pandas 数据帧中的字符串类型的时间增量

问题描述

我有一个数据框df，它的第一列是timedelta64

I have a dataframe df and its first column is timedelta64

df.info():

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 686 entries, 0 to 685
Data columns (total 6 columns):
0    686 non-null timedelta64[ns]
1    686 non-null object
2    686 non-null object
3    686 non-null object
4    686 non-null object
5    686 non-null object

例如，如果我是print(df[0][2])，它将给我0 days 05:01:11.但是，我不想提交0 days.我只希望打印05:01:11.有人可以教我该怎么做吗?非常感谢！

If I print(df[0][2]), for example, it will give me 0 days 05:01:11. However, I don't want the 0 days filed. I only want 05:01:11 to be printed. Could someone teaches me how to do this? Thanks so much!

推荐答案

可以通过以下方式实现:

It is possible by:

df['duration1'] = df['duration'].astype(str).str[-18:-10]

但是解决方案并不通用，如果输入为3 days 05:01:11，它也会删除3 days.

But solution is not general, if input is 3 days 05:01:11 it remove 3 days too.

因此，解决方案仅能有效地将时间间隔缩短至不到一天.

So solution working only for timedeltas less as one day correctly.

更通用的解决方案是创建自定义格式:

N = 10
np.random.seed(11230)
rng = pd.date_range('2017-04-03 15:30:00', periods=N, freq='13.5H')
df = pd.DataFrame({'duration': np.abs(np.random.choice(rng, size=N) - 
                                 np.random.choice(rng, size=N)) })  

df['duration1'] = df['duration'].astype(str).str[-18:-10]

def f(x):
    ts = x.total_seconds()
    hours, remainder = divmod(ts, 3600)
    minutes, seconds = divmod(remainder, 60)
    return ('{}:{:02d}:{:02d}').format(int(hours), int(minutes), int(seconds)) 

df['duration2'] = df['duration'].apply(f)
print (df)

         duration duration1  duration2
0 2 days 06:00:00  06:00:00   54:00:00
1 2 days 19:30:00  19:30:00   67:30:00
2 1 days 03:00:00  03:00:00   27:00:00
3 0 days 00:00:00  00:00:00    0:00:00
4 4 days 12:00:00  12:00:00  108:00:00
5 1 days 03:00:00  03:00:00   27:00:00
6 0 days 13:30:00  13:30:00   13:30:00
7 1 days 16:30:00  16:30:00   40:30:00
8 0 days 00:00:00  00:00:00    0:00:00
9 1 days 16:30:00  16:30:00   40:30:00

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