这是我的 DF
Start-Time Running-Time Speed-Avg HR-Avg
0 2016-12-18 10:8:14 0:24:2 20 138
1 2016-12-18 10:8:14 0:24:2 20 138
2 2016-12-23 8:52:36 0:31:19 16 134
3 2016-12-23 8:52:36 0:31:19 16 134
4 2016-12-25 8:0:51 0:30:10 50 135
5 2016-12-25 8:0:51 0:30:10 50 135
6 2016-12-26 8:41:26 0:10:1 27 116
7 2016-12-26 8:41:26 0:10:1 27 116
8 2017-1-7 11:16:9 0:26:15 22 124
9 2017-1-7 11:16:9 0:26:15 22 124
10 2017-1-10 19:2:54 0:53:51 5 142
11 2017-1-10 19:2:54 0:53:51 5 142
我一直在尝试格式化此列H:M:S格式
使用
timeDF=(pd.to_datetime(cleanDF['Running-Time'],format='%H:%M:%S'))
但我一直在得到ValueError: time data ' 0:24:2' does not match format '%M:%S' (match) this error
先感谢您。
尾随空格有问题，所以需要str.strip http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.str.strip.html:
或者如果创建DataFrame从文件中read_csv http://pandas.pydata.org/pandas-docs/stable/generated/pandas.read_csv.html添加参数skipinitialspace=True:
cleanDF = pd.read_csv(file, skipinitialspace = True)
timeDF=(pd.to_datetime(cleanDF['Running-Time'].str.strip(), format='%H:%M:%S'))
print (timeDF)
0 1900-01-01 00:24:02
1 1900-01-01 00:24:02
2 1900-01-01 00:31:19
3 1900-01-01 00:31:19
4 1900-01-01 00:30:10
5 1900-01-01 00:30:10
6 1900-01-01 00:10:01
7 1900-01-01 00:10:01
8 1900-01-01 00:26:15
9 1900-01-01 00:26:15
10 1900-01-01 00:53:51
11 1900-01-01 00:53:51
Name: Running-Time, dtype: datetime64[ns]
但也许更好的方法是将其转换为时间增量to_timedelta http://pandas.pydata.org/pandas-docs/stable/generated/pandas.to_timedelta.html:
timeDF=(pd.to_timedelta(cleanDF['Running-Time'].str.strip()))
print (timeDF)
0 00:24:02
1 00:24:02
2 00:31:19
3 00:31:19
4 00:30:10
5 00:30:10
6 00:10:01
7 00:10:01
8 00:26:15
9 00:26:15
10 00:53:51
11 00:53:51
Name: Running-Time, dtype: timedelta64[ns]
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