Question

我想排除今天之前和今天2周之后的数据。我目前有这样的数据。例如今天是01/09/19

promo   date    item   discount   stock  stock/day
   a    30/08/19   ax      15%       200     50
   a    31/08/19   ax      15%       200     50
   a    01/09/19   ax      15%       200     50
   a    02/09/19   ax      15%       200     50
   a    30/08/19   bx      15%       200     50
   a    31/08/19   bx      15%       200     50
   a    01/09/19   bx      15%       200     50
   a    02/09/19   bx      15%       200     50
   a    27/08/19   cx      25%       200    200 
   a    27/08/19   dx      15%       200    200
   b    03/09/19   ex      15%       200     25
   b    04/09/19   ex      15%       200     25
   b    09/09/19   ex      15%       200     25
   b    10/09/19   ex      15%       200     25
   b    15/09/19   fx      15%       200    200
   b    16/09/19   fx      15%       200    200
   b    21/09/19   fx      15%       200    200
   b    22/09/19   fx      15%       200    200
   b    25/09/19   fx      15%       200    200

因此数据将被排除，最终数据将如下所示。

promo   date    item   discount   stock  stock/day
   a    01/09/19   ax      15%       200     50
   a    02/09/19   ax      15%       200     50
   a    01/09/19   bx      15%       200     50
   a    02/09/19   bx      15%       200     50
   b    03/09/19   ex      15%       200     25
   b    04/09/19   ex      15%       200     25
   b    09/09/19   ex      15%       200     25
   b    10/09/19   ex      15%       200     25

该怎么做？提前非常感谢

Answer 1

首先将date转换为标准格式，然后仅保留那些在范围内的行

todays_date <- as.Date("01/09/19", "%d/%m/%y")
df$date1 <- as.Date(df$date, "%d/%m/%y")
df[df$date1 >= todays_date & df$date1 < (todays_date + 14), ]

#   promo     date item discount stock stock.day      date1
#3      a 01/09/19   ax      15%   200        50 2019-09-01
#4      a 02/09/19   ax      15%   200        50 2019-09-02
#7      a 01/09/19   bx      15%   200        50 2019-09-01
#8      a 02/09/19   bx      15%   200        50 2019-09-02
#11     b 03/09/19   ex      15%   200        25 2019-09-03
#12     b 04/09/19   ex      15%   200        25 2019-09-04
#13     b 09/09/19   ex      15%   200        25 2019-09-09
#14     b 10/09/19   ex      15%   200        25 2019-09-10

如果以后不需要，可以删除date1列。

如果您更喜欢链条和包裹，我们也可以

library(dplyr)
library(lubridate)

df %>%
  mutate(date1 = dmy(date)) %>%
  filter(date1 >= todays_date & date1 < (todays_date + 14))

如何根据日期排除数据？

1 个答案: