Python 中的字典分组和聚合列表

Question

我有一个需要在 Python 中聚合的字典列表：

data = [{"startDate": 123, "endDate": 456, "campaignName": "abc", "campaignCfid": 789, "budgetImpressions": 10}, 
{"startDate": 123, "endDate": 456, "campaignName": "abc", "campaignCfid": 789, "budgetImpressions": 50}, 
{"startDate": 456, "endDate": 789, "campaignName": "def", "campaignCfid": 123, "budgetImpressions": 80}]

我希望根据budgetImpressions 进行汇总。

所以最终的结果应该是：

data = [{"startDate": 123, "endDate": 456, "campaignName": "abc", "campaignCfid": 789, "budgetImpressions": 60}, 
{"startDate": 456, "endDate": 789, "campaignName": "def", "campaignCfid": 123, "budgetImpressions": 80}]

请注意，具有特定活动名称的每个条目将始终具有相同的对应活动 Cfid、开始日期和结束日期。

这可以在 Python 中完成吗？我试过使用 itertools 没有太大成功。使用 Pandas 会是更好的方法吗？

Answer 1

只是为了证明有时python非常适合在以下方面做这种事情：

In [11]: from collections import Counter
         from itertools import groupby

In [12]: data = [{"startDate": 123, "endDate": 456, "campaignName": "abc", "campaignCfid": 789, "budgetImpressions": 10}, {"startDate": 123, "endDate": 456, "campaignName": "abc", "campaignCfid": 789, "budgetImpressions": 50}, {"startDate": 456, "endDate": 789, "campaignName": "def", "campaignCfid": 123, "budgetImpressions": 80}]

In [13]: g = groupby(data, lambda x: x.pop('campaignName'))

In [14]: d = {}
         for campaign, campaign_data in g:
             c = Counter()
             for row in campaign_data: c.update(row)
             d[campaign] = c  # if you want a dict rather than Counter, return dict(c) here

In [15]: d
Out[15]:
{'abc': Counter({'campaignCfid': 1578, 'endDate': 912, 'startDate': 246, 'budgetImpressions': 60}),
 'def': Counter({'endDate': 789, 'startDate': 456, 'campaignCfid': 123, 'budgetImpressions': 80})}

如果您已经拥有此列表/字典集合，那么将其提升到 DataFrame 没有任何意义，留在纯 python 中通常更便宜。

Answer 2

是的，使用熊猫。这很棒。您可以使用该groupby功能并按总和进行聚合，然后将输出转换为字典列表（如果这正是您想要的）。

import pandas as pd

data = [{"startDate": 123, "endDate": 456, "campaignName": 'abc',
         "campaignCfid": 789, "budgetImpressions": 10},
        {"startDate": 123, "endDate": 456, "campaignName": 'abc',
         "campaignCfid": 789, "budgetImpressions": 50},
        {"startDate": 456, "endDate": 789, "campaignName": 'def',
         "campaignCfid": 123, "budgetImpressions": 80}]

df = pd.DataFrame(data)

grouped = df.groupby(['startDate', 'endDate', 'campaignCfid',
                      'campaignName']).agg(sum)

print grouped.reset_index().to_dict('records')

这打印：

[{'startDate': 123L, 'campaignCfid': 789L, 'endDate': 456L, 'budgetImpressions': 60L, 'campaignName': 'abc'}, {'startDate': 456L, 'campaignCfid': 123L, 'endDate': 789L, 'budgetImpressions': 80L, 'campaignName': 'def'}]