How do I add a group to my regex?
Here is my regex: (?<=code )(\d+)
Here is my code:
rsize= re.compile(r'(?<=code )(\d+)')
code = rsize.search(codeblock).group("code")
How come when I run the code I get the error: IndexError: no such group ? How do I write this regex to create a group named code?
EDIT
I read the responses, but, my question is, how exactly do I append that to my regex?
The "named group" syntax is a little bit different:
(?P<name>group)
Example:
>>> import re
>>>
>>> s = "1234 extract the numbers"
>>> pattern = re.compile(r'(?P<code>\d+)')
>>> pattern.search(s).group("code")
'1234'
A named group in Python's re syntax is defined as (?P<name>...) where name is the name of the group and ... is the pattern the group matches.
So if your goal is to create a named group 'code' that matches a set of digits, you'd want:
(?P<code>\d+)
Related
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'm using a Python script to read data from our corporate instance of JIRA. There is a value that is returned as a string and I need to figure out how to extract one bit of info from it. What I need is the 'name= ....' and I just need the numbers from that result.
<class 'list'>: ['com.atlassian.greenhopper.service.sprint.Sprint#6f68eefa[id=30943,rapidViewId=10468,state=CLOSED,name=2016.2.4 - XXXXXXXXXX,startDate=2016-05-26T08:50:57.273-07:00,endDate=2016-06-08T20:59:00.000-07:00,completeDate=2016-06-09T07:34:41.899-07:00,sequence=30943]']
I just need the 2016.2.4 portion of it. This number will not always be the same either.
Any thoughts as how to do this with RE? I'm new to regular expressions and would appreciate any help.
A simple regular expression can do the trick: name=([0-9.]+).
The primary part of the regex is ([0-9.]+) which will search for any digit (0-9) or period (.) in succession (+).
Now, to use this:
import re
pattern = re.compile('name=([0-9.]+)')
string = '''<class 'list'>: ['com.atlassian.greenhopper.service.sprint.Sprint#6f68eefa[id=30943,rapidViewId=10468,state=CLOSED,name=2016.2.4 - XXXXXXXXXX,startDate=2016-05-26T08:50:57.273-07:00,endDate=2016-06-08T20:59:00.000-07:00,completeDate=2016-06-09T07:34:41.899-07:00,sequence=30943]']'''
matches = pattern.search(string)
# Only assign the value if a match is found
name_value = '' if not matches else matches.group(1)
Use a capturing group to extract the version name:
>>> import re
>>> s = 'com.atlassian.greenhopper.service.sprint.Sprint#6f68eefa[id=30943,rapidViewId=10468,state=CLOSED,name=2016.2.4 - XXXXXXXXXX,startDate=2016-05-26T08:50:57.273-07:00,endDate=2016-06-08T20:59:00.000-07:00,completeDate=2016-06-09T07:34:41.899-07:00,sequence=30943]'
>>> re.search(r"name=([0-9.]+)", s).group(1)
'2016.2.4'
where ([0-9.]+) is a capturing group matching one or more digits or dots, parenthesis define a capturing group.
A non-regex option would involve some splitting by ,, = and -:
>>> l = [item.split("=") for item in s.split(",")]
>>> next(value[1] for value in l if value[0] == "name").split(" - ")[0]
'2016.2.4'
This, of course, needs testing and error handling.
Regex to retrieve the last portion of a string:
https://play.google.com/store/apps/details?id=com.lima.doodlejump
I'm looking to retrieve the string followed by id=
The following regex didn't seem to work in python
sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
re.search("id=(.*?)", sampleURL).group(1)
The above should give me an output:
com.lima.doodlejump
Is my search group right?
Your regular expression
(.*?)
will not work because, it will match between zero and unlimited times, as few times as possible (becasue of the ?). So, you have the following choices of RegEx
(.*) # Matches the rest of the string
(.*?)$ # Matches till the end of the string
But, you don't need RegEx at all here, simply split the string like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
print data.split("id=", 1)[-1]
Output
com.lima.doodlejump
If you really have to use RegEx, you can do like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
import re
print re.search("id=(.*)", data).group(1)
Output
com.lima.doodlejump
I'm surprised that nobody has mentioned urlparse yet...
>>> s = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> urlparse.urlparse(s)
ParseResult(scheme='https', netloc='play.google.com', path='/store/apps/details', params='', query='id=com.lima.doodlejump', fragment='')
>>> urlparse.parse_qs(urlparse.urlparse(s).query)
{'id': ['com.lima.doodlejump']}
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id']
['com.lima.doodlejump']
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id'][0]
'com.lima.doodlejump'
The HUGE advantage here is that if the url query string gets more components then it could easily break the other solutions which rely on a simple str.split. It won't confuse urlparse however :).
Just split it in the place you want:
id = url.split('id=')[1]
If you print id, you'll get:
com.lima.doodlejump
Regex isn't needed here :)
However, in case there are multiple id=s in your string, and you only wanted the last one:
id = url.split('id=')[-1]
Hope this helps!
This works:
>>> import re
>>> sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> re.search("id=(.+)", sampleURL).group(1)
'com.lima.doodlejump'
>>>
Instead of capturing non-greedily for zero or more characters, this code captures greedily for one or more.
I have 2 strings that contains the following:
name = 'Kalvo'
info = 'PC1:\nKalvo (Read)(Write)\nKL27 (Read)(Write)'
Now what I want achieve here is to search the info for the word found in name and print out everything after name.
Lets say I'm searching the string info for string name and it should the print out:
Kalvo (Read)(Write)
I tried using re.search and re.findall but I can't get them to work.
Help is much appreciated.
Br,
Toby
You can use str.format to insert the name in the Regex pattern. Then, using .*, you can get any characters after it. See a demonstration below:
>>> from re import findall
>>> name = 'Kalvo'
>>> info = 'PC1:\nKalvo (Read)(Write)\nKL27 (Read)(Write)'
>>> findall("{}.*".format(name), info)[0]
'Kalvo (Read)(Write)'
>>>
I'm using Python to parse a file in search for e-mail addresses, but I can't figure out what the syntax for alternative regexps should be. Here's the code:
addresses = []
pattern = '(\w+)#(\w+\.com)|(\w+)#(it.\w+\.com)'
for line in file:
matches = re.findall(pattern,line)
for m in matches:
address = '%s#%s' % m
addresses.append(address)
So I want to find addresses that look like john#company.com or john#it.company.com, but the above code doesn't work because either the first two groups are empty or the last two groups are empty. What is the correct solution? I need to use groups to store the user name (before #) and server name (after #) separately.
EDIT: Matching email adresses is only an example. What I'm trying to find out is how to match different regexps that have only one thing in common - they match two groups.
(\w+)#((?:it\.)?\w+\.com)
You want to capture the part after the # whether it's example.com or it.example.com, so you put both options inside the same capture group. But since they share a similar format, you can condense (it\.\w+\.com|\w+\.com) to just ((it\.)?\w+\.com)
The (?: ) makes that parens a non-capturing group, so it won't take part in your matched groups. There will be one match for the first (\w+), and one match for the whole ((?:it\.)?\w+\.com) after the #. That's two matches total, plus the default group-0 match for the full string.
EDIT: To answer your new question, all you have to do is follow the grouping I used, but stop before you condense it.
If your test cases are:
1) example#abcdef
2) example#123456
You could write your regex as such: (\w+)#([a-zA-Z]+|\d+), which would always have the part before the # in group 1, and the part after in group 2. Notice that there are only two pairs of parens, and the |("or") operator appears inside of the second parens group.
I once found here a well written email regex, it was build for extracting a wide range of valid email adresses from a generic string, so it should also be able to do what you're looking for.
Example:
>>> email_regex = re.compile("""((([a-zA-Z0-9!\#\$%&'*+\-\/=?^_`{|}~]+|"([a-zA-Z0-9!\#\$%&'*+\-\/=?^_`{|}~(),:;<>#\[\]\.]|\\[ \\"])*")\.)*([a-zA-Z0-9!\#\$%&'*+\-\/=?^_`{|}~]+|"([a-zA-Z0-9!\#\$%&'*+\-\/=?^_`{|}~(),:;<>#\[\]\.]|\\[ \\"])*"))#((([a-zA-Z0-9]([a-zA-Z0-9]*(\-[a-zA-Z0-9]*)*)?\.)*[a-zA-Z]+|\[((0?\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])\.){3}(0?\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])\]|\[[Ii][Pp][vV]6(:[0-9a-fA-F]{0,4}){6}\]))""")
>>>
>>> m = email_regex.search('john#it.company.com')
>>> m.group(0)
'john#it.company.com'
>>> m.group(1)
'john'
>>> m.group(7)
'it.company.com'
>>>
>>> n = email_regex.search('john#company.com')
>>> n.group(0)
'john#company.com'
>>> n.group(1)
'john'
>>> n.group(7)
'company.com'