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grep option to print the first match of each member of a list?

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# 1  
Old 11-22-2011
grep option to print the first match of each member of a list?
Hello,
How to grep only the first match of each (unique) member of a list from the file?
Say member.list contains:
Code:
member1
member2
member3

and table.tab which is sorted by the first 2nd and then 3rd column.
Code:
member1 1.2 234
member1 1.1 234
member2 3.3 111
member2 2.3 222
member2 2.3 111
member3 1.0 123
member3 1.0 134

and output is:
Code:
member1 1.2 234
member2 3.3 111
member3 1.0 123

Option -m or -c seems not what I need. Thanks a lot!

Yifang
# 2  
Old 11-22-2011
grep has no recall, and can't do logical statements.

how about awk, which has variables for recall and can do logical statements:

Code:
awk 'BEGIN { while(getline<"datafile") M[$1]=1 }; { if(M[$1]==1) { print; M[$1]=2 } }' listfile

# 3  
Old 11-22-2011
I thought there might be a option for grep. Of course this awk work great except the print---I need the whole row of the first match, which can be fixed I think.
Thank you very much.
# 4  
Old 11-29-2011
In what way doesn't it work? 'print' ought to print the entire row.
# 5  
Old 11-29-2011
Try this...
Code:
while read line; do grep -m 1 $line table.tab; done < member.lists

btw, Corona's awk solution works for me...

--ahamed
# 6  
Old 11-29-2011
swapped the order of the two files!
Sorry!
After your last email I double checked the script and found out the mistake I made was the swapped order of the two files, which should be:
Code:
awk 'BEGIN { while(getline<"listfile") M[$1]=1 }; { if(M[$1]==1) { print; M[$1]=2 } }' datafile

Thanks a lot!
Yifang
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