bash - How to delete the first column ( which is in fact row names) from a data file in linux?

Question

I have data file with many thousands columns and rows. I want to delete the first column which is in fact the row counter. I used this command in linux:

cut -d " " -f 2- input.txt > output.txt

but nothing changed in my output. Does anybody knows why it does not work and what should I do?

This is what my input file looks like:

col1 col2 col3 col4 ...
     1 0 0 0 1
     2 0 1 0 1
     3 0 1 0 0
     4 0 0 0 0 
     5 0 1 1 1 
     6 1 1 1 0
     7 1 0 0 0 
     8 0 0 0 0
     9 1 0 0 0
     10 1 1 1 1
     11 0 0 0 1
    .
    .
    .

I want my output look like this:

col1 col2 col3 col4 ...
0 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0 
0 1 1 1 
1 1 1 0
1 0 0 0 
0 0 0 0
1 0 0 0
1 1 1 1
0 0 0 1
.
.
.

I also tried the sed command:

 sed '1d' input.file > output.file

But it deletes the first row not the first column.

Could anybody guide me?

Answer 1

idiomatic use of cut will be

cut -f2- input > output

if you delimiter is tab ("").

Or, simply with awk magic (will work for both space and tab delimiter)

 awk '{$1=""}1' input | awk '{$1=$1}1' > output

first awk will delete field 1, but leaves a delimiter, second awk removes the delimiter. Default output delimiter will be space, if you want to change to tab, add -vOFS="" to the second awk.

UPDATED

Based on your updated input the problem is the initial spaces that cut treats as multiple columns. One way to address is to remove them first before feeding to cut

sed 's/^ *//' input | cut -d" " -f2- > output

or use the awk alternative above which will work in this case as well.