How to remove '_' from the output in python code?

Use rstrip()

val = "sad_"
print(val.rstrip('_'))
Output: sad

Description

rstip() Returns a copy of the string with right trailing characters removed.

Alternatively print(val[:-1]) will give same result in this case.

Try pattern r'_?\d*.csv$'.

Ex:

import re
from itertools import groupby

file = ["meta_data_02154.csv", "meta_data_021694.csv", "meta_data_loop_02365.csv", "meta_data_loops_0256365.csv", "output.csv"]

f = [list(i) for j, i in groupby(file, lambda a : re.split(r'\d*.csv$', a)[0])]
print(f)

for pattern in f:
    #print(pattern)
    print((re.split(r'_?\d*.csv$', pattern[0]))[0])

You may use print((re.split(r'\d*.csv$', pattern[0]))[0].rstrip('_') but you might as well use a better regex and .search instead of split.

I'm not sure what you used groupby for.

import re

file = ["meta_data_02154.csv", "meta_data_021694.csv", "meta_data_loop_02365.csv", "meta_data_loops_0256365.csv", "output.csv"]

for pattern in file:
    print(re.search(r'(.+)\d*.csv$', pattern).group(1))

Outputs

meta_data_02154
meta_data_021694
meta_data_loop_02365
meta_data_loops_0256365
output

You could use a one-liner, just splitting the filenames:

file = ["meta_data_02154.csv", "meta_data_021694.csv", "meta_data_loop_02365.csv", "meta_data_loops_0256365.csv", "output.csv"]
filePatterns = set([f.rsplit('_', 1)[0].rsplit('.csv')[0] for f in file])
print(filePatterns)

Prints:

{'meta_data_loops', 'meta_data', 'meta_data_loop', 'output'}