Adding a Column of values in a tab delimited file

How can I add a Column of values in a file which has a certain number of rows.
I have a input file like this:

Input file:

SPATA17 1   217947738
LYPLAL1 1   219383905
FAM47E  4   77192838
SHROOM3 4   77660162
SHROOM3 4   77660731
SHROOM3 4   77662248

Output file:

SPATA17 1   217947738 file1
LYPLAL1 1   219383905 file1
FAM47E  4   77192838  file1
SHROOM3 4   77660162  file1
SHROOM3 4   77660731  file1
SHROOM3 4   77662248  file1

In this case, I want to add a Column of values, upto the number of rows in the file.The value remains consistent, such as “file1”.

The reason is I have 100 of those files.I don’t want to open each file and paste a column.
Also is there any way to automate this, by going in a directory and adding a column of values.
The value comes from the filename, which has to be added in each row of the file in the last/first column.

Answers:

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Method 1

You can use a one-liner loop like this:

for f in file1 file2 file3; do sed -i "s/$/t$f/" $f; done

For each file in the list, this will use sed to append to the end of each line a tab and the filename.

Explanation:

• Using the -i flag with sed to perform a replacement in-place, overwriting the file
• Perform a substitution with s/PATTERN/REPLACEMENT/. In this example PATTERN is $, the end of the line, and REPLACEMENT is t (= a TAB), and $f is the filename, from the loop variable. The s/// command is within double-quotes so that the shell can expand variables.

Method 2

Come on why you guys recommend those powerful tools when there’s paste command!

$ cat a
A
B
C
D
$ cat b
1
2
3
4
$ paste a b
A   1
B   2
C   3
D   4

With a little trickery, you could use paste for the OP’s purpose. However, it will not replace the files inplace:

for f in file1 file2 file3; do 
    paste $f <(yes $f | head -n $(cat $f | wc -l)) > $f.new
done

This will paste the respective filename as the last column of each file into new file filename.new

Method 3

You can use awk:

awk '{print $0, FILENAME}' file1 file2 file3 ...

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0