Python分组依据

分两步完成。首先，创建字典。

>>> 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)...

然后，将该字典转换为预期的格式。

>>> [{'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'}]

使用itertools.groupby也可以，但是它要求输入首先被排序。

>>> 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'}]

请注意，这两个都不遵守键的原始顺序。如果需要保留订单，则需要一个OrderedDict。

>>> 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'], 'type': 'NOT'}, {'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}]