Finding the mean values of a field using awk? - awk

Question

This is what I am trying to do:
find the mean values for x,y,z for the HETATM records , the x value are the 7 field, the y values are the 8 field, and z values are the 9 field.
I am trying to do this using this file http://pastebin.com/EqA2SUMy
Here is the sample
HETATM 1756 O HOH A 501 -0.923 10.560 127.393 1.00 16.58 O
HETATM 1757 O HOH A 502 9.272 22.148 134.167 1.00 15.08 O
HETATM 1758 O HOH A 503 0.109 20.243 112.094 1.00 20.74 O
HETATM 1759 O HOH A 504 -3.930 10.522 125.779 1.00 20.79 O
HETATM 1760 O HOH A 505 -0.759 36.323 88.018 1.00 17.42 O
HETATM 1761 O HOH A 506 -4.645 51.936 81.852 1.00 21.43 O
HETATM 1762 O HOH A 507 -3.900 17.103 128.596 1.00 14.08 O
HETATM 1763 O HOH A 508 6.834 21.053 135.062 1.00 16.98 O
Can anyone show me how to do a script for this.
(this part is related to a comment viewers can ignore)
ATOM 214 OE2 GLU A 460 -2.959 24.000 103.360 1.00 32.19 O
ATOM 215 N ARG A 461 -5.878 28.748 106.473 1.00 22.68 N
ATOM 216 CA ARG A 461 -6.553 30.043 106.524 1.00 24.34 C
ATOM 217 C ARG A 461 -5.583 31.176 106.219 1.00 22.42 C
ATOM 218 O ARG A 461 -5.918 32.121 105.497 1.00 25.07 O
ATOM 219 CB ARG A 461 -7.222 30.272 107.887 1.00 24.53 C
ATOM 220 CG ARG A 461 -8.425 29.394 108.150 1.00 26.38

Answer 1

$ awk '{for (i=1;i<=3;i++) sum[i]+=$(i+6)}
END{if (NR) for (i=1;i in sum;i++) print sum[i]/NR}' file
0.25725
23.736
116.62
The if (NR) is necessary to avoid a divide by zero error on an empty file.
If #jaypal is correct and you need to select just the input lines containing HETATM then change it to:
awk '/HETATM/{++nr; for (i=1;i<=3;i++) sum[i]+=$(i+6)}
END{if (nr) for (i=1;i in sum;i++) print sum[i]/nr}' file

Answer 2

It's not rocket science. (Updated to catch only HETATM records — a trivial change; you can use more exacting regexes if you need to. However, it is also necessary to count which records match and use that count, not NR, since you're ignoring many records, in general.)
awk '/HETATM/ { sum7 += $7; sum8 += $8; sum9 += $9; count++ }
END { if (count > 0)
printf("avg(x) = %f, avg(y) = %f, avg(z) = %f\n",
sum7/count, sum8/count, sum9/count)
}'
And yes, you could put it all on one line but it wouldn't be as readable.
I can't answer for why it produced zeros for you; when run on the data from the question, wobbly line starts and all, it produced the output:
avg(x) = 0.257250, avg(y) = 23.736000, avg(z) = 116.620125
If you think there is a possibility of empty input (or, at least, no HETATM records in the input) and an error message is not acceptable, then you can protect the printing action with if (count > 0) or equivalent (added to script). You can generate your own preferred output if ``count` is zero.