02-15-2009
I modified this code, but i do not have access to the asorti function....what can I proceed would it work if I do not sort and just use 'count of' array for my loop control?
awk '{y[$2];i[$1];c[$1,$2]+=$2;d[$1,$2]++}END{ny=asorti(y);ni=asorti(i);printf "ID";for(j=1;j<=ny;j++); printf ";%s", y[j];print"";for(x=1;x<=ni;x++){
printf i[x];
for(j=1;j<=ny;j++);printf ";%s", c[i[x],y[j]]/d[i[x],y[j]] ""
print""
}}' file
Thanks.
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JOIN(1) General Commands Manual JOIN(1)
NAME
join - relational database operator
SYNOPSIS
join [-an] [-e s] [-o list] [-tc] file1 file2
DESCRIPTION
Join forms, on the standard output, a join of the two relations specified by the lines of file1 and file2. If file1 is `-', the standard
input is used.
File1 and file2 must be sorted in increasing ASCII collating sequence on the fields on which they are to be joined, normally the first in
each line.
There is one line in the output for each pair of lines in file1 and file2 that have identical join fields. The output line normally con-
sists of the common field, then the rest of the line from file1, then the rest of the line from file2.
Fields are normally separated by blank, tab or newline. In this case, multiple separators count as one, and leading separators are dis-
carded.
These options are recognized:
-an In addition to the normal output, produce a line for each unpairable line in file n, where n is 1 or 2.
-e s Replace empty output fields by string s.
-o list
Each output line comprises the fields specified in list, each element of which has the form n.m, where n is a file number and m is a
field number.
-tc Use character c as a separator (tab character). Every appearance of c in a line is significant.
SEE ALSO
sort(1), comm(1), awk(1).
BUGS
With default field separation, the collating sequence is that of sort -b; with -t, the sequence is that of a plain sort.
The conventions of join, sort, comm, uniq, look and awk(1) are wildly incongruous.
7th Edition April 29, 1985 JOIN(1)