python pandas datetime.time - datetime.time
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python pandas datetime.time - datetime.time
提问by James Bond
i have a dataframe which contains two columns of datetime.time items. something like
我有一个包含两列 datetime.time 项目的数据框。就像是
col1 col2
02:10:00.008209 02:08:38.053145
02:10:00.567054 02:08:38.053145
02:10:00.609842 02:08:38.053145
02:10:00.728153 02:08:38.053145
02:10:02.394408 02:08:38.053145
how can i generate a col3 which is the differences between col1 and col2? (preferablly in microseconds)?
我如何生成一个 col3，它是 col1 和 col2 之间的差异？（最好以微秒为单位）？
I searched around but I cannot find a solution here. Does anyone know?
我四处搜索，但在这里找不到解决方案。有人知道吗？
Thanks!
谢谢！
回答by HYRY
don't use datetime.time, use timedelta:
不要使用datetime.time，使用timedelta：
import pandas as pd
import io
data = """col1 col2
02:10:00.008209 02:08:38.053145
02:10:00.567054 02:08:38.053145
02:10:00.609842 02:08:38.053145
02:10:00.728153 02:08:38.053145
02:10:02.394408 02:08:38.053145"""
df = pd.read_table(io.BytesIO(data), delim_whitespace=True)
df2 = df.apply(pd.to_timedelta)
diff = df2.col1 - df2.col2
diff.astype("i8")/1e9
the output is different in seconds:
输出以秒为单位不同：
0 81.955064
1 82.513909
2 82.556697
3 82.675008
4 84.341263
dtype: float64
To convert time dataframe to timedelta dataframe:
要将时间数据帧转换为 timedelta 数据帧：
df.applymap(time.isoformat).apply(pd.to_timedelta)
回答by unutbu
Are you sure you want a DataFrame of datetime.timeobjects? There is hardly an operation you can perform conveniently on these guys especially when wrapped in a DataFrame.
你确定你想要一个datetime.time对象的 DataFrame吗？几乎没有任何操作可以方便地对这些家伙执行，尤其是在包装在 DataFrame 中时。
It might be better to have each column store an int representing the total number of microseconds.
让每列存储一个表示总微秒数的 int 可能会更好。
You can convert dfto a DataFrame storing microseconds like this:
您可以转换df为存储微秒的数据帧，如下所示：
In [71]: df2 = df.applymap(lambda x: ((x.hour*60+x.minute)*60+x.second)*10**6+x.microsecond)
In [72]: df2
Out[72]:
col1 col2
0 7800008209 7718053145
1 7800567054 7718053145
And from there, it is easy to get the result you desire:
从那里，很容易得到你想要的结果：
In [73]: df2['col1']-df2['col2']
Out[73]:
0 81955064
1 82513909
dtype: int64
回答by Acorbe
pandasconverts datetimeobjects to np.datetime64objects, whose differences are np.timedelta64objects.
pandas将datetime对象转换为np.datetime64对象，它们的区别在于np.timedelta64对象。
Consider this
考虑这个
In [30]: df
Out[30]:
0 1
0 2014-02-28 13:30:19.926778 2014-02-28 13:30:47.178474
1 2014-02-28 13:30:29.814575 2014-02-28 13:30:51.183349
I can consider the column-wise difference by
我可以考虑按列的差异
df[0] - df[1]
Out[31]:
0 -00:00:27.251696
1 -00:00:21.368774
dtype: timedelta64[ns]
and hence I can apply timedelta64conversions. For microseconds
因此我可以应用timedelta64转换。对于微秒
(df[0] - df[1]).apply(lambda x : x.astype('timedelta64[us]')) #no actual difference when displayed
or microseconds as integers
或微秒作为整数
(df[0] - df[1]).apply(lambda x : x.astype('timedelta64[us]').astype('int'))
0 -27251696000
1 -21368774000
dtype: int64
EDIT:As suggessted by @Jeff, the last expressions can be shortened as
编辑：正如@Jeff 所建议的，最后一个表达式可以缩短为
(df[0] - df[1]).astype('timedelta64[us]')
and
和
(df[0] - df[1]).astype('timedelta64[us]').astype('int')
for pandas >= .13.
对于Pandas >= .13。
