pandas 总计数不同 [英] Pandas aggregate count distinct

问题描述

假设我有一个用户活动日志，我想生成一个总持续时间和每天唯一身份用户数量的报告.

Let's say I have a log of user activity and I want to generate a report of total duration and the number of unique users per day.

import numpy as np
import pandas as pd
df = pd.DataFrame({'date': ['2013-04-01','2013-04-01','2013-04-01','2013-04-02', '2013-04-02'],
    'user_id': ['0001', '0001', '0002', '0002', '0002'],
    'duration': [30, 15, 20, 15, 30]})

汇总持续时间非常简单:

Aggregating duration is pretty straightforward:

group = df.groupby('date')
agg = group.aggregate({'duration': np.sum})
agg
            duration
date
2013-04-01        65
2013-04-02        45

我想做的是对持续时间求和并同时计算不重复次数，但我似乎找不到count_distinct的等效项:

What I'd like to do is sum the duration and count distincts at the same time, but I can't seem to find an equivalent for count_distinct:

agg = group.aggregate({ 'duration': np.sum, 'user_id': count_distinct})

这有效，但是肯定有更好的方法，不是吗?

This works, but surely there's a better way, no?

group = df.groupby('date')
agg = group.aggregate({'duration': np.sum})
agg['uv'] = df.groupby('date').user_id.nunique()
agg
            duration  uv
date
2013-04-01        65   2
2013-04-02        45   1

我在想，我只需要提供一个将Series对象的不同项目的计数返回到聚合函数的函数，但是我对各种库的了解并不多.另外，似乎groupby对象已经知道了这些信息，所以我不是要重复努力吗?

I'm thinking I just need to provide a function that returns the count of distinct items of a Series object to the aggregate function, but I don't have a lot of exposure to the various libraries at my disposal. Also, it seems that the groupby object already knows this information, so wouldn't I just be duplicating effort?

推荐答案

其中一个怎么样?

>>> df
         date  duration user_id
0  2013-04-01        30    0001
1  2013-04-01        15    0001
2  2013-04-01        20    0002
3  2013-04-02        15    0002
4  2013-04-02        30    0002
>>> df.groupby("date").agg({"duration": np.sum, "user_id": pd.Series.nunique})
            duration  user_id
date                         
2013-04-01        65        2
2013-04-02        45        1
>>> df.groupby("date").agg({"duration": np.sum, "user_id": lambda x: x.nunique()})
            duration  user_id
date                         
2013-04-01        65        2
2013-04-02        45        1

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