如何将时间转换为DataFrame Python中的相应插槽

希望我正确理解了您的想法。

首先，将您的time列转换为timedelta64：

df['time']=pd.to_timedelta(df['time'])

如果只想在每x小时内转换时间，答案很简单：

import numpy as np
x = 5
ranges = [(i,i+x) for i in range(0,24,x)]
df['slot'] = df['time'].apply(lambda x:[t<= (x/np.timedelta64(1, 'h')) <T for t,T in ranges].index(True)+1)

结果：

+----+--------+-----------------+--------+
|    | name   | time            |   slot |
|----+--------+-----------------+--------|
|  0 | a      | 0 days 18:15:10 |      4 |
|  1 | b      | 0 days 06:16:03 |      2 |
|  2 | c      | 0 days 11:51:01 |      3 |
+----+--------+-----------------+--------+

而且，如果您想要自定义灵活的时间范围（可能与* h：m：s`*一样精确），则应该：

ranges = [('0:0:0','12:0:0'),('12:0:0','23:59:59')]
ranges = [pd.to_timedelta(i) for i in ranges]
df['slot'] = df['time'].apply(lambda x:[t<=x<T for t,T in ranges].index(True)+1)

结果：

+----+--------+-----------------+--------+
|    | name   | time            |   slot |
|----+--------+-----------------+--------|
|  0 | a      | 0 days 18:15:10 |      2 |
|  1 | b      | 0 days 06:16:03 |      1 |
|  2 | c      | 0 days 11:51:01 |      1 |
+----+--------+-----------------+--------+

回答来源：stackoverflow