Python group by

Do it in 2 steps. First, create a dictionary.

>>> input = [('11013331', 'KAT'), ('9085267', 'NOT'), ('5238761', 'ETH'), ('5349618', 'ETH'), ('11788544', 'NOT'), ('962142', 'ETH'), ('7795297', 'ETH'), ('7341464', 'ETH'), ('9843236', 'KAT'), ('5594916', 'ETH'), ('1550003', 'ETH')]
>>> from collections import defaultdict
>>> res = defaultdict(list)
>>> for v, k in input: res[k].append(v)
...

Then, convert that dictionary into the expected format.

>>> [{'type':k, 'items':v} for k,v in res.items()]
[{'items': ['9085267', '11788544'], 'type': 'NOT'}, {'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}, {'items': ['11013331', '9843236'], 'type': 'KAT'}]

It is also possible with itertools.groupby but it requires the input to be sorted first.

>>> sorted_input = sorted(input, key=itemgetter(1))
>>> groups = groupby(sorted_input, key=itemgetter(1))
>>> [{'type':k, 'items':[x[0] for x in v]} for k, v in groups]
[{'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}, {'items': ['11013331', '9843236'], 'type': 'KAT'}, {'items': ['9085267', '11788544'], 'type': 'NOT'}]

Note both of these do not respect the original order of the keys. You need an OrderedDict if you need to keep the order.

>>> from collections import OrderedDict
>>> res = OrderedDict()
>>> for v, k in input:
...   if k in res: res[k].append(v)
...   else: res[k] = [v]
... 
>>> [{'type':k, 'items':v} for k,v in res.items()]
[{'items': ['11013331', '9843236'], 'type': 'KAT'}, {'items': ['9085267', '11788544'],

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