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Sum Column Value

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# 8  
Old 07-18-2014
Nice solution Don, and It only took around 70 seconds to sum a 15 million record test file I built here.

Funny thing is my datafile didn't have any value with more that 2 digits of precision yet the awk.out still disagreed:

Code:
$ time ./don
370831916729.73
370831916729.73

real    0m46.250s
user    1m11.509s
sys     0m0.248s
$ cat awk.out 
3.70832e+11
370831916730.30

The data.txt wasn't anything special I made it like this:

Code:
for((i=0;i<15000000;i++)) {
    printf "1 2 3 %d%04d.%02d\n" $((RANDOM%5)) $((RANDOM%9999)) $((RANDOM % 99))
    [ $((i%1000)) -eq 0 ] && printf "\r%'d" $i >&2
} > data.txt

Last edited by Chubler_XL; 07-18-2014 at 01:09 AM..
# 9  
Old 07-18-2014
Quote:
Originally Posted by Chubler_XL
Nice solution Don, and It only took around 70 seconds to sum a 15 million record test file I built here.

Funny thing is my datafile didn't have any value with more that 2 digits of precision yet the awk.out still disagreed:

Code:
$ time ./don
370831916729.73
370831916729.73

real    0m46.250s
user    1m11.509s
sys     0m0.248s
$ cat awk.out 
3.70832e+11
370831916730.30

The data.txt wasn't anything special I made it like this:

Code:
for((i=0;i<15000000;i++)) {
    printf "1 2 3 %d%04d.%02d\n" $((RANDOM%5)) $((RANDOM%9999)) $((RANDOM % 99))
    [ $((i%1000)) -eq 0 ] && printf "\r%'d" $i >&2
} > data.txt

There is a big difference between precision and significant digits. There are 14 significant digits in 370,831,916,729.73. And, as I said, many decimal values do not have exact representations in IEEE 754 double precision floating point format. For example, the script:
Code:
printf ".1\n.2\n.33\n.5\n"|awk '{printf("%.20f %.20f %.20f\n", $1, 2*$1, $1 +2*$1)}'

produces the output:
Code:
0.10000000000000000555 0.20000000000000001110 0.30000000000000004441
0.20000000000000001110 0.40000000000000002220 0.60000000000000008882
0.33000000000000001554 0.66000000000000003109 0.98999999999999999112
0.50000000000000000000 1.00000000000000000000 1.50000000000000000000

which shows that .1, .2, and .33 do not have exact representations as floating point numbers while .5 is exactly representable. Adding up 15 million values when many of those values have small errors may produce a sum with a significant difference from adding up the exact decimal numbers.
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