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

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# 1  
Old 10-07-2011
Printing entire field, if at least one row is matching by AWK
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.
Code:
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.
Code:
ddd
987
0.650
9864
4324

How can I solve this by awk or sed?
Your help is appreciated.
Last edited by radoulov; 10-07-2011 at 08:06 AM.. Reason: Code tags.
# 2  
Old 10-07-2011
I imagine there are much neater ways, but this seems to work:
Code:
 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:
Code:
 awk -vMYCOL="ddd" '{if (NR == 1) { for (i=1;i<=NF;i++) { if ($i == MYCOL) { mycolnum = i } } } { if (mycolnum) { print $mycolnum }}}' file1

Last edited by CarloM; 10-07-2011 at 08:02 AM.. Reason: Output the header
This User Gave Thanks to CarloM For This Post:
Chulamakuri
# 3  
Old 10-07-2011
try this,
Code:
awk '{for(i=1;i<=NF;i++) {if(NR==1 && $i~/ddd/) {print $i;a=i;next}if(a>0){print $a;next}}}'   filename

This User Gave Thanks to pravin27 For This Post:
Chulamakuri
# 4  
Old 10-07-2011
If Perl is acceptable (this should handle multiple matching columns):

Code:
perl -lane'
  $. == 1 and @f = grep $F[$_] =~ /ddd/, 0..@F;
  @f and print join "\t", @F[@f]
  ' infile

# 5  
Old 10-09-2011
Code:
pattern=ddd
awk -v p=$pattern 'NR==1{for(i=1;i<=NF;i++) if ($i~p) c=i}{print $c}' infile

# 6  
Old 10-09-2011
Code:
awk -v var=ddd 'NR==1{for(i=0;++i<=NF;)if($i==var){_=i}}{$0=$_}1' file

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