以小时和分钟计算两列之间的Pandas DataFrame时差

Question

我在数据框中有两列和迄今为止

当我尝试添加新的列差异时，使用

查找两个日期之间的差异

df['diff'] = df['todate'] - df['fromdate']

如果超过24小时，我会在几天内得到差异列。

2014-01-24 13:03:12.050000,2014-01-26 23:41:21.870000,"2 days, 10:38:09.820000"
2014-01-27 11:57:18.240000,2014-01-27 15:38:22.540000,03:41:04.300000
2014-01-23 10:07:47.660000,2014-01-23 18:50:41.420000,08:42:53.760000

如何仅在小时和分钟内转换结果，忽略天数甚至秒数。

Answer 1

Pandas时间戳差异返回datetime.timedelta对象。这可以通过使用* as_type *方法轻松转换为小时，如此

import pandas
df = pandas.DataFrame(columns=['to','fr','ans'])
df.to = [pandas.Timestamp('2014-01-24 13:03:12.050000'), pandas.Timestamp('2014-01-27 11:57:18.240000'), pandas.Timestamp('2014-01-23 10:07:47.660000')]
df.fr = [pandas.Timestamp('2014-01-26 23:41:21.870000'), pandas.Timestamp('2014-01-27 15:38:22.540000'), pandas.Timestamp('2014-01-23 18:50:41.420000')]
(df.fr-df.to).astype('timedelta64[h]')

屈服，

0    58
1     3
2     8
dtype: float64

Answer 2

由于上面的.astype()解决方案对我不起作用，这让我感到疯狂。但我发现了另一种方式。没有时间或其他任何东西，但可能会为其他人工作：

t1 = pd.to_datetime('1/1/2015 01:00')
t2 = pd.to_datetime('1/1/2015 03:30')

print pd.Timedelta(t2 - t1).seconds / 3600.0

...如果你想要几个小时。或者：

print pd.Timedelta(t2 - t1).seconds / 60.0

...如果你想要分钟。

Answer 3

• 如何将我的结果转换为仅几小时又几分钟？
  • 接受的答案仅返回days + hours。 不包括分钟。
• 要提供具有小时和分钟数的列（如hh:mm或x hours y minutes，则需要进行其他计算和字符串格式设置。
• 此答案说明了如何使用timedelta数学来获得总小时数或总分钟数，并且比使用.astype('timedelta64[h]')更快
• Pandas Time Deltas User Guide
• Pandas Time series / date functionality User Guide
• python timedelta objects：请参阅支持的操作。

import pandas as pd

# test data from OP, with values already in a datetime format
data = {'to_date': [pd.Timestamp('2014-01-24 13:03:12.050000'), pd.Timestamp('2014-01-27 11:57:18.240000'), pd.Timestamp('2014-01-23 10:07:47.660000')],
        'from_date': [pd.Timestamp('2014-01-26 23:41:21.870000'), pd.Timestamp('2014-01-27 15:38:22.540000'), pd.Timestamp('2014-01-23 18:50:41.420000')]}

# test dataframe; the columns must be in a datetime format; use pandas.to_datetime if needed
df = pd.DataFrame(data)

# add a timedelta column if wanted. It's added here for information only
# df['time_delta_with_sub'] = df.from_date.sub(df.to_date)  # also works
df['time_delta'] = (df.from_date - df.to_date)

# create a column with timedelta as total hours, as a float type
df['tot_hour_diff'] = (df.from_date - df.to_date) / pd.Timedelta(hours=1)

# create a colume with timedelta as total minutes, as a float type
df['tot_mins_diff'] = (df.from_date - df.to_date) / pd.Timedelta(minutes=1)

# display(df)
                  to_date               from_date             time_delta  tot_hour_diff  tot_mins_diff
0 2014-01-24 13:03:12.050 2014-01-26 23:41:21.870 2 days 10:38:09.820000      58.636061    3518.163667
1 2014-01-27 11:57:18.240 2014-01-27 15:38:22.540 0 days 03:41:04.300000       3.684528     221.071667
2 2014-01-23 10:07:47.660 2014-01-23 18:50:41.420 0 days 08:42:53.760000       8.714933     522.896000

其他方法

• 其他资源.total_seconds()中的播客中有一个注释项是在核心开发人员休假期间添加并合并的，因此不会获得批准。
  • 这也是为什么没有其他.total_xx方法的原因。

# convert the entire timedelta to seconds
# this is the same as td / timedelta(seconds=1)
(df.from_date - df.to_date).dt.total_seconds()
[out]:
0    211089.82
1     13264.30
2     31373.76
dtype: float64

# get the number of days
(df.from_date - df.to_date).dt.days
[out]:
0    2
1    0
2    0
dtype: int64

# get the seconds for hours + minutes + seconds, but not days
# note the difference from total_seconds
(df.from_date - df.to_date).dt.seconds
[out]:
0    38289
1    13264
2    31373
dtype: int64

其他资源

• Talk Python to Me: Episode #271: Unlock the mysteries of time, Python's datetime that is!
  • Timedelta从31分钟开始
  • 根据Python核心开发者Paul Ganssle和python dateutil的维护者：
    • 使用(df.from_date - df.to_date) / pd.Timedelta(hours=1)
    • 请勿使用(df.from_date - df.to_date).dt.total_seconds() / 3600
      • pandas.Series.dt.total_seconds
      • .dt accessor
• Real Python: Using Python datetime to Work With Dates and Times
• dateutil模块提供了对标准datetime模块的强大扩展。

%%timeit测试

import pandas as pd

# dataframe with 2M rows
data = {'to_date': [pd.Timestamp('2014-01-24 13:03:12.050000'), pd.Timestamp('2014-01-27 11:57:18.240000')], 'from_date': [pd.Timestamp('2014-01-26 23:41:21.870000'), pd.Timestamp('2014-01-27 15:38:22.540000')]}
df = pd.DataFrame(data)
df = pd.concat([df] * 1000000).reset_index(drop=True)

%%timeit
(df.from_date - df.to_date) / pd.Timedelta(hours=1)
[out]:
43.1 ms ± 1.05 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit
(df.from_date - df.to_date).astype('timedelta64[h]')
[out]:
59.8 ms ± 1.29 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)