Fastest way to count three items referenced by one of them (using python)?

Question

I have a multi column (13 columns) space separated file (some 5 million+ lines), going like this:

 1. W5 403  407 P Y 2 2 PR 22  PNIYR 22222 12.753 13.247
 2. W5 404  408 N V 2 2 PR 22  PIYYR 22222 13.216 13.247
 3. W3 274  276 E G 1 1 EG 11  EPG 121 6.492 6.492
 4. W3 275  277 P R 2 1 PR 21  PGR 211 6.365 7.503
 5. W3 276  278 G Y 1 1 GY 11  GRY 111 5.479 5.479
 6. W3 46  49 G L 1 1 GY 11  GRY 111 5.176 5.176
 7. W4 47  50 D K 1 1 DK 11  DILK 1111 4.893 5.278
 8. W4 48  51 I K 1 1 IK 11  ILKK 1111 4.985 5.552

etc., etc.,

I'm interested in 2 of these columns (col 8 & 11) and want to count the number of occurrences of particular pairs (col 8) with the strings that follow (in col 11).

Ex., reference key 'GY' : # of occurrences of '111' : 2 key 'PR' : # of occurrences of '22222': 2 key 'DK' : # of occurrences of '1111' :1 key 'EG' : # of occurrences of '121': 1

I have a dict based basic implementation of it.

countshash={}
for l in bigtable:
          cont = l.split()
          if cont[7] not in countshash: countshash[cont[7]] = {}
          if cont[11] not in countshash[cont[7]]: countshash[cont[7]][cont[10]] = 0
          countshash[cont[7]][cont[10]]+= 1;

I also have a simple awk based counting (which is super-fast) but was wondering about an efficient & faster way to do this in python. Thanks for your inputs.

Answer 1

I'm not sure if this will help with speed, but you are creating a ton of defaultdict-like objects, which I think you can make a bit more readable:

from collections import defaultdict

countshash = defaultdict(lambda: defaultdict(int))

for l in bigtable:
    cont = l.split()
    countshash[cont[7]][cont[10]] += 1

Answer 2

from collections import Counter
Counter(tuple(row.split()[8:12:3]) for row in bigtable)

using itemgetter is more flexible and may be more efficient than slicing

from operator import itemgetter
ig = itemgetter(8, 11)
Counter(ig(row.split()) for row in bigtable)

using imap can make things a tiny big faster too

from itertools import imap
Counter(imap(ig, imap(str.split, bigtable)))