Pandas - 将累积值转换为实际值

Question

假设我的数据框看起来像这样：

date,site,country_code,kind,ID,rank,votes,sessions,avg_score,count
2017-03-20,website1,US,0,84,226,0.0,15.0,3.370812,53.0
2017-03-21,website1,US,0,84,214,0.0,15.0,3.370812,53.0
2017-03-22,website1,US,0,84,226,0.0,16.0,3.370812,53.0
2017-03-23,website1,US,0,84,234,0.0,16.0,3.369048,54.0
2017-03-24,website1,US,0,84,226,0.0,16.0,3.369048,54.0
2017-03-25,website1,US,0,84,212,0.0,16.0,3.369048,54.0
2017-03-26,website1,US,0,84,228,0.0,16.0,3.369048,54.0
2017-02-15,website2,AU,1,91,144,4.0,148.0,4.727272,521.0
2017-02-16,website2,AU,1,91,144,3.0,147.0,4.727272,524.0
2017-02-17,website2,AU,1,91,100,4.0,148.0,4.727272,524.0
2017-02-18,website2,AU,1,91,118,6.0,149.0,4.727272,527.0
2017-02-19,website2,AU,1,91,114,4.0,151.0,4.727272,529.0

最后的count 列是累积计数。 我需要做的是找到特定的实际计数 date+site+country+kind+ID 元组，这将导致：

date,site,country_code,kind,ID,rank,votes,sessions,avg_score,count
2017-03-20,website1,US,0,84,226,0.0,15.0,3.370812,0.0
2017-03-21,website1,US,0,84,214,0.0,15.0,3.370812,0.0
2017-03-22,website1,US,0,84,226,0.0,16.0,3.370812,0.0
2017-03-23,website1,US,0,84,234,0.0,16.0,3.369048,1.0
2017-03-24,website1,US,0,84,226,0.0,16.0,3.369048,0.0
2017-03-25,website1,US,0,84,212,0.0,16.0,3.369048,0.0
2017-03-26,website1,US,0,84,228,0.0,16.0,3.369048,0.0
2017-02-15,website2,AU,1,91,144,4.0,148.0,4.727272,0.0
2017-02-16,website2,AU,1,91,144,3.0,147.0,4.727272,3.0
2017-02-17,website2,AU,1,91,100,4.0,148.0,4.727272,0.0
2017-02-18,website2,AU,1,91,118,6.0,149.0,4.727272,3.0
2017-02-19,website2,AU,1,91,114,4.0,151.0,4.727272,2.0

我知道这将涉及到groupby 电话，但我不知道除此之外还能做什么。假设元组的第一个实例的计数为0。 任何帮助都会很棒。谢谢

Answer 1

使用groupby + diff，cumsum 的倒数。

cols = ['site', 'country_code', 'kind', 'ID']
df['count'] = df.groupby(cols)['count'].diff().fillna(0)

print(df['count'])
0     0.0
1     0.0
2     0.0
3     1.0
4     0.0
5     0.0
6     0.0
7     0.0
8     3.0
9     0.0
10    3.0
11    2.0
Name: count, dtype: float64

感谢 MaxU 指出错误！