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Replace last row of a column in bash/awk/sed

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# 1  
Old 02-17-2011
Replace last row of a column in bash/awk/sed
Hi,

I've got a file with 3 columns which ends like this:
Code:
...
1234  345  1400
5287  733  1400
8472  874  1400
9317  726  1400

I want to replace the last row of the last column with the value 0. So my new file will end:

Code:
...
1234  345  1400
5287  733  1400
8472  874  1400
9317  726  0

Thanks for your help
Moderator's Comments:
Mod Comment
Please use code tags when posting data and code samples!
Last edited by vgersh99; 02-17-2011 at 12:23 PM.. Reason: code tags, please!
# 2  
Old 02-17-2011
Code:
sed '$s/[0-9][0-9]*$/0/' myFile

This User Gave Thanks to vgersh99 For This Post:
jhunter87
# 3  
Old 02-17-2011
Vgersh99, if you remove the first number, it also works as '*' means zero or more occurrences:
Code:
sed '$s/[0-9]*$/0/' input_file

This User Gave Thanks to Shell_Life For This Post:
jhunter87
# 4  
Old 02-17-2011
Quote:
Originally Posted by Shell_Life
Vgersh99, if you remove the first number, it also works as '*' means zero or more occurrences:
Code:
sed '$s/[0-9]*$/0/' input_file

true, but....
Code:
...
1234 345 1400
5287 733 1400
8472 874 1400
9317 726

results in:
Code:
...
1234 345 1400
5287 733 1400
8472 874 1400
9317 726 0

which probably isn't a concern for the OP....
# 5  
Old 02-17-2011
You are right, as the OP says:
Code:
I've got a file with 3 columns which ends like this:

Cheers.
# 6  
Old 02-17-2011
Thanks a lot guys!
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