I would like to find the first occurrence of a string, after a certain other string, and between a certain pattern. The documents that I am parsing are not xml, but have similar rules of start/end. and example of what I would be looking at:
...b.filext", "{xxxxx-xxx-xxx-xxx-xxxxxxxxxxx}"
and I am trying to get the string between { and }, right after an occurrence of a vcxproj, every time it happens in the document.
I tried the following, but I get a list of None:
my_list=[]
for line in text.split('.vcxproj'):
if '{' in line:
my_list.append(re.match( r"(?<=\{)(.*?)(?=\})", line))
I have tried to alter my expression but no success. Help ? Thank you.
re.match matches only at the beginning of the string. Try re.search. Alternatively, replace the loop with re.findall.
How about:
re.findall(r"vcxproj.*?\{(.*?)\}", text)
Related
I'm trying to find any text between a '>' character and a new line, so I came up with this regex:
result = re.search(">(.*)\n", text).group(1)
It works perfectly with only one result, such as:
>test1
(something else here)
Where the result, as intended, is
test1
But whenever there's more than one result, it only shows the first one, like in:
>test1
(something else here)
>test2
(something else here)
Which should give something like
test1\ntest2
But instead just shows
test1
What am I missing? Thank you very much in advance.
re.search only returns the first match, as documented:
Scan through string looking for the first location where the regular
expression pattern produces a match, and return a corresponding
MatchObject instance.
To find all the matches, use findall.
Return all non-overlapping matches of pattern in string, as a list of
strings. The string is scanned left-to-right, and matches are returned
in the order found.
Here's an example from the shell:
>>> import re
>>> re.findall(">(.*)\n", ">test1\nxxx>test2\nxxx")
['test1', 'test2']
Edit: I just read your question again and realised that you want "test1\ntest2" as output. Well, just join the list with \n:
>>> "\n".join(re.findall(">(.*)\n", ">test1\nxxx>test2\nxxx"))
'test1\ntest2'
You could try:
y = re.findall(r'((?:(?:.+?)(?:(?=[\n\r][^\n\r])\n|))+)', text)
Which returns ['t1\nt2\nt3'] for 't1\nt2\nt3\n'. If you simply want the string, you can get it by:
s = y[0]
Although it seems much larger than your initial code, it will give you your desired string.
Explanation -
((?:(?:.+?)(?:(?=[\n\r][^\n\r])\n|))+) is the regex as well as the match.
(?:(?:.+?)(?:(?=[\n\r][^\n\r])\n|)) is the non-capturing group that matches any text followed by a newline, and is repeatedly found one-or-more times by the + after it.
(?:.+?) matches the actual words which are then followed by a newline.
(?:(?=[\n\r][^\n\r])\n|) is a non-capturing conditional group which tells the regex that if the matched text is followed by a newline, then it should match it, provided that the newline is not followed by another newline or carriage return
(?=[\n\r][^\n\r]) is a positive look-ahead which ascertains that the text found is followed by a newline or carriage return, and then some non-newline characters, which combined with the \n| after it, tells the regex to match a newline.
Granted, after typing this big mess out, the regex is pretty long and complicated, so you would be better off implementing the answers you understand, rather than this answer, which you may not. However, this seems to be the only one-line answer to get the exact output you desire.
I am trying to delete all URLS in a big csv-file and replace it with the string "URL" (a so-called equivalence token). The code does what I want, but it clumps/concatenates some rows together in one row.
That means that the original csv has 63.000 rows and the output csv only 55000. That is not what I want. How can I replace links with this token and leave all columns separated?
#links are replaced with links
import re
with open('data_feat1.csv',"r", encoding="utf-8") as oldfile2, open('data_feat2.csv', 'w',encoding="utf-8") as newfile2:
for line in oldfile2:
line=re.sub(r"http\S+", r"URL", line) #replaces links with "URL"
newfile2.write(line)
newfile2.close()
The solution was to add a ' to "URL":
line=re.sub(r"http\S+", r'URL"', line) #replaces links with "URL"
I do not know why it worked, but it did!
the reason why it works is because it is using a regular expression to search for http.
the re modulehandles regular expressions. re.sub will replace the matched regular expression with the second argument (URL" in your case).
what that regular expression is doing is searching for http and everything after it. "everything after it" is designated by the \S+ characters which is saying "everything that goes until white space"
Take a look at Pythex. It'll be a good place to learn how regular expression work in Python
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 need to build a program that can read multiple lines of code, and extract the right information from each line.
Example text:
no matches
one match <'found'>
<'one'> match <found>
<'three'><'matches'><'found'>
For this case, the program should detect <'found'>, <'one'>, <'three'>, <'matches'> and <'found'> as matches because they all have "<" and "'".
However, I cannot work out a system using regex to account for multiple matches on the same line. I was using something like:
re.search('^<.*>$')
But if there are multiple matches on one line, the extra "'<" and ">'" are taken as part of the .*, without counting them as separate matches. How do I fix this?
This works -
>>> r = re.compile(r"\<\'.*?\'\>")
>>> r.findall(s)
["<'found'>", "<'one'>", "<'three'>", "<'matches'>", "<'found'>"]
Use findall instead of search:
re.findall( r"<'.*?'>", str )
You can use re.findall and match on non > characters inside of the angle brackets:
>>> re.findall('<[^>]*>', "<'three'><'matches'><'found'>")
["<'three'>", "<'matches'>", "<'found'>"]
Non-greedy quantifier '?' as suggested by anubhava is also an option.
I have a couple email addresses, 'support#company.com' and '1234567#tickets.company.com'.
In perl, I could take the To: line of a raw email and find either of the above addresses with
/\w+#(tickets\.)?company\.com/i
In python, I simply wrote the above regex as'\w+#(tickets\.)?company\.com' expecting the same result. However, support#company.com isn't found at all and a findall on the second returns a list containing only 'tickets.'. So clearly the '(tickets\.)?' is the problem area, but what exactly is the difference in regular expression rules between Perl and Python that I'm missing?
The documentation for re.findall:
findall(pattern, string, flags=0)
Return a list of all non-overlapping matches in the string.
If one or more groups are present in the pattern, return a
list of groups; this will be a list of tuples if the pattern
has more than one group.
Empty matches are included in the result.
Since (tickets\.) is a group, findall returns that instead of the whole match. If you want the whole match, put a group around the whole pattern and/or use non-grouping matches, i.e.
r'(\w+#(tickets\.)?company\.com)'
r'\w+#(?:tickets\.)?company\.com'
Note that you'll have to pick out the first element of each tuple returned by findall in the first case.
I think the problem is in your expectations of extracted values. Try using this in your current Python code:
'(\w+#(?:tickets\.)?company\.com)'
Two problems jump out at me:
You need to use a raw string to avoid having to escape "\"
You need to escape "."
So try:
r'\w+#(tickets\.)?company\.com'
EDIT
Sample output:
>>> import re
>>> exp = re.compile(r'\w+#(tickets\.)?company\.com')
>>> bool(exp.match("s#company.com"))
True
>>> bool(exp.match("1234567#tickets.company.com"))
True
There isn't a difference in the regexes, but there is a difference in what you are looking for. Your regex is capturing only "tickets." if it exists in both regexes. You probably want something like this
#!/usr/bin/python
import re
regex = re.compile("(\w+#(?:tickets\.)?company\.com)");
a = [
"foo#company.com",
"foo#tickets.company.com",
"foo#ticketsacompany.com",
"foo#compant.org"
];
for string in a:
print regex.findall(string)