Okay so in Python, I'm trying to search for the pattern "comma, space, any lowercase character", but I cant get a regular expression that seems to work. The whole regular expressions thing is pretty new to me and I have no idea what I'm doing. I was able to search for a "number, space, any character using "[1-9]+ [a-zA-z]", but I'm not sure how to search for the pattern mentioned above. The picture included is an example of what pattern I am trying to search for in the text file.
Thanks,
Schulzy
A Regex expression that would work is
, [a-z]
the comma and space are matched exactly, and the '[]' is a group, where anything in the group could be matched. you want any lowercase char's, so we put [a-z] for any character between lowercase a to z.
Related
Could someone help me on regex to match German words/sentences in
python? It does not work on jupyter notebook. I tried same in jsfiddle
it works fine. I tried using this below script but does not work
import re
pattern = re.compile(r'\[^a-zA-Z0-9äöüÄÖÜß]\\', re.UNICODE)
print(pattern.search(text))
Your expression will always fail:
\[^a-zA-Z0-9äöüÄÖÜß]\\
Broken down, you require
[ # literally
^ # start of the line / text
a-z # literally, etc.
The problem is that you require a [ literally right before the start of a line which can never be true (either there's nothing or a newline). So in the end, either remove the backslash to get a proper character class as in:
[^a-zA-Z0-9äöüÄÖÜß]+
But this will surely not match the words you're looking for (quite the opposite). So either use something as simple as \w+ or the solution proposed by #Wiktor in the comments section.
The square brackets define a range of characters you want to look for, however the '^' negates these characters if it appears within the character class.
If you want to specify the beginning of the line you need to put the '^' before the brackets.
Also you need to add a multiplier behind the class to search for more than just one character in this case:
r'^[a-zA-Z0-9äöüÄÖÜß]+'
One ore more characters contained in the brackets are matched as long as they are not seperated by any other character not listed between '[]'
Here's the link to the official documentation
I want a regex that doesn't match a string if contains the word page, and match if it's not contain.
^https?.+/(event|news)/.+(?!page).+$ this is the regex I'm currently using, so I want it to not match with, e.g. https://www.foosite.com/news/foopath/page/10, but it does. Where did I made a mistake?
The double .+ expressions should imply that there should be some string around the page string, and (?!page) should imply there must not be a string like page between them. What's wrong with this expression? Thanks, and sorry for poor grammar.
Your problem is that .+(?!page).+ will match foopath/page/10 because the first .+ match can end at the 1 in 10, and the second can match from there until $. Instead, just assert there is no combination of characters plus the word page after (event|news)/:
^https?.+/(event|news)/(?!.*page)
Demo on regex101
If you want more than just a match/nomatch decision, you can capture the entire matching string with this regex:
^https?.+/(event|news)/(?!.*page).*$
Demo on regex101
You might be looking for
^https?.+/(event|news)/(?:(?!page).)+$
See a demo on regex101.com.
Matching is usually way easier in regex than excluding.
I would rather match your excluded words and invert the logic on the if-clause.
if(!re.match(...
I am trying to split text of clinical trials into a list of fields. Here is an example doc: https://obazuretest.blob.core.windows.net/stackoverflowquestion/NCT00000113.txt. Desired output is of the form: [[Date:<date>],[URL:<url>],[Org Study ID:<id>],...,[Keywords:<keywords>]]
I am using re.split(r"\n\n[^\s]", text) to split at paragraphs that start with a character other than space (to avoid splitting at the indented paragraphs within a field). This is all good, except the resulting fields are all (except the first field) missing their first character. Unfortunately, it is not possible to use string.partition with a regex.
I can add back the first characters by finding them using re.findall(r"\n\n[^\s]", text), but this requires a second iteration through the entire text (and seems clunky).
I am thinking it makes sense to use re.findall with some regex that matches all fields, but I am getting stuck. re.findall(r"[^\s].+\n\n") only matches the single line fields.
I'm not so experienced with regular expressions, so I apologize if the answer to this question is easily found elsewhere. Thanks for the help!
You may use a positive lookahead instead of a negated character class:
re.split(r"\n\n(?=\S)", text)
Now, it will only match 2 newlines if they are followed with a non-whitespace char.
Also, if there may be 2 or more newlines, you'd better use a {2,} limiting quantifier:
re.split(r"\n{2,}(?=\S)", text)
See the Python demo and a regex demo.
You want a lookahead. You also might want it to be more flexible as far as how many newlines / what newline characters. You might try this:
import re
r = re.compile(r"""(\r\n|\r|\n)+(?=\S)""")
l = r.split(text)
though this does seem to insert \r\n characters into the list... Hmm.
UPDATED
I want to find a string within a big text
..."img good img two_apple.txt"
Want to extract the two_apples.txt from a text, but it can change to one_apple, three_apple..so on...
When I try to use lookbehinds, it matches text all the way from the beginning.
You are mis-using lookarounds. Looks like you dont even NEED a lookaround:
pattern = r'src="images/(.+?.png")'
should work for you. As my comment suggests though, using regex is not recommended for parsing HTML/XML style documents but you do you.
EDIT - accommodate your edit:
Now that I understand your problem more, I can see why you would want to use a look-around. However, since you are looking for a file name, you know there aren't going to be any spaces in the name, so you can just ensure that your capturing token does not include spaces:
pattern = r'src="img (\w+?.png")'
^ ensure there is a space HERE because of how your text is
\w - \w is equivalent to [a-zA-Z0-9_] (any letters, numbers or underscore)
This removes the greediness of capture the first 'img ' string that pops up and ensures your capture group doesnt have any spaces.
by using \w, I am assuming you are only expecting _ and letter characters. to include anything else, make your own character group with [any characters you want to capture in here]
" ([^ ]+_apple\.txt)"
Starts with a space, ends with _apple.txt. The middle bit is anything-except-a-space which stops it matching "good img two". Parentheses to capture the bit you care about.
Try it here: https://regex101.com/r/wO7lG3/2
Hi I'm new to regexes.
I have a string that I want to match any number of A-Z a-z 0-9 - and _
I've tried the following in python however it always matches, even the empty space. Can someone tell me why that is?
re.match(r'[A-Za-z0-9_-]+', 'gfds9 41.-=,434')
Your regex matches one or more of those characters. Your text starts with one or more of those characters, hence it matches. If you want it to only match those characters then you have to match them from the beginning to the end of the text.
re.match(r'^[A-Za-z0-9_-]+$', 'gfds9 41.-=,434')
Try the alternative for it maybe it will work for you:
[\w-]+
EDIT:
Although the initial regex you provided also works for me.