[python] Python group by

The following function will quickly (no sorting required) group tuples of any length by a key having any index:

# given a sequence of tuples like [(3,'c',6),(7,'a',2),(88,'c',4),(45,'a',0)],
# returns a dict grouping tuples by idx-th element - with idx=1 we have:
# if merge is True {'c':(3,6,88,4),     'a':(7,2,45,0)}
# if merge is False {'c':((3,6),(88,4)), 'a':((7,2),(45,0))}
def group_by(seqs,idx=0,merge=True):
    d = dict()
    for seq in seqs:
        k = seq[idx]
        v = d.get(k,tuple()) + (seq[:idx]+seq[idx+1:] if merge else (seq[:idx]+seq[idx+1:],))
        d.update({k:v})
    return d

In the case of your question, the index of key you want to group by is 1, therefore:

group_by(input,1)

gives

{'ETH': ('5238761','5349618','962142','7795297','7341464','5594916','1550003'),
 'KAT': ('11013331', '9843236'),
 'NOT': ('9085267', '11788544')}

which is not exactly the output you asked for, but might as well suit your needs.