正则表达式 python - 在列表中查找在第二个字符“_”到字符“.”之间具有相同数字的匹配项。

Question

I have the following list:

list_paths=imgs/foldeer/img_ABC_21389_1.tif.tif,
imgs/foldeer/img_ABC_15431_10.tif.tif,
imgs/foldeer/img_GHC_561321_2.tif.tif,
imgs_foldeer/img_BCL_871125_21.tif.tif,
...

I want to be able to run a for loop to match string with specific number,which is the number between the second occurance of "_" to the ".tif.tif" , for example, when number is 1, the string to be matched is "imgs/foldeer/img_ABC_21389_1.tif.tif", for number 2, the match string will be "imgs/foldeer/img_GHC_561321_2.tif.tif".

For that, I wanted to use regex expression. Based on this answer, I have tested this regex expression on Regex101:

[^\r\n_]+\.[^\r\n_]+\_([0-9])

But this doesn't match anything, and also doesn't make sure that it will take the exact number, so if number is 1, it might also select items with number 10.

My end goal is to be able to match items in the list that have the request number between the 2nd occurrence of "_" to the first occirance of ".tif", using regex expression, looking for help with the regex expression.

EDIT: The output should be the whole path and not only the number.

Answer 1

I'll show you something working and equally ugly as regex which I hate:

data = ["imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_15431_10.tif.tif",
"imgs/foldeer/img_GHC_561321_2.tif.tif",
"imgs_foldeer/img_BCL_871125_21.tif.tif"]

numbers = [int(x.split("_",3)[-1].split(".")[0]) for x in data]

• First split gives ".tif.tif"
• extract the last element
• split again by the dot this time, take the first element (thats your number as a string), cast it to int

Please keep in mind it's gonna work only for the format you provided, no flexibility at all in this solution (on the other hand regex doesn't give any neither)

Answer 2

without regex if allowed.

import re
s= 'imgs/foldeer/img_ABC_15431_10.tif.tif'
last =s[s.rindex('_')+1:]
print(re.findall(r'\d+', last)[0])

Gives #

10

Answer 3

[0-9]*(?=\.tif\.tif)

This regex expression uses a lookahead to capture the last set of numbers (what you're looking for)

Answer 4

Try this:

import re

s = '''imgs/foldeer/img_ABC_21389_1.tif.tif
imgs/foldeer/img_ABC_15431_10.tif.tif
imgs/foldeer/img_GHC_561321_2.tif.tif
imgs_foldeer/img_BCL_871125_21.tif.tif'''



number = 1
res1 = re.findall(f".*_{number}\.tif.*", s)

number = 21
res21 = re.findall(f".*_{number}\.tif.*", s)


print(res1)
print(res21)

Results

['imgs/foldeer/img_ABC_21389_1.tif.tif']
['imgs_foldeer/img_BCL_871125_21.tif.tif']

Answer 5

Your pattern [^\r\n_]+\.[^\r\n_]+\_([0-9]) does not match anything, because you are matching an underscore \_ (note that you don't have to escape it) after matching a dot, and that does not occur in the example data.

Then you want to match a digit, but the available digits only occur before any of the dots.

In your question, the numbers that you are referring to are after the 3rd occurrence of the _

---

What you could do to get the path(s) is to make the number a variable for the number you want to find:

^\S*?/(?:[^\s_/]+_){3}\d+\.tif\b[^\s/]*$

Explanation

• \S*? Match optional non whitespace characters, as few as possible
• / Match literally
• (?:[^\s_/]+_){3} Match 3 times (non consecutive) _
• \d+ Match 1+ digits
• \.tif\b[^\s/]* Match .tif followed by any char except /
• $ End of string

See a regex demo and a Python demo .

Example using a list comprehension to return all paths for the given number:

import re

number = 10
pattern = rf"^\S*?/(?:[^\s_/]+_){{3}}{number}\.tif\b[^\s/]*$"

list_paths = [
     "imgs/foldeer/img_ABC_21389_1.tif.tif",
     "imgs/foldeer/img_ABC_15431_10.tif.tif",
     "imgs/foldeer/img_GHC_561321_2.tif.tif",
     "imgs_foldeer/img_BCL_871125_21.tif.tif",
     "imgs_foldeer/img_BCL_871125_21.png.tif"
]

res = [lp for lp in list_paths if re.search(pattern, lp)]
print(res)

Output

['imgs/foldeer/img_ABC_15431_10.tif.tif']