I have a problem where I want to match any number of German words inside [] braces, ignoring the case. The expression should only match spaces and words, nothing else i.e no punctuation marks or parenthesis
E.g :
The expression ['über das thema schreibt'] should be matched with ['Über', 'das', 'Thema', 'schreibt']
I have one list with items of the former order and another with the latter order, as long as the words are same, they both should match.
The code I tried with is -
regex = re.findall('[(a-zA-Z_äöüÄÖÜß\s+)]', str(term))
or
re.findall('[(\S\s+)]', str(term))
But they are not working. Kindly help me find a solution
In the simplest form using \w+ works for finding words (needs Unicode flag for non-ascii chars), but since you want them to be within the square brackets (and quotes I assume) you'd need something a bit complex
\[(['\"])((\w+\s?)+)\1\]
\[ and \] are used to match the square brackets
['\"] matches either quote and the \1 makes sure the same quote is one the other end
\w+ captures 1 word. The \s? is for an optional space.
The whole string is in the second group which you can split to get the list
import re
text = "['über das thema schreibt']"
regex = re.compile("\[(['\"])((\w+\s?)+)['\"]\]", flags=re.U)
match = regex.match(text)
if match:
print(match.group(2).split())
(slight edit as \1 did not seem to work in the terminal for me)
I found the easiest solution to it :
for a, b in zip(list1, list2):
reg_a = re.findall('[(\w\s+)]', str(a).lower())
reg_b = re.findall('[(\w\s+)]', str(b).lower())
if reg_a == reg_b:
return True
else
return False
Updated based on comments to match each word. This simply ignores spaces, single quotes and square braces
import re
text = "['über das thema schreibt']"
re.findall("([a-zA-Z_äöüÄÖÜß]+)", str(text))
# ['über', 'das', 'thema', 'schreibt']
If you are solving case sensitivity issue, add the regex flaf re.IGNORECASE
like
re.findall('[(\S\s+)]', str(term),re.IGNORECASE)
You might need to consider converting them to unicode, if it did not help.
Related
So I have been trying to construct a regex that can detect the pattern {word}{.,#}{word} and seperate it into [word,',' (or '.','#'), word].
But i am not able to create one that does strict matching for this pattern and ignores everything else.
I used the following regex
r"[\w]+|[.]"
this one is doing well , but it doesnt do strict matching, as in if (,, # or .) characters dont occur in text, it will still give me words, which i dont want.
I would like to have a regex which strictly matches the above pattern and gives me the splits(using re.findall) and if not returns the whole word as it is.
Please Note: word on either side of the {,.#} , both words are not strictly to be present but atleast one should be present
Some example text for reference:
no.16 would give me ['no','.','16']
#400 would give me ['#,'400']
word1.word2 would give me ['word1','.','word2']
Looking forward to some help and assistance from all regex gurus out there
EDIT:
I forgot to add this. #viktor's version works as needed with only one problem, It ignores ALL other words during re.findall
eg. ONE TWO THREE #400 with the viktor's regex gives me ['','#','400']
but what was expected was ['ONE','TWO','THREE','#',400]
this can be done with NLTK or spacy, but use of those is a limitation.
I suggest using
(\w+)?([.,#])((?(1)\w*|\w+))
See the regex demo.
Details
(\w+)? - An optional group #1: one or more word chars
([.,#]) - Group #2: ., , or #
((?(1)\w*|\w+)) - Group #3: if Group 1 matched, match zero or more word chars (the word is optional on the right side then), else, match one or more word chars (there must be a word on the right side of the punctuation chars since there is no word before them).
See the Python demo:
import re
pattern = re.compile(r'(\w+)?([.,#])((?(1)\w*|\w+))')
strings = ['no.16', '#400', 'word1.word2', 'word', '123']
for s in strings:
print(s, ' -> ', pattern.findall(s))
Output:
no.16 -> [('no', '.', '16')]
#400 -> [('', '#', '400')]
word1.word2 -> [('word1', '.', 'word2')]
word -> []
123 -> []
The answer to your edit is
if re.search(r'\w[.,#]|[.,#]\w', text):
print( re.findall(r'[.,#]|[^\s.,#]+', text) )
If there is a word char, then any of the three punctuation symbols, and then a word char again in the input string, you can find and extract all occurrences of the [.,#]|[^\s.,#]+ pattern, namely a ., , or #, or one or more occurrences of any one or more chars other than whitespace, ., , and #.
I hope this code will solve your problem if you want to split the string by any of the mentioned special characters:
a='no.16'
b='#400'
c='word1.word2'
lst=[a, b, c]
for elem in lst:
result= re.split('(\.|#|,)',elem)
while('' in result):
result.remove('')
print(result)
You could do something like this:
import re
str = "no.16"
pattern = re.compile(r"(\w+)([.|#])(\w+)")
result = list(filter(None, pattern.split(str)))
The list(filter(...)) part is needed to remove the empty strings that split returns (see Python - re.split: extra empty strings that the beginning and end list).
However, this will only work if your string only contains these two words separated by one of the delimiters specified by you. If there is additional content before or after the pattern, this will also be returned by split.
for string "//div[#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings']", I want to find "#..'...'" like "#id~'objectnavigator-card-list'" or "#class~'outbound-alert-settings'". But when I use regex ((#.+)\~(\'.*?\')), it find "#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings'". So how to modify the regex to find the string successfully?
Use non-capturing, non greedy, modifiers on the inner brackets and search for not the terminating character, e.g.:
re.findall(r"((?:#[^\~]+)\~(?:\'[^\]]*?\'))", test)
On your test string returns:
["#id~'objectnavigator-card-list'", "#class~'outbound-alert-settings'"]
Limit the characters you want to match between the quotes to not match the quote:
>>> re.findall(r'#[a-z]+~\'[-a-z]*\'', x)
I find it's much easier to look for only the characters I know are going to be in a matching section rather than omitting characters from more permissive matches.
For your current test string's input you can try this pattern:
import re
a = "//div[#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings']"
# find everything which begins by '#' and neglect ']'
regex = re.compile(r'(#[^\]]+)')
strings = re.findall(regex, a)
# Or simply:
# strings = re.findall('(#[^\\]]+)', a)
print(strings)
Output:
["#id~'objectnavigator-card-list'", "#class~'outbound-alert-settings'"]
I have a string in Python:
Tt = "This is a <\"string\">string, It should be <\"changed\">changed to <\"a\">a nummber."
print Tt
'This is a <"string">string, It should be <"changed">changed to <"a">a nummber.'
You see the some words repeat in this part <\" \">.
My question is, how to delete those repeated parts (delimited with the named characters)?
The result should be like:
'This is a string, It should be changed to a nummber.'
Use regular expressions:
import re
Tt = re.sub('<\".*?\">', '', Tt)
Note the ? after *. It makes the expression non-greedy,
so it tries to match so few symbols between <\" and \"> as possible.
The Solution of James will work only in cases when the delimiting substrings
consist only from one character (< and >). In this case it is possible to use negations like [^>]. If you want to remove a substring delimited with character sequences (e.g. with begin and end), you should use non-greedy regular expressions (i.e. .*?).
I'd use a quick regular expression:
import re
Tt = "This is a <\"string\">string, It should be <\"changed\">changed to <\"a\">a number."
print re.sub("<[^<]+>","",Tt)
#Out: This is a string, It should be changed to a nummber.
Ah - similar to Igor's post, he beat my by a bit. Rather than making the expression non-greedy, I don't match an expression if it contains another start tag "<" in it, so it will only match a start tag that's followed by an end tag ">".
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.
How to match the following i want all the names with in the single quotes
This hasn't been much that much of a twist and turn's to 'Tom','Harry' and u know who..yes its 'rock'
How to extract the name within the single quotes only
name = re.compile(r'^\'+\w+\'')
The following regex finds all single words enclosed in quotes:
In [6]: re.findall(r"'(\w+)'", s)
Out[6]: ['Tom', 'Harry', 'rock']
Here:
the ' matches a single quote;
the \w+ matches one or more word characters;
the ' matches a single quote;
the parentheses form a capture group: they define the part of the match that gets returned by findall().
If you only wish to find words that start with a capital letter, the regex can be modified like so:
In [7]: re.findall(r"'([A-Z]\w*)'", s)
Out[7]: ['Tom', 'Harry']
I'd suggest
r = re.compile(r"\B'\w+'\B")
apos = r.findall("This hasn't been much that much of a twist and turn's to 'Tom','Harry' and u know who..yes its 'rock'")
Result:
>>> apos
["'Tom'", "'Harry'", "'rock'"]
The "negative word boundaries" (\B) prevent matches like the 'n' in words like Rock'n'Roll.
Explanation:
\B # make sure that we're not at a word boundary
' # match a quote
\w+ # match one or more alphanumeric characters
' # match a quote
\B # make sure that we're not at a word boundary
^ ('hat' or 'caret', among other names) in regex means "start of the string" (or, given particular options, "start of a line"), which you don't care about. Omitting it makes your regex work fine:
>>> re.findall(r'\'+\w+\'', s)
["'Tom'", "'Harry'", "'rock'"]
The regexes others have suggested might be better for what you're trying to achieve, this is the minimal change to fix your problem.
Your regex can only match a pattern following the start of the string. Try something like: r"'([^']*)'"