further: ^>.*
I am currently having trouble removing the end of strings using regex. I have tried using .partition with unsuccessful results. I am now trying to use regex unsuccessfully. All the strings follow the format of some random words **X*.* Some more words. Where * is a digit and X is a literal X. For Example 21X2.5. Everything after this dynamic string should be removed. I am trying to use re.sub('\d\d\X\d.\d', string). Can someone point me in the right direction with regex and how to split the string?
The expected output should read:
some random words 21X2.5
Thanks!
Use following regex:
re.search("(.*?\d\dX\d\.\d)", "some random words 21X2.5 Some more words").groups()[0]
Output:
'some random words 21X2.5'
Your regex is not correct. The biggest problem is that you need to escape the period. Otherwise, the regex treats the period as a match to any character. To match just that pattern, you can use something like:
re.findall('[\d]{2}X\d\.\d', 'asb12X4.4abc')
[\d]{2} matches a sequence of two integers, X matches the literal X, \d matches a single integer, \. matches the literal ., and \d matches the final integer.
This will match and return only 12X4.4.
It sounds like you instead want to remove everything after the matched expression. To get your desired output, you can do something like:
re.split('(.*?[\d]{2}X\d\.\d)', 'some random words 21X2.5 Some more words')[1]
which will return some random words 21X2.5. This expression pulls everything before and including the matched regex and returns it, discarding the end.
Let me know if this works.
To remove everything after the pattern, i.e do exactly as you say...:
s = re.sub(r'(\d\dX\d\.\d).*', r'\1', s)
Of course, if you mean something else than what you said, something different will be needed! E.g if you want to also remove the pattern itself, not just (as you said) what's after it:
s = re.sub(r'\d\dX\d\.\d.*', r'', s)
and so forth, depending on what, exactly, are your specs!-)
I've looked thrould the forums but could not find exactly how exactly to solve my problem.
Let's say I have a string like the following:
UDK .636.32/38.082.4454.2(575.3)
and I would like to match the expression with a regex, capturing the actual number (in this case the '.636.32/38.082.4454.2(575.3)').
There could be some garbage characters between the 'UDK' and the actual number, and characters like '.', '/' or '-' are valid parts of the number. Essentially the number is a sequence of digits separated by some allowed characters.
What I've came up with is the following regex:
'UDK.*(\d{1,3}[\.\,\(\)\[\]\=\'\:\"\+/\-]{0,3})+'
but it does not group the '.636.32/38.082.4454.2(575.3)'! It leaves me with nothing more than a last digit of the last group (3 in this case).
Any help would be greatly appreciated.
First, you need a non-greedy .*?.
Second, you don't need to escape some chars in [ ].
Third, you might just consider it as a sequence of digits AND some allowed characters? Why there is a \d{1,3} but a 4454?
>>> re.match(r'UDK.*?([\d.,()\[\]=\':"+/-]+)', s).group(1)
'.636.32/38.082.4454.2(575.3)'
Not so much a direct answer to your problem, but a general regexp tip: use Kodos (http://kodos.sourceforge.net/). It is simply awesome for composing/testing out regexps. You can enter some sample text, and "try out" regular expressions against it, seeing what matches, groups, etc. It even generates Python code when you're done. Good stuff.
Edit: using Kodos I came up with:
UDK.*?(?P<number>[\d/.)(]+)
as a regexp which matches the given example. Code that Kodos produces is:
import re
rawstr = r"""UDK.*?(?P<number>[\d/.)(]+)"""
matchstr = """UDK .636.32/38.082.4454.2(575.3)"""
# method 1: using a compile object
compile_obj = re.compile(rawstr)
match_obj = compile_obj.search(matchstr)
# Retrieve group(s) by name
number = match_obj.group('number')