将字符串转回日期时间时间增量 [英] Turn a string back into a datetime timedelta

问题描述

我的 Pandas 数据框中的一列表示我使用 datetime 计算的时间增量，然后导出到 csv 并读回 Pandas 数据框中.现在该列的 dtype 是 object 而我希望它是 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！

A column in my pandas data frame represents a time delta that I calculated with datetime then exported into a csv and read back into a pandas data frame. Now the column's dtype is object whereas I want it to be a timedelta so I can perform a groupby function on the dataframe. Below is what the strings look like. Thanks!

  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，我最好尝试编写一个 for 循环来迭代我的 Pandas 数据帧中的特定列:

Update, my best attempt at writing a for-loop to iterate over a specific column in my pandas dataframe:

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)

推荐答案

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]

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