有没有办法根据字典键 map 字典列表？

Question

我有一个看起来像的字典列表

[{'n': 3, 'topic': 4, 'shared_via_twitter': 'N'}, {'n': 72, 'topic': 1, 'shared_via_twitter': 'Y'}, {'n': 46, 'topic': 2, 'shared_via_twitter': 'N'}, {'n': 36, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 5, 'topic': 4, 'shared_via_twitter': 'N'}, {'n': 29, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 102, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 13, 'topic': 8, 'shared_via_twitter': 'Y'}, {'n': 52, 'topic': 2, 'shared_via_twitter': 'N'}, {'n': 24, 'topic': 5, 'shared_via_twitter': 'N'}]

我希望能够将具有相同topic和shared_via_twitter值的所有字典合并为一个，并更新新字典的n值以反映总和。

例如，如果我有

[{'n': 3, 'topic': 1, 'shared_via_twitter': 'N'}, {'n': 7, 'topic': 1, 'shared_via_twitter': 'N'}]

然后我想要{'n': 10, 'topic': 1, 'shared_via_twitter': 'N'}作为结果。

我正在考虑使用 map 之类的东西

def xs(x):
    # {'n': 3, 'topic': 4, 'shared_via_twitter': 'N'}
    # {'n': 7, 'topic': 4, 'shared_via_twitter': 'N'}
    if x['topic'] == v['topic'] and x['shared_via_twitter'] == v['shared_via_twitter']:
        x['n']+=v['n']
        v = dict(x)
        return x

g = map(xs, rows)

但这显然看起来不干净/工作。 任何想法将不胜感激。

Answer 1

这不是映射操作。 目前尚不清楚您尝试在做什么，确切地说，因为v没有定义。 但基本上， map不是最好的工具，因为它将 function 应用于每个元素，您希望按您的键分组并将"n"的值与总和聚合。 使用字典分组习语：

>>> data = [{'n': 3, 'topic': 4, 'shared_via_twitter': 'N'}, {'n': 72, 'topic': 1, 'shared_via_twitter': 'Y'}, {'n': 46, 'topic': 2, 'shared_via_twitter': 'N'}, {'n': 36, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 5, 'topic': 4, 'shared_via_twitter': 'N'}, {'n': 29, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 102, 'topic': 2, 'shared_via_twitter': 'Y'}, {'n': 13, 'topic': 8, 'shared_via_twitter': 'Y'}, {'n': 52, 'topic': 2, 'shared_via_twitter': 'N'}, {'n': 24, 'topic': 5, 'shared_via_twitter': 'N'}]
>>> grouper = {}
>>> for d in data:
...     key = d['topic'], d['shared_via_twitter']
...     grouper[key] = grouper.get(key, 0) + d['n']
...
>>> grouper
{(4, 'N'): 8, (1, 'Y'): 72, (2, 'N'): 98, (2, 'Y'): 167, (8, 'Y'): 13, (5, 'N'): 24}

您可以使用以下方法将其转换为最终形式：

>>> [dict(topic=t, shared_via_twitter=s, n=n) for (t, s), n in grouper.items()]
[{'topic': 4, 'shared_via_twitter': 'N', 'n': 8}, {'topic': 1, 'shared_via_twitter': 'Y', 'n': 72}, {'topic': 2, 'shared_via_twitter': 'N', 'n': 98}, {'topic': 2, 'shared_via_twitter': 'Y', 'n': 167}, {'topic': 8, 'shared_via_twitter': 'Y', 'n': 13}, {'topic': 5, 'shared_via_twitter': 'N', 'n': 24}]

Answer 2

如果我正确理解您想要做什么，请尝试collections.defaultdict或collections.Counter ：

import collections

totals = collections.Counter()
for d in data:
   totals[d['topic'], d['shared_via_twitter']] += d['n']

这会将其保留在{(1, 'N'): 10}类的结构中，您可以按原样使用它，也可以将其转换为 list-of-dicts 形式：

converted = [
    {'n': n, 'topic': topic, 'shared_via_twitter': shared_via_twitter}
    for (topic, shared_via_twitter), n in totals.items()
]