如何根据日期条件创建新列？答案

【问题标题】：How can I create a new column, based on date condition?如何根据日期条件创建新列？
【发布时间】：2022-11-12 03:44:56
【问题描述】：

我有两个数据框。第一个遵循以下结构：

                   time transactID itemID order salesPrice        day
 1: 2018-01-01 00:01:56    2278968    450     1      17.42 2018-01-01
 2: 2018-01-01 00:01:56    2278968     83     1       5.19 2018-01-01
 3: 2018-01-01 00:07:11    2255797   7851     2      20.47 2018-01-01
 4: 2018-01-01 00:09:24    2278968    450     1      17.42 2018-01-01
 5: 2018-01-01 00:09:24    2278968     83     1       5.19 2018-01-01
 6: 2018-01-01 00:39:26    2257125   9375     1      31.02 2018-01-01
 7: 2018-01-01 00:51:59    2278968    450     1      17.42 2018-01-01
 8: 2018-01-01 00:51:59    2278968     83     1       5.19 2018-01-01
 9: 2018-01-01 00:51:59    2278968     19     1      77.64 2018-01-01
10: 2018-01-01 00:51:59    2278968    297     1      43.53 2018-01-01
...

第二个具有这种结构：

   day       
   <date>    
 1 NA        
 2 2018-07-13
 3 2018-07-01
 4 2018-07-02
 5 2018-07-05
 6 2018-07-11
 7 2018-07-04
 8 2018-07-08
 9 2018-07-09
10 2018-06-30
11 2018-07-06
12 2018-07-10
13 2018-07-03
14 2018-07-12
15 2018-07-07

我的目标是创建一个名为 prom 的新列，用于标记匹配的日期。例如，我在第一列和第二列之间有一个匹配项天2018-07-13，那么它应该在当天的新列中标记，例如用星号 (*)。不匹配的行应该不填充""。

我尝试的是这样的（没有工作，只是为了我想要完成的想法）：

df1$prom <- ifelse(df1$day %in% df2$day, "*","")

复制代码（不包括匹配日期）：

第一个df

structure(list(time = structure(c(1514764916, 1514764916, 1514765231, 
                                  1514765364, 1514765364, 1514767166, 1514767919, 1514767919, 1514767919, 
                                  1514767919, 1514767919, 1514767919, 1514768104, 1514768214, 1514768214, 
                                  1514768214, 1514768214, 1514768214, 1514768214, 1514770106), class = c("POSIXct", 
                                                                                                         "POSIXt"), tzone = "UTC"), transactID = c(2278968L, 2278968L, 
                                                                                                                                                   2255797L, 2278968L, 2278968L, 2257125L, 2278968L, 2278968L, 2278968L, 
                                                                                                                                                   2278968L, 2278968L, 2278968L, 2255111L, 2278968L, 2278968L, 2278968L, 
                                                                                                                                                   2278968L, 2278968L, 2278968L, 2255111L), itemID = c(450L, 83L, 
                                                                                                                                                                                                       7851L, 450L, 83L, 9375L, 450L, 83L, 19L, 297L, 295L, 109L, 2049L, 
                                                                                                                                                                                                       19L, 83L, 295L, 297L, 450L, 109L, 4322L), order = c(1L, 1L, 2L, 
                                                                                                                                                                                                                                                           1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                                                                                                                                                                                                                                                           1L), salesPrice = c(17.42, 5.19, 20.47, 17.42, 5.19, 31.02, 17.42, 
                                                                                                                                                                                                                                                                               5.19, 77.64, 43.53, 37.79, 5.8, 35.75, 77.64, 5.19, 37.79, 43.53, 
                                                                                                                                                                                                                                                                               17.42, 5.8, 22.86), day = structure(c(17532, 17532, 17532, 17532, 
                                                                                                                                                                                                                                                                                                                     17532, 17532, 17532, 17532, 17532, 17532, 17532, 17532, 17532, 
                                                                                                                                                                                                                                                                                                                     17532, 17532, 17532, 17532, 17532, 17532, 17532), class = "Date")), row.names = c(NA, 
                                                                                                                                                                                                                                                                                                                                                                                                       -20L), class = c("data.table", "data.frame"))

第二个df

structure(list(day = structure(c(NA, 17725, 17713, 17714, 17717, 
                                 17723, 17716, 17720, 17721, 17712, 17718, 17722, 17715, 17724, 
                                 17719), class = "Date")), row.names = c(NA, -15L), class = c("tbl_df", 
                                                                                              "tbl", "data.frame"))

【问题讨论】：

标签： r

【解决方案1】：

不确定这是否是您要查找的内容。你能告诉我们你期望的结果如何吗？

df2$prom <- "*"

library(tidyverse)
left_join(df1, df2) %>%
  mutate(prom = replace_na(prom, ""))

【讨论】：