将字符串转回日期时间 timedelta

Question

我的 pandas 数据框中的一列表示我用 datetime 计算的时间增量，然后导出到 csv 并读回 pandas 数据框。现在该列的 dtype 是对象，而我希望它是一个 timedelta，这样我就可以在数据帧上执行 groupby 函数。下面是字符串的样子。谢谢！

  0 days 00:00:57.416000
  0 days 00:00:12.036000
  0 days 16:46:23.127000  
 49 days 00:09:30.813000  
 50 days 00:39:31.306000  
 55 days 12:39:32.269000
-1 days +22:03:05.256000

更新，我最好的尝试是编写一个 for 循环来迭代我的 pandas 数据帧中的特定列：

def delta(i):
    days, timestamp = i.split(" days ")
    timestamp = timestamp[:len(timestamp)-7]
    t = datetime.datetime.strptime(timestamp,"%H:%M:%S") + 
    datetime.timedelta(days=int(days))
    delta = datetime.timedelta(days=t.day, hours=t.hour, 
    minutes=t.minute, seconds=t.second)
    delta.total_seconds()

data['diff'].map(delta)

Answer 1

您可以这样做，循环遍历 CSV 中的每个值来代替字符串日期：

stringdate = "2 days 00:00:57.416000"
days_v_hms = string1.split('days')
hms = days_v_hms[1].split(':')
dt = datetime.timedelta(days=int(days_v_hms[0]), hours=int(hms[0]), minutes=int(hms[1]), seconds=float(hms[2]))

干杯！

Answer 2

import datetime

#Parse your string
days, timestamp = "55 days 12:39:32.269000".split(" days ")
timestamp = timestamp[:len(timestamp)-7]

#Generate datetime object
t = datetime.datetime.strptime(timestamp,"%H:%M:%S") + datetime.timedelta(days=int(days))

#Generate a timedelta
delta = datetime.timedelta(days=t.day, hours=t.hour, minutes=t.minute, seconds=t.second)

#Represent in Seconds
delta.total_seconds()

Answer 3

使用pd.to_timedelta

pd.to_timedelta(df.iloc[:, 0])

0     0 days 00:00:57.416000
1     0 days 00:00:12.036000
2     0 days 16:46:23.127000
3    49 days 00:09:30.813000
4    50 days 00:39:31.306000
5    55 days 12:39:32.269000
6   -1 days +22:03:05.256000
Name: 0, dtype: timedelta64[ns]

将字符串转回日期时间 timedelta

Turn a string back into a datetime timedelta

python

datetime

timedelta

pandas