Pandas - 多列分组移动平均

Question

我是新手，所以请耐心等待。

我的数据框格式如下：

date,name,country,tag,cat,score
2017-05-21,X,US,free,4,0.0573
2017-05-22,X,US,free,4,0.0626
2017-05-23,X,US,free,4,0.0584
2017-05-24,X,US,free,4,0.0563
2017-05-21,X,MX,free,4,0.0537
2017-05-22,X,MX,free,4,0.0640
2017-05-23,X,MX,free,4,0.0648
2017-05-24,X,MX,free,4,0.0668

我正在尝试想出一种方法，在国家/标签/类别组内找到X天移动平均值，所以我需要：

date,name,country,tag,cat,score,moving_average
2017-05-21,X,US,free,4,0.0573,0
2017-05-22,X,US,free,4,0.0626,0.0605
2017-05-23,X,US,free,4,0.0584,0.0594
2017-05-24,X,US,free,4,0.0563,and so on
...
2017-05-21,X,MX,free,4,0.0537,and so on
2017-05-22,X,MX,free,4,0.0640,and so on
2017-05-23,X,MX,free,4,0.0648,and so on
2017-05-24,X,MX,free,4,0.0668,and so on

我尝试按照需要的列进行分组，然后使用pd.rolling_mean函数，但最终得到了一堆NaN值。

df.groupby(['date', 'name', 'country', 'tag'])['score'].apply(pd.rolling_mean, 2, min_periods=2)  # window size 2

我该如何正确地执行这个操作？

- Craig

---

你不应该按date分组：df.groupby(['name', 'country', 'tag'])['score'].apply(pd.rolling_mean, 2, min_periods=2)。 - BENY

我没有重新创建你的数据，但是在我打开的一个df上，我做了这个操作：df.groupby(['col1', 'col2'])['col3'].apply(lambda x: x.rolling(window=3, center=True).mean())，它可以正常工作。rolling_mean已经被弃用，你应该使用rolling代替。 - Woody Pride

@Wen - 是的 - 谢谢（为什么这不是一个答案？） :) - Craig

@WoodyPride - 谢谢，你提到的弃用问题很有道理。 - Craig

@Craig 因为我懒得修改它成为一个答案.. :) - BENY

Answer 1

我理解为：

(df.assign(moving_score=df.groupby(['name','country','tag'], as_index=False)[['score']]
                           .rolling(2, min_periods=2).mean().fillna(0)
                           .reset_index(0, drop=True)))

输出：

         date name country   tag  cat   score  moving_score
0  2017-05-21    X      US  free    4  0.0573       0.00000
1  2017-05-22    X      US  free    4  0.0626       0.05995
2  2017-05-23    X      US  free    4  0.0584       0.06050
3  2017-05-24    X      US  free    4  0.0563       0.05735
4  2017-05-21    X      MX  free    4  0.0537       0.00000
5  2017-05-22    X      MX  free    4  0.0640       0.05885
6  2017-05-23    X      MX  free    4  0.0648       0.06440
7  2017-05-24    X      MX  free    4  0.0668       0.06580

- Scott Boston