Using regex, how can i split a string and keep it's deliminators in the returned results? I'm trying to split a string containing numbers and strings by a set of letters followed by any numerical value including '.' however it's not appearing to work correctly.
Below is my test string, im using python 2.7 and it's not producing what id expect.
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
parts = filter(None, re.split('([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, re.IGNORECASE))
print len(parts), parts
>>> 3 ['M160.394,83.962', 'L121.5,52', 'L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z']
I would expect it to give me this
>>> 10 ['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']
It should output a list of strings where each string starts with a letter, found in the original regex MLHVCSQTAZ
In your code you are passing re.IGNORECASE as 3rd argument to re.split but 3rd argument of re.split is maxsplit not flags.
re.IGNORECASE equals to 2 hence your input is split only two times.
You may use:
>>> list(filter(None, re.split(r'([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, 0, re.I)))
['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']
Or use inline mode for ignore case:
re.split(r'(?i)([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s)
I suggest using this simple re.findall code that uses almost identical regex:
parts = re.findall('(?i)[MLHVCSQTAZ][^MLHVCSQTAZ]*', s)
Reference: SRE_FLAG_IGNORECASE = 2 in lib/python2.7/sre_constants.py (thanks to comment from #vks)
You can use re.findall:
import re
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
result = re.findall('[A-Z][\.\d,]+|[A-Z]', s)
Output:
['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']
parts = filter(None, re.split('([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, flags=re.IGNORECASE))
You need to use flags.Check re.split function definition.
Default re does not support 0 width assertion split.So you can also use regex module for that.
import regex
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
print regex.split('(?=[MLHVCSQTAZ][^MLHVCSQTAZ])', s, flags=regex.IGNORECASE|regex.VERSION1)
I want to get String before last occurrence of my given sub string.
My String was,
path =
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov
my substring, 1001-1010 which will occurred twice. all i want is get string before its last occurrence.
Note: My substring is dynamic with different padding but only number.
I want,
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v
I have done using regex and slicing,
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> q = re.findall("\d*-\d*",p)
>>> q[-1].join(p.split(q[-1])[:-1])
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v'
>>>
Is their any better way to do by purely using regex?
Please Note I have tried so many eg:
regular expression to match everything until the last occurrence of /
Regex Last occurrence?
I got answer by using regex with slicing but i want to achieve by using regex alone..
Why use regex. Just use built in string methods:
path = "D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov"
index = path.rfind("1001-1010")
print(path[:index])
You can use a simple greedy match and a capture group:
(.*)1001-1010
Your match is in capture group #1
Since .* is greedy by nature, it will match longest match before matching your keyword 1001-1010.
RegEx Demo
As per comments below if keyword is not a static string then you may use this regex:
r'(.*\D)\d+-\d+'
Python Code:
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> print (re.findall(r'(.*\D)\d+-\d+', p))
['D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v']
Thanks #anubhava,
My first regex was,
.*(\d*-\d*)\/
Now i have corrected mine..
.*(\d*-\d*)
or
(.*)(\d*-\d*)
which gives me,
>>> q = re.search('.+(\d*-\d*)', p)
>>> q.group()
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v0001-1001'
>>>
(.*\D)\d+-\d+
this gives me exactly what i want...
>>> q = re.search('(.*\D)\d+-\d+', p)
>>> q.groups()
('D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v',)
>>>
I have a list of strings
['time_10', 'time_23', 'time_345', 'date_10', 'date_23', 'date_345']
I want to use regular expression to get strings that end with a specific number.
As I understand, first I have to combine all strings from the list into large string, then use form some kind of a pattern to use it for regular expression
I would be grateful if you could provide
regex(some_pattern, some_string)
that would return
['time_10', 'date_10']
or just
'time_10, date_10'
str.endswith is enough.
l = ['time_10', 'time_23', 'time_345', 'date_10', 'date_23', 'date_345']
result = [s for s in l if s.endswith('10')]
print(result)
['time_10', 'date_10']
If you insist on using regex,
import re
result = [s for s in l if re.search('10$', s)]
I'm trying to find all links on a webpage in the form of "http://something" or https://something. I made a regex and it works:
L = re.findall(r"http://[^/\"]+/|https://[^/\"]+/", site_str)
But, is there a shorter way to write this? I'm repeating ://[^/\"]+/ twice, probably without any need. I tried various stuff, but it doesn't work. I tried:
L = re.findall(r"http|https(://[^/\"]+/)", site_str)
L = re.findall(r"(http|https)://[^/\"]+/", site_str)
L = re.findall(r"(http|https)(://[^/\"]+/)", site_str)
It's obvious I'm missing something here or I just don't understand python regexes enough.
You are using capturing groups, and .findall() alters behaviour when you use those (it'll only return the contents of capturing groups). Your regex can be simplified, but your versions will work if you use non-capturing groups instead:
L = re.findall(r"(?:http|https)://[^/\"]+/", site_str)
You don't need to escape the double quote if you use single quotes around the expression, and you only need to vary the s in the expression, so s? would work too:
L = re.findall(r'https?://[^/"]+/', site_str)
Demo:
>>> import re
>>> example = '''
... "http://someserver.com/"
... "https://anotherserver.com/with/path"
... '''
>>> re.findall(r'https?://[^/"]+/', example)
['http://someserver.com/', 'https://anotherserver.com/']
I need to find the value of "taxid" in a large number of strings similar to one given below. For this particular string, the 'taxid' value is '9606'. I need to discard everything else. The "taxid" may appear anywhere in the text, but will always be followed by a ":" and then number.
score:0.86|taxid:9606(Human)|intact:EBI-999900
How to write regular expression for this in python.
>>> import re
>>> s = 'score:0.86|taxid:9606(Human)|intact:EBI-999900'
>>> re.search(r'taxid:(\d+)', s).group(1)
'9606'
If there are multiple taxids, use re.findall, which returns a list of all matches:
>>> re.findall(r'taxid:(\d+)', s)
['9606']
for line in lines:
match = re.match(".*\|taxid:([^|]+)\|.*",line)
print match.groups()