计算 DataFrame Pandas 中“时间”行之间的差异
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Calculate difference between 'times' rows in DataFrame Pandas
提问by Pragnya Srinivasan
My DataFrame is in the Form:
我的数据帧采用以下形式：
TimeWeek TimeSat TimeHoli
0 6:40:00 8:00:00 8:00:00
1 6:45:00 8:05:00 8:05:00
2 6:50:00 8:09:00 8:10:00
3 6:55:00 8:11:00 8:14:00
4 6:58:00 8:13:00 8:17:00
5 7:40:00 8:15:00 8:21:00
I need to find the time difference between each row in TimeWeek , TimeSat and TimeHoli, the output must be
我需要找到 TimeWeek 、 TimeSat 和 TimeHoli 中每一行之间的时差，输出必须是
TimeWeekDiff TimeSatDiff TimeHoliDiff
00:05:00 00:05:00 00:05:00
00:05:00 00:04:00 00:05:00
00:05:00 00:02:00 00:04:00
00:03:00 00:02:00 00:03:00
00:02:00 00:02:00 00:04:00
I tried using (d['TimeWeek']-df['TimeWeek'].shift().fillna(0), it throws an error:
我尝试使用(d['TimeWeek']-df['TimeWeek'].shift().fillna(0)，它抛出一个错误：
TypeError: unsupported operand type(s) for -: 'str' and 'str'
Probably because of the presence of ':' in the column. How do I resolve this?
可能是因为列中存在“:”。我该如何解决？
回答by Alexander
It looks like the error is thrown because the data is in the form of a string instead of a timestamp. First convert them to timestamps:
看起来抛出错误是因为数据是字符串形式而不是时间戳。首先将它们转换为时间戳：
df2 = df.apply(lambda x: [pd.Timestamp(ts) for ts in x])
They will contain today's date by default, but this shouldn't matter once you difference the time (hopefully you don't have to worry about differencing 23:55 and 00:05 across dates).
默认情况下，它们将包含今天的日期，但是一旦您区分时间，这应该无关紧要（希望您不必担心跨日期区分 23:55 和 00:05）。
Once converted, simply difference the DataFrame:
转换后，只需区分 DataFrame：
>>> df2 - df2.shift()
TimeWeek TimeSat TimeHoli
0 NaT NaT NaT
1 00:05:00 00:05:00 00:05:00
2 00:05:00 00:04:00 00:05:00
3 00:05:00 00:02:00 00:04:00
4 00:03:00 00:02:00 00:03:00
5 00:42:00 00:02:00 00:04:00
Depending on your needs, you can just take rows 1+ (ignoring the NaTs):
根据您的需要，您可以只取第 1+ 行（忽略 NaT）：
(df2 - df2.shift()).iloc[1:, :]
or you can fill the NaTs with zeros:
或者您可以用零填充 NaT：
(df2 - df2.shift()).fillna(0)
回答by jwilner
Forget everything I just said. Pandas has great timedelta parsing.
忘记我刚才说的一切。Pandas 有很好的 timedelta 解析。
df["TimeWeek"] = pd.to_timedelta(df["TimeWeek"])
(d['TimeWeek']-df['TimeWeek'].shift().fillna(pd.to_timedelta("00:00:00"))
回答by S.Sreeram
>>> import pandas as pd
>>> df = pd.DataFrame({'TimeWeek': ['6:40:00', '6:45:00', '6:50:00', '6:55:00', '7:40:00']})
>>> df["TimeWeek_date"] = pd.to_datetime(df["TimeWeek"], format="%H:%M:%S")
>>> print df
TimeWeek TimeWeek_date
0 6:40:00 1900-01-01 06:40:00
1 6:45:00 1900-01-01 06:45:00
2 6:50:00 1900-01-01 06:50:00
3 6:55:00 1900-01-01 06:55:00
4 7:40:00 1900-01-01 07:40:00
>>> df['TimeWeekDiff'] = (df['TimeWeek_date'] - df['TimeWeek_date'].shift().fillna(pd.to_datetime("00:00:00", format="%H:%M:%S")))
>>> print df
TimeWeek TimeWeek_date TimeWeekDiff
0 6:40:00 1900-01-01 06:40:00 06:40:00
1 6:45:00 1900-01-01 06:45:00 00:05:00
2 6:50:00 1900-01-01 06:50:00 00:05:00
3 6:55:00 1900-01-01 06:55:00 00:05:00
4 7:40:00 1900-01-01 07:40:00 00:45:00
