New to regex.
Aim- To match a whole word which might have either '.' or '-' with it at the end. I want to keep it for the .start() and .end() position calculation.
txt = "The indian in. Spain."
pattern = "in."
x = re.search(r"\b" + pattern + r"\b" , txt)
print(x.start(), x.end())
I want the position for 'in.' word, as highlighted "The indian in. Spain.". The expression I have used gives error for a Nonetype object.
What would be the expression to match the '.' in the above code? Same if '-' is present instead of '.'
There are two issues here.
In regex . is special. It means "match one of any character". However, you are trying to use it to match a regular period. (It will indeed match that, but it will also match everything else.) Instead, to match a period, you need to use the pattern \.. And to change that to match either a period or a hyphen, you can use a class, like [-.].
You are using \b at the end of your pattern to match the word boundary, but \b is defined as being the boundary between a word character and a non-word character, and periods and spaces are both non-word characters. This means that Python won't find a match. Instead, you could use a lookahead assertion, which will match whatever character you want, but won't consume the string.
Now, to match a whole word - any word - you can do something like \w+, which matches one or more word characters.
Also, it is quite possible that there won't be a match anyway, so you should check whether a match occurred using an if statement or a try statement. Putting it all together:
txt = "The indian in. Spain."
pattern = r"\w+[-.]"
x = re.search(r"\b" + pattern + r"(?=\W)", txt)
if x:
print(x.start(), x.end())
Edit
There is one problem with the lookahead assertion above - it won't match the end of the string. This means that if your text is The rain in Spain. then it won't match Spain., as there is no non-word character following the final period.
To fix this, you can use a negative lookahead assertion, which matches when the following text does not include the pattern, and also does not consume the string.
x = re.search(r"\b" + pattern + r"(?!\w)", txt)
This will match when the character after the word is anything other than a word character, including the end of the string.
Related
I am basically trying to match string pattern(wildcard match)
Please carefully look at this -
*(star) - means exactly one word .
This is not a regex pattern...it is a convention.
So,if there patterns like -
*.key - '.key.' is preceded by exactly one word(word containing no dots)
*.key.* - '.key.' is preceded and succeeded by exactly one word having no dots
key.* - '.key' preceeds exactly one word .
So,
"door.key" matches "*.key"
"brown.door.key" doesn't match "*.key".
"brown.key.door" matches "*.key.*"
but "brown.iron.key.door" doesn't match "*.key.*"
So, when I encounter a '*' in pattern, I have replace it with a regex so that it means it is exactly one word.(a-zA-z0-9_).Can anyone please help me do this in python?
To convert your pattern to a regexp, you first need to make sure each character is interpreted literally and not as a special character. We can do that by inserting a \ in front of any re special character. Those characters can be obtained through sre_parse.SPECIAL_CHARS.
Since you have a special meaning for *, we do not want to escape that one but instead replace it by \w+.
Code
import sre_parse
def convert_to_regexp(pattern):
special_characters = set(sre_parse.SPECIAL_CHARS)
special_characters.remove('*')
safe_pattern = ''.join(['\\' + c if c in special_characters else c for c in pattern ])
return safe_pattern.replace('*', '\\w+')
Example
import re
pattern = '*.key'
r_pattern = convert_to_regexp(pattern) # '\\w+\\.key'
re.match(r_pattern, 'door.key') # Match
re.match(r_pattern, 'brown.door.key') # None
And here is an example with escaped special characters
pattern = '*.(key)'
r_pattern = convert_to_regexp(pattern) # '\\w+\\.\\(key\\)'
re.match(r_pattern, 'door.(key)') # Match
re.match(r_pattern, 'brown.door.(key)') # None
Sidenote
If you intend looking for the output pattern with re.search or re.findall, you might want to wrap the re pattern between \b boundary characters.
The conversion rules you are looking for go like this:
* is a word, thus: \w+
. is a literal dot: \.
key is and stays a literal string
plus, your samples indicate you are going to match whole strings, which in turn means your pattern should match from the ^ beginning to the $ end of the string.
Therefore, *.key becomes ^\w+\.key$, *.key.* becomes ^\w+\.key\.\w+$, and so forth..
Online Demo: play with it!
^ means a string that starts with the given set of characters in a regular expression.
$ means a string that ends with the given set of characters in a regular expression.
\s means a whitespace character.
\S means a non-whitespace character.
+ means 1 or more characters matching given condition.
Now, you want to match just a single word meaning a string of characters that start and end with non-spaced string. So, the required regular expression is:
^\S+$
You could do it with a combination of "any characters that aren't period" and the start/end anchors.
*.key would be ^[^.]*\.key, and *.key.* would be ^[^.]*\.key\.[^.]*$
EDIT: As tripleee said, [^.]*, which matches "any number of characters that aren't periods," would allow whitespace characters (which of course aren't periods), so using \w+, "any number of 'word characters'" like the other answers is better.
I'm looking for a Python regex that can match 'didn't' and returns only the character that is immediately preceded by an apostrophe, like 't, but not the 'd or t' at the beginning and end.
I have tried (?=.*\w)^(\w|')+$ but it only matches the apostrophe at the beginning.
Some more examples:
'I'm' should only match 'm and not 'I
'Erick's' should only return 's and not 'E
The text will always start and end with an apostrophe and can include apostrophes within the text.
To match an apostrophe inside a whole string = match it anwyhere but at the start/end of the string:
(?!^)'(?!$)
See the regex demo.
Often, the apostophe is searched only inside a word (but in fact, a pair of words where the second one is shortened), then you may use
\b'\b
See this regex demo. Here, the ' is preceded and followed with a word boundary, so that ' could be preceded with any word, letter or _ char. Yes, _ char and digits are allowed to be on both sides.
If you need to match a ' only between two letters, use
(?<=[A-Za-z])'(?=[A-Za-z]) # ASCII only
(?<=[^\W\d_])'(?=[^\W\d_]) # Any Unicode letters
See this regex demo.
As for this current question, here is a bunch of possible solutions:
import re
s = "'didn't'"
print(s.strip("'")[s.strip("'").find("'")+1])
print(re.search(r'\b\'(\w)', s).group(1))
print(re.search(r'\b\'([^\W\d_])', s).group(1))
print(re.search(r'\b\'([a-z])', s, flags=re.I).group(1))
print(re.findall(r'\b\'([a-z])', "'didn't know I'm a student'", flags=re.I))
The s.strip("'")[s.strip("'").find("'")+1] gets the character after the first ' after stripping the leading/trailing apostrophes.
The re.search(r'\b\'(\w)', s).group(1) solution gets the word (i.e. [a-zA-Z0-9_], can be adjusted from here) char after a ' that is preceded with a word char (due to the \b word boundary).
The re.search(r'\b\'([^\W\d_])', s).group(1) is almost identical to the above solution, it only fetches a letter character as [^\W\d_] matches any char other than a non-word, digit and _.
Note that the re.search(r'\b\'([a-z])', s, flags=re.I).group(1) solution is next to identical to the above one, but you cannot make it Unicode aware with re.UNICODE.
The last re.findall(r'\b\'([a-z])', "'didn't know I'm a student'", flags=re.I) just shows how to fetch multiple letter chars from a string input.
I am learning Regular Expressions, so apologies for a simple question.
I want to select the words that have a '-' (minus sign) in it but not at the beginning and not at the end of the word
I tried (using findall):
r'\b-\b'
for
str = 'word semi-column peace'
but, of course got only:
['-']
Thank you!
What you actually want to do is a regex like this:
\w+-\w+
What this means is find a alphanumeric character at least once as indicated by the utilization of '+', then find a '-', following by another alphanumeric character at least once, again, as indicated by the '+' again.
str is a built in name, better not to use it for naming
st = 'word semi-column peace'
# \w+ word - \w+ word after -
print(re.findall(r"\b\w+-\w+\b",st))
['semi-column']
a '-' (minus sign) in it but not at the beginning and not at the end of the word
Since "-" is not a word character, you can't use word boundaries (\b) to prevent a match from words with hyphens at the beggining or end. A string like "-not-wanted-" will match both \b\w+-\w+\b and \w+-\w+.
We need to add an extra condition before and after the word:
Before: (?<![-\w]) not preceded by either a hyphen nor a word character.
After: (?![-\w]) not followed by either a hyphen nor a word character.
Also, a word may have more than 1 hyphen in it, and we need to allow it. What we can do here is repeat the last part of the word ("hyphen and word characters") once or more:
\w+(?:-\w+)+ matches:
\w+ one or more word characters
(?:-\w+)+ a hyphen and one or more word characters, and also allows this last part to repeat.
Regex:
(?<![-\w])\w+(?:-\w+)+(?![-\w])
regex101 demo
Code:
import re
pattern = re.compile(r'(?<![-\w])\w+(?:-\w+)+(?![-\w])')
text = "-abc word semi-column peace -not-wanted- one-word dont-match- multi-hyphenated-word"
result = re.findall(pattern, text)
ideone demo
You can also use the following regex:
>>> st = "word semi-column peace"
>>> print re.findall(r"\S+\-\S+", st)
['semi-column']
You can try something like this: Centering on the hyphen, I match until there is a white space in either direction from the hyphen I also make check to see if the words are surrounded by hyphens (e.g -test-cats-) and if they are I make sure not to include them. The regular expression should also work with findall.
st = 'word semi-column peace'
m = re.search(r'([^ | ^-]+-[^ | ^-]+)', st)
if m:
print m.group(1)
I'm having trouble matching a string with regexp (I'm not that experienced with regexp). I have a string which contains a forward slash after each word and a tag. An example:
led/O by/O Timothy/PERSON R./PERSON Geithner/PERSON ,/O the/O president/O of/O the/O New/ORGANIZATION
In those strings, I am only interested in all strings that precede /PERSON. Here's the regexp pattern that I came up with:
(\w)*\/PERSON
And my code:
match = re.findall(r'(\w)*\/PERSON', string)
Basically, I am matching any word that comes before /PERSON. The output:
>>> reg
['Timothy', '', 'Geithner']
My problem is that the second match, matched to an empty string as for R./PERSON, the dot is not a word character. I changed my regexp to:
match = re.findall(r'(\w|.*?)\/PERSON', string)
But the match now is:
['led/O by/O Timothy', ' R.', ' Geithner']
It is taking everything prior to the first /PERSON which includes led/O by/O instead of just matching Timothy. Could someone please help me on how to do this matching, while including a full stop as an abbreviation? Or at least, not have an empty string match?
Thanks,
Match everything but a space character ([^ ]*). You also need the star (*) inside the capture:
match = re.findall(r'([^ ]*)\/PERSON', string)
Firstly, (\w|.) matches "a word character, or any character" (dot matches any character which is why you're getting those spaces).
Escaping this with a backslash will do the trick: (\w|\.)
Second, as #Ionut Hulub points out you may want to use + instead of * to ensure you match something but Regular Expressions work on the principle of "leftmost, longest" so it'll always try to match the longest part that it can before the slash.
If you want to match any non-whitespace character you can use \S instead of (\w|\.), which may actually be what you want.
Example;
X=This
Y=That
not matching;
ThisWordShouldNotMatchThat
ThisWordShouldNotMatch
WordShouldNotMatch
matching;
AWordShouldMatchThat
I tried (?<!...) but seems not to be easy :)
^(?!This).*That$
As a free-spacing regex:
^ # Start of string
(?!This) # Assert that "This" can't be matched here
.* # Match the rest of the string
That # making sure we match "That"
$ # right at the end of the string
This will match a single word that fulfills your criteria, but only if this word is the only input to the regex. If you need to find words inside a string of many other words, then use
\b(?!This)\w*That\b
\b is the word boundary anchor, so it matches at the start and at the end of a word. \w means "alphanumeric character. If you also want to allow non-alphanumerics as part of your "word", then use \S instead - this will match anything that's not a space.
In Python, you could do words = re.findall(r"\b(?!This)\w*That\b", text).