Question

以下代码生成df：

import pandas as pd
import random
from datetime import timedelta

def randomTime():

    rtime = int(random.random()*86400)

    hours   = int(rtime/3600)
    minutes = int((rtime - hours*3600)/60)
    seconds = rtime - hours*3600 - minutes*60

    time_string = '%02d:%02d:%02d' % (hours, minutes, seconds)
    return time_string

time = [randomTime() for _ in range(8)]

k = 5
N = 8

d = ({'Time' : (time),
    'Events' : ['ABC','DEF','GHI','JKL','ABC','DEF','GHI','JKL'],
    'Number1' : ['xx','xx',1,'xx','xx','xx',2,'xx'],
    'Number2' : ['xx',1,'xx',1,'xx',2,'xx',2]})

df = pd.DataFrame(data=d)

我正在尝试为df中选定的时间戳行添加时间。我试图改变最后4个时间戳来增加3个小时。目前，它正在制作字符串。

df.iloc[4:8,3] = pd.TimedeltaIndex(df.iloc[4:8,3]) + timedelta(hours=3)

print(df)

Output:  
Events Number1 Number2            Time
0    ABC      xx      xx        14:25:51
1    DEF      xx       1        10:02:32
2    GHI       1      xx        01:23:32
3    JKL      xx       1        07:27:42
4    ABC      xx      xx  74325000000000
5    DEF      xx       2  38992000000000
6    GHI       2      xx  19158000000000
7    JKL      xx       2  26746000000000

我不确定这是否因为此处的计算中添加了日期？我不确定我做错了什么。

Answer 1

您的Time列是一列字符串。我建议将整个列转换为timedelta 首先，然后再添加3小时。否则，你最终会混合dtypes（字符串和timedelta），并且会发生奇怪的事情。

df['Time'] = pd.to_timedelta(df.Time, errors='coerce')
df.iloc[4:8, 3] += pd.Timedelta(hours=3)

df

  Events Number1 Number2            Time
0    ABC      xx      xx 0 days 13:51:35
1    DEF      xx       1 0 days 23:14:11
2    GHI       1      xx 0 days 02:16:28
3    JKL      xx       1 0 days 05:25:40
4    ABC      xx      xx 0 days 15:40:15
5    DEF      xx       2 0 days 04:05:26
6    GHI       2      xx 0 days 15:48:06
7    JKL      xx       2 1 days 00:12:30

为pandas DataFrame中的行子集添加时间

1 个答案: