Question

我有一个像这样的数据帧df1，其中starttime和endtime是datetime对象。

StartTime EndTime

9:08 9:10
9:10 9:35
9:35 9:55
9:55 10:10
10:10 10:20

如果endtime.hour与startime.hour不同，我想分一次这样的时间

StartTime EndTime

9:08 9:10

9:10 9:55

9:55 10:00

10:00 10:10

10:10 10:20

基本上在现有数据框df1中插入一行。我看了很多例子，但还没弄明白怎么做。如果我的问题不明确，请告诉我。

由于

Answer 1

这就是你想做的......

# load your data into a DataFrame
data="""StartTime EndTime
9:08 9:10
9:10 9:35
9:35 9:55
9:55 10:10
10:10 10:20
"""
from StringIO import StringIO # import from io for Python 3
df = pd.read_csv(StringIO(data), header=0, sep=' ', index_col=None)

# convert strings to Pandas Timestamps (we will ignore the date bit) ...
import datetime as dt
df.StartTime = [dt.datetime.strptime(x, '%H:%M') for x in df.StartTime]
df.EndTime = [dt.datetime.strptime(x, '%H:%M') for x in df.EndTime]

# assumption - all intervals are less than 60 minutes
#            - ie. no multi-hour intervals

# add rows
dfa = df[df.StartTime.dt.hour != df.EndTime.dt.hour].copy()
dfa.EndTime = [dt.datetime.strptime(str(x), '%H') for x in dfa.EndTime.dt.hour]

# play with the start hour ...
df.StartTime = df.StartTime.where(df.StartTime.dt.hour == df.EndTime.dt.hour,
    other = [dt.datetime.strptime(str(x), '%H') for x in df.EndTime.dt.hour])

# bring back together and sort
df = pd.concat([df, dfa], axis=0) #top/bottom
df = df.sort('StartTime')

# convert the Timestamps to times for easy reading
df.StartTime = [x.time() for x in df.StartTime]
df.EndTime = [x.time() for x in df.EndTime]

和收益

In [40]: df
Out[40]: 
  StartTime   EndTime
0  09:08:00  09:10:00
1  09:10:00  09:35:00
2  09:35:00  09:55:00
3  09:55:00  10:00:00
3  10:00:00  10:10:00
4  10:10:00  10:20:00

按小时Python分割时间

1 个答案: