我有以下Python 3.4脚本示例。它执行以下操作：
这是脚本：
import numpy as np
import pandas as pd
# Create dataframe consisting of id, date and two categories (gender and age)
tempDF = pd.DataFrame({ 'id': [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],
'date': ["02/04/2015 02:34","06/04/2015 12:34","09/04/2015 23:03","12/04/2015 01:00","15/04/2015 07:12","21/04/2015 12:59","29/04/2015 17:33","04/05/2015 10:44","06/05/2015 11:12","10/05/2015 08:52","12/05/2015 14:19","19/05/2015 19:22","27/05/2015 22:31","01/06/2015 11:09","04/06/2015 12:57","10/06/2015 04:00","15/06/2015 03:23","19/06/2015 05:37","23/06/2015 13:41","27/06/2015 15:43"],
'gender': ["male","female","female","male","male","female","female",np.nan,"male","male","female","male","female","female","male","female","male","female",np.nan,"male"],
'age': ["young","old","old","old","old","old",np.nan,"old","old","young","young","old","young","young","old",np.nan,"old","young",np.nan,np.nan]})
# Convert date to datetime
tempDF['date'] = pd.to_datetime(tempDF['date'])
# Create groupby object based on two categorical variables
tempGroupby = tempDF.sort_values(['gender','age','id']).groupby(['gender','age'])
# Count number in each group and merge with original dataframe to create 'count' column
tempCountsDF = tempGroupby['id'].count().reset_index(drop=False)
tempCountsDF = tempCountsDF.rename(columns={'id': 'count'})
tempDF = tempDF.merge(tempCountsDF, on=['gender','age'])
# Calculate difference between consecutive rows in each group. (First row in each
# group should have date difference = NaT)
tempGroupby = tempDF.sort_values(['gender','age','id']).groupby(['gender','age'])
tempDF['diff'] = tempGroupby['date'].diff()
print(tempDF)
此脚本生成以下输出：
age date gender id count diff
0 young 2015-02-04 02:34:00 male 1 2 NaT
1 young 2015-10-05 08:52:00 male 10 2 243 days 06:18:00
2 old 2015-06-04 12:34:00 female 2 3 NaT
3 old 2015-09-04 23:03:00 female 3 3 92 days 10:29:00
4 old 2015-04-21 12:59:00 female 6 3 -137 days +13:56:00
5 old 2015-12-04 01:00:00 male 4 6 NaT
6 old 2015-04-15 07:12:00 male 5 6 -233 days +06:12:00
7 old 2015-06-05 11:12:00 male 9 6 51 days 04:00:00
8 old 2015-05-19 19:22:00 male 12 6 -17 days +08:10:00
9 old 2015-04-06 12:57:00 male 15 6 -44 days +17:35:00
10 old 2015-06-15 03:23:00 male 17 6 69 days 14:26:00
11 young 2015-12-05 14:19:00 female 11 4 NaT
12 young 2015-05-27 22:31:00 female 13 4 -192 days +08:12:00
13 young 2015-01-06 11:09:00 female 14 4 -142 days +12:38:00
14 young 2015-06-19 05:37:00 female 18 4 163 days 18:28:00
这正是我所期待的。但是，它似乎依赖于两次创建groupby对象（以完全相同的方式）。如果第二个groupby定义被注释掉，它似乎导致diff列中的输出非常不同：
import numpy as np
import pandas as pd
# Create dataframe consisting of id, date and two categories (gender and age)
tempDF = pd.DataFrame({ 'id': [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],
'date': ["02/04/2015 02:34","06/04/2015 12:34","09/04/2015 23:03","12/04/2015 01:00","15/04/2015 07:12","21/04/2015 12:59","29/04/2015 17:33","04/05/2015 10:44","06/05/2015 11:12","10/05/2015 08:52","12/05/2015 14:19","19/05/2015 19:22","27/05/2015 22:31","01/06/2015 11:09","04/06/2015 12:57","10/06/2015 04:00","15/06/2015 03:23","19/06/2015 05:37","23/06/2015 13:41","27/06/2015 15:43"],
'gender': ["male","female","female","male","male","female","female",np.nan,"male","male","female","male","female","female","male","female","male","female",np.nan,"male"],
'age': ["young","old","old","old","old","old",np.nan,"old","old","young","young","old","young","young","old",np.nan,"old","young",np.nan,np.nan]})
# Convert date to datetime
tempDF['date'] = pd.to_datetime(tempDF['date'])
# Create groupby object based on two categorical variables
tempGroupby = tempDF.sort_values(['gender','age','id']).groupby(['gender','age'])
# Count number in each group and merge with original dataframe to create 'count' column
tempCountsDF = tempGroupby['id'].count().reset_index(drop=False)
tempCountsDF = tempCountsDF.rename(columns={'id': 'count'})
tempDF = tempDF.merge(tempCountsDF, on=['gender','age'])
# Calculate difference between consecutive rows in each group. (First row in each
# group should have date difference = NaT)
# ****** THIS TIME THE FOLLOWING GROUPBY DEFINITION IS COMMENTED OUT *****
# tempGroupby = tempDF.sort_values(['gender','age','id']).groupby(['gender','age'])
tempDF['diff'] = tempGroupby['date'].diff()
print(tempDF)
而且，这次输出非常不同（而不是我想要的）
age date gender id count diff
0 young 2015-02-04 02:34:00 male 1 2 NaT
1 young 2015-10-05 08:52:00 male 10 2 NaT
2 old 2015-06-04 12:34:00 female 2 3 92 days 10:29:00
3 old 2015-09-04 23:03:00 female 3 3 NaT
4 old 2015-04-21 12:59:00 female 6 3 -233 days +06:12:00
5 old 2015-12-04 01:00:00 male 4 6 -137 days +13:56:00
6 old 2015-04-15 07:12:00 male 5 6 NaT
7 old 2015-06-05 11:12:00 male 9 6 NaT
8 old 2015-05-19 19:22:00 male 12 6 51 days 04:00:00
9 old 2015-04-06 12:57:00 male 15 6 243 days 06:18:00
10 old 2015-06-15 03:23:00 male 17 6 NaT
11 young 2015-12-05 14:19:00 female 11 4 -17 days +08:10:00
12 young 2015-05-27 22:31:00 female 13 4 -192 days +08:12:00
13 young 2015-01-06 11:09:00 female 14 4 -142 days +12:38:00
14 young 2015-06-19 05:37:00 female 18 4 -44 days +17:35:00
（在我的现实剧本中，结果似乎有些不稳定，有时候它会起作用，有时则不然。但在上面的剧本中，不同的输出似乎始终如一。）
为什么有必要在使用.diff（）函数之前立即重新创建基于相同数据帧的groupby对象（尽管添加了一个额外的列）？这对我来说似乎非常危险。
答案 0 :(得分：2)
不一样，索引已经改变了。例如：
tempDF.loc[1].id # before
10
tempDF.loc[1].id # after
2
因此，如果您使用旧tempGroupby计算tempDF，然后在执行此操作时更改tempDF中的索引：
tempDF['diff'] = tempGroupby['date'].diff()
索引与您期望的不匹配。您要为每一行分配与旧tempDF中具有该索引的行相对应的差异。