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Count occurrences in first column

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# 1  
Old 05-19-2015
Count occurrences in first column
input

Code:
amex-11	10	abc
amex-11	20	bcn
amed-12	1	abc

I tried something like this.

Code:
awk '{h[$1]++}; END { for(k in h) print k, h[k] }' rm1

output
Code:
amex-11 1 10 abc
amex-11 1 20 bcn
amed-12 2 1 abc

Note: The second column represents the occurrences. amex-11 is first one and amed-12 is the second one. ....
# 2  
Old 05-19-2015
Not quite sure what you mean by "The second column represents the occurrences"

Perhaps this:

Code:
awk '{print $1,++c[$3],$2,$3}' infile

Last edited by Chubler_XL; 05-19-2015 at 06:25 PM..
# 3  
Old 05-19-2015
yes almost. more detailed example for clarification.

input
Code:
amex-11	10	abc
amex-11	20	bcn
amed-12	1	abc
amex-12	10	abc
amex-12	20	bcn
amed-13	1	abc

ouput shouldbe

Code:
amex-11	1 10	abc
amex-11	1 20	bcn
amed-12	2 1	abc
amex-12	3 10	abc
amex-12	3 20	bcn
amed-13	4 1	abc

# 4  
Old 05-19-2015
How about one of these two:

Code:
awk '{c[$1];$1=$1 OFS length(c)}1' infile


Code:
awk '
!($1 in c){c[$1]=++U}
$1=$1 OFS c[$1]' infile

This User Gave Thanks to Chubler_XL For This Post:
quincyjones
# 5  
Old 05-20-2015
Note: using length() to determine the number of elements in an array is a non-standard extension that only works in GNU awk ( and may work in BSD awk as an undocumented feature)...
# 6  
Old 05-20-2015
This one should work.

Code:
awk 'BEGIN{h[" "]=0;max=0}{ind=-1;for(k in h)if(k==$1){ind=h[k];break}if(ind==-1){ind=++max;h[$1]=ind}print $1,ind,$2,$3}' infile

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