Chris Larson

而这可以通过 re ,使用 datetime.strptime 在@liamhawkins的回答中[ https://stackoverflow.com/a/54600322/214150 ]如果你总是在处理格式化的日期,最好是这样。

此外,您还可以通过简单地使用 string method ( .split() )以及结果列表的基本切片。例如:

import re

word = 'Sat Apr 18 23:22:15 PDT 2009'

# Example using re.
rehour = re.findall('(\d+):\d+:\d+', word)
print('rehour:', *rehour)

# Example using string.split() and slicing.
somedate = word.split(' ')
somehour = somedate[3][:2]
print('somedate:', somedate)
print('somehour:', somehour)

假设您的示例中有日期格式,下面将从使用您的时间戳生成的DateTime对象生成一周中的小时:

from datetime import datetime

timestamp = '2/19/2019 17:33'
timestamp_object = datetime.strptime(timestamp, '%m/%d/%Y %H:%M')

hour_of_week = (timestamp_object.timetuple().tm_wday * 24 + 24) - (24 - timestamp_object.timetuple().tm_hour)

print('Hour of the week: {}'.format(hour_of_week))

输出:

 Hour of the week: 41

这将为您提供在上下文中应用所需的工具。像这样:

df['Date'] = df['Date'].apply(pd.to_datetime)

df['Hour of Week'] = ((df['Date'].dt.dayofweek) * 24 + 24) - (24 - df['Date'].dt.hour)
df['Day of Week'] = df['Date'].dt.dayofweek
df['Hour of Day'] = df['Date'].dt.hour
df = df[['Date', 'Day of Week', 'Hour of Day', 'Hour of Week', 'Price of X', 'Price of Y']]

请注意,如果您确实希望从1小时开始一天,而不是从零开始,则需要执行以下操作:

df['Hour of Week'] = ((df['Date'].dt.dayofweek) * 24 + 24) - (24 - df['Date'].dt.hour) + 1

和

df['Hour of Day'] = df['Date'].dt.hour + 1

以免混淆。

就像下面要抵消的第一天 1 :

df['Day of Week'] = df['Date'].dt.dayofweek + 1

只要记住 Hour of Day 价值 df['Date'].dt.hour 开始于 0 当你做比较的时候。因此,我建议使用 零 作为你的标准。同样适用于 Day of Week .

用 Hour of Week 开始于 零 ,您的数据帧如下所示:

                Date  Day of Week  Hour of Day  Hour of Week  Price of X  Price of Y
0 2011-01-01 00:30:00            5            0           120          19         172
1 2011-01-01 01:00:00            5            1           121          98         105
2 2011-01-01 01:30:00            5            1           121          40          -6
3 2011-01-01 02:00:00            5            2           122          68         120
4 2011-01-01 02:30:00            5            2           122          16         182
5 2011-01-01 03:00:00            5            3           123          23         187
6 2011-01-01 03:30:00            5            3           123          94         200
7 2019-02-18 00:01:00            0            0             0          23         187
8 2019-02-18 23:15:00            0           23            23          23         187
9 2019-02-24 23:59:00            6           23           167          23         187

天和小时增加1:

df['Hour of Week'] = ((df['Date'].dt.dayofweek) * 24 + 24) - (24 - df['Date'].dt.hour) + 1
df['Day of Week'] = df['Date'].dt.dayofweek + 1
df['Hour of Day'] = df['Date'].dt.hour + 1

你得到:

                 Date  Day of Week  Hour of Day  Hour of Week  Price of X  Price of Y
0 2011-01-01 00:30:00            6            1           121          19         172
1 2011-01-01 01:00:00            6            2           122          98         105
2 2011-01-01 01:30:00            6            2           122          40          -6
3 2011-01-01 02:00:00            6            3           123          68         120
4 2011-01-01 02:30:00            6            3           123          16         182
5 2011-01-01 03:00:00            6            4           124          23         187
6 2011-01-01 03:30:00            6            4           124          94         200
7 2019-02-18 00:01:00            1            1             1          23         187
8 2019-02-18 23:15:00            1           24            24          23         187
9 2019-02-24 23:59:00            7           24           168          23         187