Python 熊猫聚合计数不同
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Pandas aggregate count distinct
提问by dave
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.
假设我有一个用户活动日志，我想生成一份总持续时间和每天唯一用户数的报告。
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]})
Aggregating duration is pretty straightforward:
聚合持续时间非常简单：
group = df.groupby('date')
agg = group.aggregate({'duration': np.sum})
agg
duration
date
2013-04-01 65
2013-04-02 45
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:
我想做的是同时对持续时间和不同的计数求和，但我似乎找不到 count_distinct 的等效项：
agg = group.aggregate({ 'duration': np.sum, 'user_id': count_distinct})
This works, but surely there's a better way, no?
这有效，但肯定有更好的方法，不是吗？
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
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?
我想我只需要提供一个函数，该函数将 Series 对象的不同项的计数返回给聚合函数，但我没有太多接触可供我使用的各种库。此外，似乎 groupby 对象已经知道这些信息，所以我不会只是重复工作吗？
采纳答案by DSM
How about either of:
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
回答by Ricky McMaster
'nunique' is an option for .agg() since pandas 0.20.0, so:
'nunique' 是 .agg() 的一个选项，因为 pandas 0.20.0，所以：
df.groupby('date').agg({'duration': 'sum', 'user_id': 'nunique'})
回答by user6903745
Just adding to the answers already given, the solution using the string "nunique"seems much faster, tested here on ~21M rows dataframe, then grouped to ~2M
只是添加到已经给出的答案，使用字符串的解决方案"nunique"似乎要快得多，在 ~21M 行数据帧上测试，然后分组到 ~2M
%time _=g.agg({"id": lambda x: x.nunique()})
CPU times: user 3min 3s, sys: 2.94 s, total: 3min 6s
Wall time: 3min 20s
%time _=g.agg({"id": pd.Series.nunique})
CPU times: user 3min 2s, sys: 2.44 s, total: 3min 4s
Wall time: 3min 18s
%time _=g.agg({"id": "nunique"})
CPU times: user 14 s, sys: 4.76 s, total: 18.8 s
Wall time: 24.4 s
