Printing entire field, if at least one row is matching by AWK

Shell Programming and Scripting
1

Dear all,
I have been trying to print an entire field, if the first line of the field is matching.
For example, my input looks something like this.

aaa      ddd       zzz
123      987       126
24        0.650    985
354      9864     0.32
0.333   4324      000

I am looking for a pattern, for example "ddd", which is in the second field and first row. Then I need to print the entire field, output should look like below.

ddd
987
0.650
9864
4324

How can I solve this by awk or sed?
Your help is appreciated.

2

I imagine there are much neater ways, but this seems to work:

 awk -vMYCOL="ddd" '{if (NR == 1) { for (i=1;i<=NF;i++) { if ($i == MYCOL) { mycolnum = i } } } { print $mycolnum }}' file1

EDIT: Changed a bit to output the header line.

Also, if there are no matching columns it just prints everything - if you want no output you'd need an additional if:

 awk -vMYCOL="ddd" '{if (NR == 1) { for (i=1;i<=NF;i++) { if ($i == MYCOL) { mycolnum = i } } } { if (mycolnum) { print $mycolnum }}}' file1
1 Like
3

try this,

awk '{for(i=1;i<=NF;i++) {if(NR==1 && $i~/ddd/) {print $i;a=i;next}if(a>0){print $a;next}}}'   filename
1 Like
4

If Perl is acceptable (this should handle multiple matching columns):

perl -lane'
  $. == 1 and @f = grep $F[$_] =~ /ddd/, 0..@F;
  @f and print join "\t", @F[@f]
  ' infile
5 
pattern=ddd
awk -v p=$pattern 'NR==1{for(i=1;i<=NF;i++) if ($i~p) c=i}{print $c}' infile
6 
awk -v var=ddd 'NR==1{for(i=0;++i<=NF;)if($i==var){_=i}}{$0=$_}1' file