I’m trying to use python and regex to get the last set of integers in a filename (string) Which the method does what i need, however I want to also return the inverse or remaining parts of the regex. How can i do that?
Here is the regex ([0-9]+|#+)(?!.*([0-9]+|#+))
JavaScript
x
25
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1
import re2
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values = [4
'image.0001',5
'image###',6
'###image###',7
'image001',8
'image_001',9
'001',10
'0001.image',11
'001image',12
'001_image',13
'image',14
'01_image01',15
'03_image01',16
]17
18
pattern = '([0-9]+|#+|@+)'19
regex = '{0}(?!.*{0})'.format(pattern)20
21
for v in values:22
result = re.search(regex, v)23
if result:24
print result.groups()25
Currently it is returning…. ('01', None) I’d like it to return something like ('image', '0001')
Updated
Optionally is there a way to split the strings by groups of numbers…for example
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'image.0001' > ['image.', '0001']2
'image###' > ['image', '###']3
'###image###' > ['###', 'image', '###']4
'image001' > ['image', '001']5
'image_001' > ['image_', '001']6
'001' > ['001']7
'0001.image' > ['0001', '.image']8
'001image' > ['001', 'image']9
'001_image' > ['001', '_image']10
'image' > ['image']11
'01_image01' > ['01', '_image', '01']12
'03_image01' > ['03', '_image', '01']13
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Answer
EDIT:
Use
JavaScript
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re.findall(r'd+|#+|@+|[^#@d]+', v)2
See proof.
Explanation
JavaScript
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20
1
--------------------------------------------------------------------------------2
d+ digits (0-9) (1 or more times (matching3
the most amount possible))4
--------------------------------------------------------------------------------5
| OR6
--------------------------------------------------------------------------------7
#+ '#' (1 or more times (matching the most8
amount possible))9
--------------------------------------------------------------------------------10
| OR11
--------------------------------------------------------------------------------12
@+ '@' (1 or more times (matching the most13
amount possible))14
--------------------------------------------------------------------------------15
| OR16
--------------------------------------------------------------------------------17
[^#@d]+ any character except: '#', '@', digits (0-18
9) (1 or more times (matching the most19
amount possible))20
ORIGINAL:
Use re.split, add capturing group to keep captured part inside the result:
JavaScript
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import re2
3
values = [4
'image.0001',5
'image###',6
'###image###',7
'image001',8
'image_001',9
'001',10
'0001.image',11
'001image',12
'001_image',13
'image',14
'01_image01',15
'03_image01',16
]17
18
pattern = '[0-9]+|#+|@+'19
regex = re.compile(r'({0})(?!.*(?:{0}))'.format(pattern))20
for v in values:21
print(regex.split(v))22
See Python proof
Results:
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['image.', '0001', '']2
['image', '###', '']3
['###image', '###', '']4
['image', '001', '']5
['image_', '001', '']6
['', '001', '']7
['', '0001', '.image']8
['', '001', 'image']9
['', '001', '_image']10
['image']11
['01_image', '01', '']12
['03_image', '01', '']13
See regex proof.
Explanation
JavaScript
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1
--------------------------------------------------------------------------------2
( group and capture to 1:3
--------------------------------------------------------------------------------4
[0-9]+ any character of: '0' to '9' (1 or more5
times (matching the most amount6
possible))7
--------------------------------------------------------------------------------8
| OR9
--------------------------------------------------------------------------------10
#+ '#' (1 or more times (matching the most11
amount possible))12
--------------------------------------------------------------------------------13
| OR14
--------------------------------------------------------------------------------15
@+ '@' (1 or more times (matching the most16
amount possible))17
--------------------------------------------------------------------------------18
) end of 119
--------------------------------------------------------------------------------20
(?! look ahead to see if there is not:21
--------------------------------------------------------------------------------22
.* any character except n (0 or more times23
(matching the most amount possible))24
--------------------------------------------------------------------------------25
(?: group, but do not capture:26
--------------------------------------------------------------------------------27
[0-9]+ any character of: '0' to '9' (1 or28
more times (matching the most amount29
possible))30
--------------------------------------------------------------------------------31
| OR32
--------------------------------------------------------------------------------33
#+ '#' (1 or more times (matching the34
most amount possible))35
--------------------------------------------------------------------------------36
| OR37
--------------------------------------------------------------------------------38
@+ '@' (1 or more times (matching the39
most amount possible))40
--------------------------------------------------------------------------------41
) end of grouping42
--------------------------------------------------------------------------------43
) end of look-ahead44