How to delete the first column ( which is in fact row names) from a data file in linux? - 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 ("\t").
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="\t" 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.

Answer 2

#Karafka I had CSV files so I added the "," separator (you can replace with yours
cut -d"," -f2- input.csv > output.csv
Then, I used a loop to go over all files inside the directory
# files are in the directory tmp/
for f in tmp/*
do
name=`basename $f`
echo "processing file : $name"
#kepp all column excep the first one of each csv file
cut -d"," -f2- $f > new/$name
#files using the same names are stored in directory new/
done

Answer 3

You can use cut command with --complement option:
cut -f1 -d" " --complement input.file > output.file
This will output all columns except the first one.

Answer 4

As #karakfa notes, it looks like it's the leading whitespace which is causing your issues.
Here's a sed oneliner to do the job (that will account for spaces or tabs):
sed -i.bak "s|^[ \t]\+[0-9]\+[ \t]\+||" input.txt
Explanation:
-i edit existing file in place
.bak backup original file and add .bak file extension (can use whatever you like)
s substitute
| separator (easiest character to read as sed separator IMO)
^ start match at start of the line
[ \t] match space or tab
\+ match one or more times (escape required so sed does not interpret '+' literally)
[0-9] match any number 0 - 9
As noted; the input.txt file will be edited in place. The original content of input.txt will be saved as input.txt.bak. Use just -i instead if you don't want a backup of the original file.
Also, if you know that they are definitely leading spaces (not tabs), you could shorten it to this:
sed -i.bak "s|^ \+[0-9]\+[ \t]\+||" input.txt

Answer 5

You can also achieve this with grep:
grep -E -o '[[:digit:]]([[:space:]][[:digit:]]){3}$' input.txt
Which assumes single character digit and space columns. To accommodate a variable number of spaces and digits you can do:
grep -E -o '[[:digit:]]+([[:space:]]+[[:digit:]]+){3}$' input.txt
If your grep supports the -P flag (--perl-regexp) you can do:
grep -P -o '\d+(\s+\d+){3}$' input.txt
And here are a few options if you are using GNU sed:
sed 's/^\s\+\w\+\s\+//' input.txt
sed 's/^\s\+\S\+\s\+//' input.txt
sed 's/^\s\+[0-9]\+\s\+//' input.txt
sed 's/^\s\+[[:digit:]]\+\s\+//' input.txt
Note that the grep regexes are matching the parts that we want to keep while the sed regexes are matching the parts we want to remove.