Question

我有一个包含以下列的数据框：user_id，product_id，created_at和remove_at。我想添加一个布尔列“ is_switch”，如果对于给定的用户，created_at的时间戳是在timedelta内（比如说1秒），而该用户组中任何其他行的remove_at则为true。如何在不遍历每一行的情况下执行此操作，或者这是执行该操作的适当方法？

我试图编写一个将与.apply结合使用的自定义函数，该函数将在每个用户组上运行，但是我不确定如何一次比较行与所有其他行。

# Code to create sample data frame. 
# the below are just timestamps that are within a second of each other.

import datetime

a = datetime.datetime.now()
a2 = a-datetime.timedelta(seconds=1)
b = datetime.datetime.now()-datetime.timedelta(days=4)
b2 = b-datetime.timedelta(seconds=1)
c = datetime.datetime.now()-datetime.timedelta(days=40)
c2 = c - datetime.timedelta(seconds=1)
d = datetime.datetime.now()-datetime.timedelta(days=30)
d2 = d - datetime.timedelta(seconds=1)
e = datetime.datetime.now()-datetime.timedelta(days=60)
e2 = e - datetime.timedelta(seconds=1)
f = datetime.datetime.now()-datetime.timedelta(days=100)
g = datetime.datetime.now()-datetime.timedelta(days=99)

df = pd.DataFrame(
{"user_id" : [0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4],
"product_id" : [100, 101, 102, 101, 102, 104, 105, 106, 107, 105, 106, 107],
"created_at" : [a, a, b, c, d, c, f, f, e2, f, f, d],
"removed_at" : ['NaT', b2, 'NaT', d2, 'NaT', 'NaT', e, g, 'NaT', e2, g, b]},
index = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
df

print(df)

产生此：


        user_id  product_id                 created_at                 removed_at
0         0         100 2019-08-04 09:15:05.200981                        NaT
1         1         101 2019-08-04 09:15:05.200981 2019-07-31 09:15:04.201063
2         1         102 2019-07-31 09:15:05.201063                        NaT
3         2         101 2019-06-25 09:15:05.201121 2019-07-05 09:15:04.201179
4         2         102 2019-07-05 09:15:05.201179                        NaT
5         2         104 2019-06-25 09:15:05.201121                        NaT
6         3         105 2019-04-26 09:15:05.201290 2019-06-05 09:15:05.201235
7         3         106 2019-04-26 09:15:05.201290 2019-04-27 09:15:05.201324
8         3         107 2019-06-05 09:15:04.201235                        NaT
9         4         105 2019-04-26 09:15:05.201290 2019-06-05 09:15:04.201235
10        4         106 2019-04-26 09:15:05.201290 2019-04-27 09:15:05.201324
11        4         107 2019-07-05 09:15:05.201179 2019-07-31 09:15:05.201063

所以我目前有这样的东西：

group_by_user = df.groupby('user_id')

def calculate_is_switch(grp):
    # What goes here? how can i do it without iterating over each row?

# group_by_user.apply(calculate_is_switch)

我想添加“ is_switch”列，因此输出为：

    user_id  product_id                 created_at                 removed_at  \
0         0         100 2019-08-04 09:15:05.200981                        NaT   
1         1         101 2019-08-04 09:15:05.200981 2019-07-31 09:15:04.201063   
2         1         102 2019-07-31 09:15:05.201063                        NaT   
3         2         101 2019-06-25 09:15:05.201121 2019-07-05 09:15:04.201179   
4         2         102 2019-07-05 09:15:05.201179                        NaT   
5         2         104 2019-06-25 09:15:05.201121                        NaT   
6         3         105 2019-04-26 09:15:05.201290 2019-06-05 09:15:05.201235   
7         3         106 2019-04-26 09:15:05.201290 2019-04-27 09:15:05.201324   
8         3         107 2019-06-05 09:15:04.201235                        NaT   
9         4         105 2019-04-26 09:15:05.201290 2019-06-05 09:15:04.201235   
10        4         106 2019-04-26 09:15:05.201290 2019-04-27 09:15:05.201324   
11        4         107 2019-07-05 09:15:05.201179 2019-07-31 09:15:05.201063   

    is_switch  
0       False  
1       False  
2        True  
3       False  
4        True  
5       False  
6       False  
7       False  
8        True  
9       False  
10      False  
11      False

Answer 1

将GroupBy.apply与自定义功能一起使用-首先用一些默认值datetime替换缺少的值，例如Timestamp.min，然后每组比较具有广播的列-所有created_at中removed_at的值，获取绝对值，比较1秒，然后每行至少返回True any：

val = pd.Timedelta(1, unit='s')

def f(x):
    y = x['created_at'].values - x['removed_at'].values[:, None]
    y = np.any((np.abs(y).astype(np.int64) <= val.value), axis=0)

    return pd.Series(y, index=x.index)

df['is_switch'] = (df.assign(removed_at = df['removed_at'].fillna(pd.Timestamp.min))
                     .groupby('user_id')
                     .apply(f)
                     .reset_index(level=0, drop=True))

print(df)
    user_id  product_id                 created_at                 removed_at  \
0         0         100 2019-08-04 16:22:39.309093                        NaT   
1         1         101 2019-08-04 16:22:39.309093 2019-07-31 16:22:38.309093   
2         1         102 2019-07-31 16:22:39.309093                        NaT   
3         2         101 2019-06-25 16:22:39.309093 2019-07-05 16:22:38.309093   
4         2         102 2019-07-05 16:22:39.309093                        NaT   
5         2         104 2019-06-25 16:22:39.309093                        NaT   
6         3         105 2019-04-26 16:22:39.309093 2019-06-05 16:22:39.309093   
7         3         106 2019-04-26 16:22:39.309093 2019-04-27 16:22:39.309093   
8         3         107 2019-06-05 16:22:38.309093                        NaT   
9         4         105 2019-04-26 16:22:39.309093 2019-06-05 16:22:38.309093   
10        4         106 2019-04-26 16:22:39.309093 2019-04-27 16:22:39.309093   
11        4         107 2019-07-05 16:22:39.309093 2019-07-31 16:22:39.309093   

    is_switch  
0       False  
1       False  
2        True  
3       False  
4        True  
5       False  
6       False  
7       False  
8        True  
9       False  
10      False  
11      False

Answer 2

单线是：

print(~df['created_at'].sub(df.groupby('user_id').transform('first')['created_at']).dt.days.between(-1, 1))

输出：

0    False
1    False
2     True
3    False
4     True
5    False
Name: created_at, dtype: bool

如何将一列中的行值与组中另一列中的所有其他行进行比较？

2 个答案: