Hi, 

I have a table like this; 

    6       6:29002062:rs7755402    0       29002062        G       A
    6       6:29004091:rs9468471    0       29004091        A       G
    6       6:29006250:rs9468473    0       29006250        A       G
    6       6:29006493:rs9461499    0       29006493        C       A
    6       6:29006844:rs7743837    0       29006844        G       A

I want to remove everything before "rs" in the second column. I know I can use egrep,

     egrep -o "(rs\S+)" file | cut -d " " -f 2 > newfile

However, then I`m left with only the rs string;

    rs7755402
    rs9468471
    rs9468473
    rs9461499
    rs7743837
    rs6919044
    rs41424052
    rs6924824
    rs6456886
    rs6456887

But I actually want the other columns too. 
Any help is greatly appreciated!
    sed 's/\t\b.*:rs/\trs/' file > newfile

Explantion:

    s/ = substitution
    \t\b = look for tab, and then a word boundary
    .*:rs/\trs = ./ matches one or more charaters. you match everything till rs and replace it by tab and rs (ie \trs)