Matching multiple arguments in a string using regex

Question

Suppose that I have following string:

mystr = """
<p>Some text and another text. </p> ![image_file_1][image_desc_1] some other text. 
<p>some text</p> 
![image_file_2][image_desc_2] and image: ![image_file_3][image_desc_3] 
test case 1: ![dont_match_1]
test case 2: [dont_match_2][dont_match_3]
finally: ![image_file_4][image_desc_4]
"""

I can get image_file_X's using the following code:

import re
re.findall('(?<=!\[)[^]]+(?=\]\[.*?\])', mystr)

I want to capture image_desc_X's but following does not work:

re.findall('(?!\[.*?\]\[)[^]]+(?=\])', mystr)

Any suggestions? If I can get both image_file's and image_desc's using one command that would be even better.

RomanPerekhrest · Accepted Answer · 2017-04-25 04:18:02Z

2

Use the following approach:

result = re.findall(r'!\[([^]]+)\]\[([^]]+)\]', mystr)
print(result)

The output:

[('image_file_1', 'image_desc_1'), ('image_file_2', 'image_desc_2'), ('image_file_3', 'image_desc_3'), ('image_file_4', 'image_desc_4')]

edited Apr 25, 2017 at 4:18

answered Apr 24, 2017 at 21:44

RomanPerekhrest

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1 Comment

revo Over a year ago

There is no need to express existence of ! with a positive lookbehind. Literal ! is enough.

Pedro Lobito · Accepted Answer · 2017-04-24 21:54:49Z

1

I guess you can use:

for match in re.finditer(r"!\[(.*?)\]\[(.*?)]", mystr):
    print match.group(1)
    print match.group(2)

output:

image_file_1
image_desc_1
image_file_2
image_desc_2
image_file_3
image_desc_3
image_file_4
image_desc_4

DEMO

edited Apr 24, 2017 at 21:54

answered Apr 24, 2017 at 21:45

Pedro Lobito

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2 Comments

revo Over a year ago

Are you sure case-insensitivity flag should be set here?

Pedro Lobito Over a year ago

It came by default...I'll remove it, tks!

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