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Question

例如，对于数据框架：

：d

我想下面的输出：

date_rng = pd.date_range(start='2024-01-01', periods=48, freq='30min')  
d = pd.DataFrame({'datetime': date_rng})
    

您可以创建一个获得一个值并将其转换为字符串，

d.head(40)
Out[186]: 
              datetime part
0  2024-01-01 00:00:00   ON
1  2024-01-01 00:30:00   ON
2  2024-01-01 01:00:00   ON
3  2024-01-01 01:30:00   ON
4  2024-01-01 02:00:00   ON
5  2024-01-01 02:30:00   ON
6  2024-01-01 03:00:00   ON
7  2024-01-01 03:30:00   ON
8  2024-01-01 04:00:00   ON
9  2024-01-01 04:30:00   ON
10 2024-01-01 05:00:00   ON
11 2024-01-01 05:30:00   ON
12 2024-01-01 06:00:00   ON
13 2024-01-01 06:30:00   ON
14 2024-01-01 07:00:00   ON
15 2024-01-01 07:30:00   ON
16 2024-01-01 08:00:00   ON
17 2024-01-01 08:30:00   ON
18 2024-01-01 09:00:00   FH
19 2024-01-01 09:30:00   FH
20 2024-01-01 10:00:00    M
21 2024-01-01 10:30:00    M
22 2024-01-01 11:00:00    M
23 2024-01-01 11:30:00    M
24 2024-01-01 12:00:00    M
25 2024-01-01 12:30:00    M
26 2024-01-01 13:00:00    M
27 2024-01-01 13:30:00    M
28 2024-01-01 14:00:00    M
29 2024-01-01 14:30:00    M
30 2024-01-01 15:00:00  SLH
31 2024-01-01 15:30:00   LH
32 2024-01-01 16:00:00   ON
33 2024-01-01 16:30:00   ON
34 2024-01-01 17:00:00   ON
35 2024-01-01 17:30:00   ON
36 2024-01-01 18:00:00   ON
37 2024-01-01 18:30:00   ON
38 2024-01-01 19:00:00   ON
39 2024-01-01 19:30:00   ON

，

ON

Answer 1

FH

小型工作代码

我将时间转换为

df['part'] = df['datetime'].apply(your_function)

并与字符串进行比较，但您也可以直接比较对象

string

---

datetime.time