Extract substring from a python string

Question

I want to extract the string before the 9 digit number below:

tmp = place1_128017000_gw_cl_mask.tif

The output should be place1

I could do this: tmp.split('_')[0] but I also want the solution to work for:

tmp = place1_place2_128017000_gw_cl_mask.tif where the result would be: place1_place2

You can assume that the number will also be 9 digits long

Try Python regular rexpressions: m = re.search("\d{9}", tmp); print(m.group()) — DarrylG
– DarrylG, Commented Jul 13, 2022 at 3:46
Have you looked at regular expressions at all? Check out re.search(), a regular expression for what you need would look like '.+(?=_\d{9}_)', that is: one or more characters before an underscore before 9 digits (followed by an underscore); check sites like regex101.com to design and test regular expressions. — Grismar
– Grismar, Commented Jul 13, 2022 at 3:46

Grismar · Accepted Answer · 2022-07-13 03:51:58Z

3

Using regular expressions and the lookahead feature of regex, this is a simple solution:

tmp = "place1_place2_128017000_gw_cl_mask.tif"
m = re.search(r'.+(?=_\d{9}_)', tmp)
print(m.group())

Result:

place1_place2

Note that the \d{9} bit matches exactly 9 digits. And the bit of the regex that is in (?= ... ) is a lookahead, which means it is not part of the actual match, but it only matches if that follows the match.

answered Jul 13, 2022 at 3:51

Grismar

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Comments

Tim Biegeleisen · Accepted Answer · 2022-07-13 03:47:00Z

3

Assuming we can phrase your problem as wanting the substring up to, but not including the underscore which is followed by all numbers, we can try:

tmp = "place1_place2_128017000_gw_cl_mask.tif"
m = re.search(r'^([^_]+(?:_[^_]+)*)_\d+_', tmp)
print(m.group(1))  # place1_place2

answered Jul 13, 2022 at 3:47

Tim Biegeleisen

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Comments

Robert · Accepted Answer · 2022-07-13 03:56:39Z

1

Use a regular expression:

import re

places = (
    "place1_128017000_gw_cl_mask.tif",
    "place1_place2_128017000_gw_cl_mask.tif",
)
pattern = re.compile("(place\d+(?:_place\d+)*)_\d{9}")
for p in places:
    matched = pattern.match(p)
    if matched:
        print(matched.group(1))

prints:

place1

place1_place2

The regex works like this (adjust as needed, e.g., for less than 9 digits or a variable number of digits):

( starts a capture
place\d+ matches "places plus 1 to many digits"
(?: starts a group, but does not capture it (no need to capture)
_place\d+ matches more "places"
) closes the group
* means zero or many times the previous group
) closes the capture
\d{9} matches 9 digits

The result is in the first (and only) capture group.

edited Jul 13, 2022 at 3:56

answered Jul 13, 2022 at 3:50

Robert

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Comments

SanV · Accepted Answer · 2022-07-13 04:45:24Z

1

Here's a possible solution without regex (unoptimized!):

def extract(s):
    result = ''
    for x in s.split('_'):
        try: x = int(x)
        except: pass
        if isinstance(x, int) and len(str(x)) == 9:
            return result[:-1]
        else:
            result += x + '_'

tmp = 'place1_128017000_gw_cl_mask.tif'
tmp2 = 'place1_place2_128017000_gw_cl_mask.tif'

print(extract(tmp))   # place1
print(extract(tmp2))  # place1_place2

answered Jul 13, 2022 at 4:45

SanV

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