I am new to python.
I want to write a regex to find a whole string using re.match.
For example,
word1=I
word2=U
str_to_match = word1+"[There could be some space in between]"+ word2[this is the end of string I want to match]"
In this case, it should match I[0 or more spaces in between]U.
It shouldn't match abI[0 or more spaces in between]U, OR abI[0 or more spaces in between]Ucd, OR
I[0 or more spaces in between]Ucd.
I know you could use boundary \b to set the word boundary but since there could be a number of spaces in between word1 and word2, the whole string is like a variable, not a fixed string, it won't work for me:
ret=re.match(r"\b"+word1+r"\s+" +word2+r"\b")
not working
Does anyone know how could I find the correct way to match this case?
Thanks!
match only matches at the beginning of the string. Which since you are using \b should not be what you want. findall will find all matching strings.
If you want to find the first occurrence and just test if it exists in the string, and not find all matches you can use search:
word1 = 'I'
word2 = 'U'
ret = re.findall(rf"\b{word1}\s*{word2}\b"," I U IU I U aI U I Ub")
print(ret)
ret = re.search(rf"\b{word1}\s*{word2}\b"," IUb .I U ")
print(ret)
Related
I want to get the content between single quotes, but only if it contains a certain word (i.e 'sample_2'). It additionally should not match ones with white space.
Input example: (The following should match and return only: ../sample_2/file and sample_2/file)
['asdf', '../sample_2/file', 'sample_2/file', 'example with space', sample_2, sample]
Right now I just have that matched the first 3 items in the list:
'(.\S*?)'
I can't seem to find the right regex that would return those containing the word 'sample_2'
If you want specific words/characters you need to have them in the regular expression and not use the '\S'. The \S is the equivalent to [^\r\n\t\f\v ] or "any non-whitespace character".
import re
teststr = "['asdf', '../sample_2/file', 'sample_2/file', 'sample_2 with spaces','example with space', sample_2, sample]"
matches = re.findall(r"'([^\s']*sample_2[^\s]*?)',", teststr)
# ['../sample_2/file', 'sample_2/file']
Based on your wording, you suggest the desired word can change. In that case, I would recommend using re.compile() to dynamically create a string which then defines the regular expression.
import re
word = 'sample_2'
teststr = "['asdf', '../sample_2/file', 'sample_2/file', ' sample_2 with spaces','example with space', sample_2, sample]"
regex = re.compile("'([^'\\s]*"+word+"[^\\s]*?)',")
matches = regex.findall(teststr)
# ['../sample_2/file', 'sample_2/file']
Also if you haven't heard of this tool yet, check out regex101.com. I always build my regular expressions here to make sure I get them correct. It gives you the references, explanation of what is happening and even lets you test it right there in the browser.
Explanation of regex
regex = r"'([^\s']*sample_2[^\s]*?)',"
Find first apostrophe, start group capture. Capture anything except a whitespace character or the corresponding ending apostrophe. It must see the letters "sample_2" before accepting any non-whitespace character. Stop group capture when you see the closing apostrophe and a comma.
Note: In python, a string " or ' prepositioned with the character 'r' means the text is compiled as a regular expression. Strings with the character 'r' also do not require double-escape '\' characters.
I am trying to write a regex that excludes square brackets and the text inside them.
My sample text looks like this: 'WordA, WordB, WordC, [WordD]'
I want to match each text item in the string except '[WordD]'. I've tried using a negative lookahead, something like... [A-Z][A-Za-z]+(?!\[[A-Z]+\]) but doing so is still matching the text inside the brackets.
Is negative lookahead the best approach? If so, where am I going wrong?
Rather than a regex, you might consider splitting by commas and then filtering by whether the word starts with [:
output = [word for word in str.split(', ') if word[0] != '[']
If you use a regex, you can match either the beginning of the string, or lookbehind for a space:
re.findall(r'(?:^|(?<= ))[A-Z][A-Za-z]+', str)
Or you could negative lookahead for ] at the end, after a word boundary:
output = re.findall(r'[A-Z][A-Za-z]+\b(?!\])', str)
This can be as simple as
(\w+),
Regex Demo
Retrieve value of Group 1 for desired result.
I'm guessing that maybe you were trying to write some expression similar to:
[A-Z][a-z]*[A-Z](?=,|$)
or,
[A-Z][a-z]+[A-Z](?=,|$)
Test
import re
regex = r"[A-Z][a-z]*[A-Z](?=,|$)"
string = """
WordA, WordB, WordC, [WordD]
WordA, WordB, WordC, [WordD], WordE
"""
print(re.findall(regex, string))
Output
['WordA', 'WordB', 'WordC', 'WordA', 'WordB', 'WordC', 'WordE']
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
I want to select select all character strings that begin with 0
x= '1,1,1075 1,0,39 2,4,1,22409 0,1,1,755,300 0,1,1,755,50'
I have
re.findall(r'\b0\S*', x)
but this returns
['0,39', '0,1,1,755,300', '0,1,1,755,50']
I want
['0,1,1,755,300', '0,1,1,755,50']
The problem is that \b matches the boundaries between digits and commas too. The simplest way might be not to use a regex at all:
thingies = [thingy for thingy in x.split() if thingy.startswith('0')]
Instead of using the boundary \b which will match between the comma and number (between any word [a-zA-Z0-9_] and non word character), you will want to match on start of string or space like (^|\s).
(^|\s)0\S*
https://regex101.com/r/Mrzs8a/1
Which will match the start of string or a space preceding the target string. But that will also include the space if present so I would suggest either trimming your matched string or wrapping the latter part with parenthesis to make it a group and then just getting group 1 from the matches like:
(?:^|\s)(0\S*)
https://regex101.com/r/Mrzs8a/2
I defined
s='f(x) has an occ of x but no y'
def italicize_math(line):
p="(\W|^)(x|y|z|f|g|h)(\W|$)"
repl=r"\1<i>\2</i>\3"
return re.sub(p,repl,line)
and made the following call:
print(italicize_math(s)
The result is
'<i>f</i>(x) has an occ of <i>x</i> but no <i>y</i>'
which is not what I expected. I wanted this instead:
'<i>f</i>(<i>x</i>) has an occ of <i>x</i> but no <i>y</i>'
Can anyone tell me why the first occurence of x was not enclosed in inside the "i" tags?
You seem to be trying to match non-alphanumeric characters (\W) when you really want a word boundary (\b):
>>> p=r"(\b)(x|y|z|f|g|h)(\b)"
>>> re.sub(p,repl,s)
'<i>f</i>(<i>x</i>) has an occ of <i>x</i> but no <i>y</i>'
Of course, ( is non alpha-numeric -- The reason your inner content doesn't match is because \W consumes a character in the match. so with a string like 'f(x)', you match the ( when you match f. Since ( was already matched, it won't match again when you try to match x. By contrast, word boundaries don't consume any characters.
Because the group construct is matching the position at the beginning of the string first and x would overlap the previous match. Also, the first and third groups are redundant since they can be replaced by word boundaries; and you can make use of a character class to combine letters.
p = r'\b([fghxyz])\b'
repl = r'<i>\1</i>'
Like previous answer mention, its because the ( char being consume when matching f thus cause subsequent x to fail the match.
beside replace with word boundary \b, you could also use lookahead regex which just do a peek and won't consume anything match inside the lookahead. Since it didn't consume anything, you don't need the \3 either
p=r"(\W|^)(x|y|z|f|g|h)(?=\W|$)"
repl=r"\1<i>\2</i>"
re.sub(p,repl,line)
How could I find all instances of a substring in a string?
For example I have the string ("%1 is going to the %2 with %3"). I need to extract all placeholders in this string (%1, %2, %3)
The current code could only find the first two because the ending is not a white space.
import re
string = "%1 is going to the %2 with %3"
r = re.compile('%(.*?) ')
m = r.finditer(string)
for y in m:
print (y.group())
Don't match on whitespace, match on a word boundary instead using \b:
r = re.compile(r'%(.*?)\b')
You may want to restrict your characters to word characters only instead of the . wildcard, and match at least one character:
r = re.compile(r'%(\w+)\b')
You don't appear to be using the capturing group either, so you could just omit that:
r = re.compile(r'%\w+\b')