使用现有的公共假期和周末列为长周末创建新列答案

【问题标题】：Creating a new column for long weekend with existing public holiday and weekend columns使用现有的公共假期和周末列为长周末创建新列
【发布时间】：2019-09-10 06:42:52
【问题描述】：

我有一个 pandas 数据框，其中包含 date、week_day、public_holiday 和weekend 列。

 weekday Date           Public_Holiday? Weekend?
   5     2015-01-10              no      yes
   0     2015-01-12              no       no
   1     2015-01-13              no       no
   2     2015-01-14              no       no
   3     2015-01-15              no       no
   4     2015-01-16              no       no
   5     2015-01-17              no      yes
   6     2015-01-18              no      yes
   0     2015-01-19              no       no
   1     2015-01-20              no       no
   2     2015-01-21              no       no
   3     2015-01-22              no       no
   4     2015-01-23              yes      no
   5     2015-01-24              no      yes
   6     2015-01-25              no      yes
   1     2015-01-27              no       no
   2     2015-01-28              no       no
   3     2015-01-29              no       no
   4     2015-01-30              no       no
   5     2015-01-31              no      yes
   0     2015-02-02              no       no
   1     2015-02-03              no       no
   2     2015-02-04              no       no
   3     2015-02-05              no       no
   4     2015-02-06              no       no
   5     2015-02-07              no      yes
   6     2015-02-08              no      yes
   0     2015-02-09              yes      no
   1     2015-02-10              no       no
   2     2015-02-11              no       no

我需要添加一个带有长周末标志的附加列。输出应如下所示。

    long_weekend  weekday   Date          Public_Holiday? Weekend?
            0        5     2015-01-10              no      yes
            0        0     2015-01-12              no       no
            0        1     2015-01-13              no       no
            0        2     2015-01-14              no       no
            0        3     2015-01-15              no       no
            0        4     2015-01-16              no       no
            0        5     2015-01-17              no      yes
            0        6     2015-01-18              no      yes
            0        0     2015-01-19              no       no
            0        1     2015-01-20              no       no
            0        2     2015-01-21              no       no
            0        3     2015-01-22              no       no
            1        4     2015-01-23              yes      no
            1        5     2015-01-24              no      yes
            1        6     2015-01-25              no      yes
            0        1     2015-01-27              no       no
            0        2     2015-01-28              no       no
            0        3     2015-01-29              no       no
            0        4     2015-01-30              no       no
            0        5     2015-01-31              no      yes
            0        0     2015-02-02              no       no
            0        1     2015-02-03              no       no
            0        2     2015-02-04              no       no
            0        3     2015-02-05              no       no
            0        4     2015-02-06              no       no
            1        5     2015-02-07              no      yes
            1        6     2015-02-08              no      yes
            1        0     2015-02-09              yes      no
            0        1     2015-02-10              no       no
            0        2     2015-02-11              no       no

常规周末不被视为长周末。仅当星期五或星期一以及在某些情况下星期四或星期二是假期时，整个系列才被视为长周末。

这是我在下面尝试过的

df['long_weekend'] = np.where((df['Public_Holiday?'] == 'yes') | (df['Weekend?'] == 'yes'), 1, 0)
df['weekday'] = df['Predicted_Date'].dt.dayofweek
df['long_weekend'] = np.where(((df['long_weekend'] == 1) & (df['weekday'] == 4)) | (df['long_weekend'] == 1) & (df['weekday'] == 0)), 'yes','no')

这给了我以下输出，甚至正常工作日为 1。

    long_weekend  weekday         Date   Public_Holiday? Weekend?
            1        5     2015-01-10              no      yes
            0        0     2015-01-12              no       no
            0        1     2015-01-13              no       no
            0        2     2015-01-14              no       no
            0        3     2015-01-15              no       no
            0        4     2015-01-16              no       no
            1        5     2015-01-17              no      yes
            1        6     2015-01-18              no      yes
            0        0     2015-01-19              no       no
            0        1     2015-01-20              no       no
            0        2     2015-01-21              no       no
            0        3     2015-01-22              no       no
            1        4     2015-01-23              yes      no
            1        5     2015-01-24              no      yes
            1        6     2015-01-25              no      yes
            0        1     2015-01-27              no       no
            0        2     2015-01-28              no       no
            0        3     2015-01-29              no       no
            0        4     2015-01-30              no       no
            1        5     2015-01-31              no      yes
            0        0     2015-02-02              no       no
            0        1     2015-02-03              no       no
            0        2     2015-02-04              no       no
            0        3     2015-02-05              no       no
            0        4     2015-02-06              no       no
            1        5     2015-02-07              no      yes
            1        6     2015-02-08              no      yes
            1        0     2015-02-09              yes      no
            0        1     2015-02-10              no       no
            0        2     2015-02-11              no       no

我怎样才能让它工作？任何帮助都会很棒。提前致谢。

【问题讨论】：

您为 df['long_weekend'] 发布的输出格式为 - 1 0 1，但在您的代码中，您将 yes no 分配给 df['long_weekend']，您能确认一下吗？跨度>

标签： python python-3.x pandas

【解决方案1】：

按照@jezraer 的非常好的解决方案，我实现了一种自动发现长周末的方法。正如 OP 要求的那样，它还考虑了周二或周四假期的情况。

该方法在本项目中可用： https://github.com/kryptonite0/python-long-weekends

import holidays as holidays_api
from long_weekends.long_weekends import spot_holiday_bridges

start = '2021-01-01'
end = '2021-12-31'
holidays = holidays_api.CH(prov='TI', years=[2020, 2021, 2022])
bridges, long_weekends = spot_holiday_bridges(
    start=start, end=end, holidays=holidays)

bridges
[Timestamp('2021-05-14 00:00:00'),
 Timestamp('2021-06-04 00:00:00'),
 Timestamp('2021-06-28 00:00:00')]


long_weekends
[Timestamp('2021-01-01 00:00:00'),
 Timestamp('2021-01-02 00:00:00'),
 Timestamp('2021-01-03 00:00:00'),
 Timestamp('2021-03-19 00:00:00'),
 Timestamp('2021-03-20 00:00:00'),
 Timestamp('2021-03-21 00:00:00'),
 Timestamp('2021-04-02 00:00:00'),
 Timestamp('2021-04-03 00:00:00'),
 Timestamp('2021-04-04 00:00:00'),
 Timestamp('2021-04-05 00:00:00'),
 Timestamp('2021-05-13 00:00:00'),
 Timestamp('2021-05-14 00:00:00'),
 Timestamp('2021-05-15 00:00:00'),
 Timestamp('2021-05-16 00:00:00'),
 Timestamp('2021-05-22 00:00:00'),
 Timestamp('2021-05-23 00:00:00'),
 Timestamp('2021-05-24 00:00:00'),
 Timestamp('2021-06-03 00:00:00'),
 Timestamp('2021-06-04 00:00:00'),
 Timestamp('2021-06-05 00:00:00'),
 Timestamp('2021-06-06 00:00:00'),
 Timestamp('2021-06-26 00:00:00'),
 Timestamp('2021-06-27 00:00:00'),
 Timestamp('2021-06-28 00:00:00'),
 Timestamp('2021-06-29 00:00:00'),
 Timestamp('2021-10-30 00:00:00'),
 Timestamp('2021-10-31 00:00:00'),
 Timestamp('2021-11-01 00:00:00')]

【讨论】：

【解决方案2】：

想法是通过shift 和cumsum 创建连续的组，并用map 和value_counts 计算组的数量，并用2 过滤更多：

long = (df['Public_Holiday?'] == 'yes') | (df['Weekend?'] == 'yes')
s = long.ne(long.shift()).cumsum()
df['long_weekend'] = np.where((s.map(s.value_counts()) > 2) & long, 1, 0)

print (df)
    weekday Predicted_Date Public_Holiday? Weekend?  long_weekend
0         5     2015-01-10              no      yes             0
1         0     2015-01-12              no       no             0
2         1     2015-01-13              no       no             0
3         2     2015-01-14              no       no             0
4         3     2015-01-15              no       no             0
5         4     2015-01-16              no       no             0
6         5     2015-01-17              no      yes             0
7         6     2015-01-18              no      yes             0
8         0     2015-01-19              no       no             0
9         1     2015-01-20              no       no             0
10        2     2015-01-21              no       no             0
11        3     2015-01-22              no       no             0
12        4     2015-01-23             yes       no             1
13        5     2015-01-24              no      yes             1
14        6     2015-01-25              no      yes             1
15        1     2015-01-27              no       no             0
16        2     2015-01-28              no       no             0
17        3     2015-01-29              no       no             0
18        4     2015-01-30              no       no             0
19        5     2015-01-31              no      yes             0
20        0     2015-02-02              no       no             0
21        1     2015-02-03              no       no             0
22        2     2015-02-04              no       no             0
23        3     2015-02-05              no       no             0
24        4     2015-02-06              no       no             0
25        5     2015-02-07              no      yes             1
26        6     2015-02-08              no      yes             1
27        0     2015-02-09             yes       no             1
28        1     2015-02-10              no       no             0
29        2     2015-02-11              no       no             0

【讨论】：