Question

consider the following list:

alist = [[1, 'AAA0'], [2, 'AAAA1'], [3, 'BB2BB2'], [4, 'A3A3'], [5, 'A3A3']]

then if I want to sort the list according to the 2nd slot but at the same time get only the items that have ‘A’ inside I use:

 from operator import itemgetter
 print str([pt for pt in sorted(alist, key=itemgetter(1)) if 'A' in pt[1]])

my question is how after sorting them alphabetically, how do I group them by string length first and then group the ones that are the same. Like being able to retrieve the 2 ‘A3A3’ only.

the

if 'A' in pt[1] 

is returning all items that contain ‘A’ not the ones with 1 ‘A’.

so how do I group those with 1 ‘A’, those with 2 ‘A’ etc?

Answer 1

You can group elements of an iterable by custom key function with itertools.groupby:

In [1]: from itertools import groupby
In [2]: alist = [[1, 'AAA0'], [2, 'AAAA1'], [3, 'BB2BB2'],
   ...: [4, 'A3A3'], [5, 'A3A3']]
In [3]: key = lambda x: x[1].count('A')
In [4]: for k, g in groupby(sorted(alist, key=key), key=key):
    print k, tuple(g)
   ...:     
0 ([3, 'BB2BB2'],)
2 ([4, 'A3A3'], [5, 'A3A3'])
3 ([1, 'AAA0'],)
4 ([2, 'AAAA1'],)