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的列

当我尝试添加新的列差异时,找到两个日期之间的差异使用

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,就像这样

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

这让我疯狂,因为.astype()上面的解决方案对我不起作用.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.要么:

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

......如果你想要分钟.

Answer 3

• 如何将结果转换为仅小时和分钟
  • 接受的答案只返回days + hours。分钟不包括在内。
• 要提供一个小时和分钟为hh:mmor的列，x hours y minutes需要额外的计算和字符串格式。
• 这个答案显示了如何使用timedelta数学将总小时数或总分钟数作为浮点数，并且比使用更快.astype('timedelta64[h]')
• Pandas 时间增量用户指南
• Pandas 时间序列/日期功能用户指南
• pythontimedelta对象：查看支持的操作。
• 以下示例数据已经是一个datetime64[ns] dtype. 要求所有相关列都使用pandas.to_datetime().

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

• 其他资源中播客的一个注意事项.total_seconds()是在核心开发人员休假时添加和合并的，并且不会被批准。
  • 这也是没有其他.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

• 和我谈谈 Python：第 271 集：解开时间的奥秘，即 Python 的日期时间！
  • Timedelta 从 31 分钟开始
  • 根据 Python 核心开发人员Paul Ganssle和Pythondateutil维护人员的说法：
    • 用 (df.from_date - df.to_date) / pd.Timedelta(hours=1)
    • 不要使用 (df.from_date - df.to_date).dt.total_seconds() / 3600
      • pandas.Series.dt.total_seconds
      • .dt 存取器
• 真正的 Python：使用 Python datetime 处理日期和时间
• 该dateutil模块为标准datetime模块提供了强大的扩展。

%%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)