How can I remove the first column in every row besides the columns in shell?

Question

This is an example of what the data I have looks like:

ID    pos    gene
1   SAMPLE1   1234     BRCA
2   SAMPLE2   2910    EGFR
3   SAMPLE3   1271    MYC

This is the desired output

ID    pos    gene
SAMPLE1   1234    BRCA
SAMPLE2   2910    EGFR
SAMPLE3   1271    MYC

I tried cut -f2- mydata.txt but that removes the entire column and I would still like to keep ID as the column name.

Answer 1

$ sed -E '2,$s/[^ ]+ +//' file

Answer 2

Using awk and sub to preserve the spaces. If the file is tab separated, forget this one:

$ awk '
FNR==1 {                    # first record
    nf=NF                   # store field count to nf
}
NF>nf {                     # if NF > nf
    for(i=1;i<=NF-nf;i++)   # using sub remove NF-nf first fields
        sub(/^[^ ]+ +/,"")
}1' file                    # output

Output:

ID    pos    gene
SAMPLE1   1234     BRCA
SAMPLE2   2910    EGFR
SAMPLE3   1271    MYC

Answer 3

How can I remove the first column in every row besides the columns in shell?

You may use this awk :

awk 'NR > 1 {sub(/^[ \t]*[^ \t]+[ \t]+/, "")} 1' file

ID    pos    gene
SAMPLE1 1234 BRCA
SAMPLE2 2910 EGFR
SAMPLE3 1271 MYC

Answer 4

head -n1 mydata.txt; tail -n +2 mydata.txt | cut -d' ' -f2-

Answer 5

A trick for skipping the first line in something that doesn't otherwise have a way to handle a header line is to use a command group where everything in it shares the same standard input and output streams. You read the first line and echo it, and then the real program works on the rest of the input:

{ IFS= read -r line && echo "$line" && cut -f2-; } < mydata.txt

Answer 6

$ awk 'NR>1{sub(/[^[:space:]]+[[:space:]]+/,"")}1' file
ID    pos    gene
SAMPLE1   1234     BRCA
SAMPLE2   2910    EGFR
SAMPLE3   1271    MYC

If your file is tab-separated AND might have empty 2nd fields or blanks in the 1st field then just change [^[:space:]]+[[:space:]]+ to ^[^\t]*[\t]