Match file names and not substrings in python regex

Question

I am trying to match a list of file names using a regex. Instead of matching just the full name, it is matching both the name and a substring of the name.

Three example files are

t0 = r"1997_06_daily.txt"
t1 = r"2010_12_monthly.txt"
t2 = r"2018_01_daily_images.txt"

I am using the regex d.

a = r"[0-9]{4}"
b = r"_[0-9]{2}_"
c = r"(daily|daily_images|monthly)"
d = r"(" + a + b + c + r".txt)"

when I run

t0 = r"1997_06_daily.txt"
t1 = r"2010_12_monthly.txt"
t2 = r"2018_01_daily_images.txt"

a = r"[0-9]{4}"
b = r"_[0-9]{2}_"
c = r"(daily|daily_images|monthly)"
d = r"(" + a + b + c + r".txt)"

for t in (t0, t1, t2):
    m = re.match(d, t)
    if m is not None:
        print(t, m.groups(), sep="\n", end="\n\n")

I get

1997_06_daily.txt
("1997_06_daily.txt", "daily")

2010_12_monthly.txt
("2010_12_monthly.txt", "monthly")

2018_01_daily_images.txt
("2018_01_daily_images.txt", "daily_images")

How can I force the regex to only return the version that includes the full file name and not the substring?

taras · Accepted Answer · 2018-07-27 19:44:53Z

1

You should make your c pattern non-capturing with '?:'

c = r"(?:daily|daily_images|monthly)"

answered Jul 27, 2018 at 19:44

taras

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Comments

Bill S. · Accepted Answer · 2018-07-27 19:46:23Z

0

This is working correctly. The issue you are seeing is how groups work in regex. Your regex c is in parentheses. Parentheses in regex signify that this match should be treated as a group. By printing m.group(), you are printing a tuple of all the groups that matched. Luckily, the first element in the group is always the full match, so just use the following:

print(t, m.groups()[0], sep="\n", end="\n\n")

answered Jul 27, 2018 at 19:46

Bill S.

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user669132 · Accepted Answer · 2018-07-27 19:54:15Z

0

I know you're only looking for regex solutions but you could easily use os module to split the extension and return index 0. Otherwise, as Bill S. stated, m.groups()[0] returns the 0th index of the regex group.

# os solution
import os

s = "1997_06_daily.txt"

os.path.splitext(s)[0]

answered Jul 27, 2018 at 19:54

user669132

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