You could try something like the following:

Code:

awk '
BEGIN { FS = "[ /\t]+"
        OFS = "    "
        s = " "
}
{       v[$1 s $2 s $3 OFS $8] += $5 }
END {   for(i in v)
                printf("%s%s%.*f\n",
                        i, OFS, 9 - int(log(v[i]) / log(10)), v[i])
}' file1 | sort -k3n,3 -k1n,1 -k2n,2 -k4,4 | sed 's# #/#;s# #/#'

As always, if you are going to run this on a Solaris/SunOS system, use /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk instead of /usr/bin/awk or /bin/awk.

By having awk use sequences of spaces, slashes, and tabs as field separators, the date field is split into month, day, and year fields as input lines are read. The subscript used for the v[] array (which contains the sum of the values in column 3 [field 5 after splitting the date field]) is the month followed by a space followed by the day of the month followed by a space followed by the year followed by four spaces followed by the contents of the 6th column (8th field after splitting the date). The END clause prints the subscript for each value found along with the sum of the values accumulated for each subscript.

Translating the slashes in the date field to spaces allows the sort command to sort the output produced by awk on the various numeric components of the date and the original contents of the alphanumeric input file's 6th column. After sorting the output, the sed command converts the 1st two spaces on the output line back to slashes thereby restoring the date field to its original format.

The above script produces the output you said you wanted in the 1st message in this thread except that the output shown in red below was rounded differently than in your example:

Code:

1/1/2013    X1    1012.909698
1/1/2013    X2    600.8333588
1/2/2013    X1    844.2973022
1/2/2013    X2    833.9300537
1/3/2013    X1    563.6917419
1/3/2013    X2    632.0749969
1/4/2013    X1    48.33055687

Note that the log() calculations in the awk printf statement are there to calculate the varying number of decimal places you showed in your desired output. That printf statement could be simplified if you were willing to accept a constant number of digits after the decimal point in the printed sums.

Alternatively, you could split the date field, sort the input into the desired output order, reform the date field in the sorted input and use the procedures outlined in the thread bakunin referenced. I haven't made any attempt to compare the efficiency of these alternative approaches.

Hope this helps,
Don