removing numeric value from a string

Question

I have a file of 2000 rows and 1 column

1007_s_at1
1007_s_at2
1007_s_at3
1007_s_at4
1007_s_at5
1007_s_at6
1007_s_at7
1007_s_at8
1007_s_at9
1007_s_at10

looks like above, I want to remove the last numeric value after "at". In principle whatever number is in the last should be truncated.

I have tried things like splitting them and then rejoioning it, but it just complicates the problem and I am far away from answer.

Could you please suggest something in bash or shell or python or perl to solve this.

An output like below is desired

   1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at
    1007_s_at

Thank you

Answer 1

使用Perl ：

perl -p -e "s/\d+$//" input.txt > output.txt

Answer 2

sed -i -e 's/[[:digit:]]*$//' filename

Answer 3

Just pass string.digits to .rstrip() to remove digits from the right-hand side of your strings:

import string
with open('inputfile') as infile, open('outputfile') as outfile:
    for line in infile:
        outfile.write(line.rstrip().rstrip(string.digits) + '\n')

Answer 4

If the only the number at the end changes you could potentially splice:

>>> a = '1007_s_at1'
>>> a[0:9]
'1007_s_at'

Answer 5

Python

Just strip all digits from the end.

>>> "1007_s_at10".rstrip('01234567890')
'1007_s_at'

Answer 6

If you are using Linux or Unix a simple one liner solution would be:

perl -i.bak -pe 's/\d+$//g' file.txt

if Windows:

perl -i.bak -pe "s/\d+$//g" file.txt

If you already know what it is doing then well and good otherwise, in very simple terms, -i switch with .bak would first create a backup of your file.txt and name it file.txt.bak .

The -p option would then loop over the entries in the file and print/save the output in file.txt after s/\\d+$//g removes the digits in the end.

Answer 7

Nobody's suggested a bash solution yet:

shopt -s extglob
while read line; do
    echo "${line%%*([0-9])}"
done < filename