R语言中如何基于m列与date列的耦合关联生成对应date4列

实现方案 #

方案1：同步生成m4和date4（推荐） #

直接复用你原有case_when的分支逻辑，在生成m4的同时定义date4的取值，从根源上保证耦合对应关系，不会出现匹配错误：

library(tidyverse)
library(lubridate)

df <- data.frame(ID = c('1','2','3'),
                 diff2.1 = c(28, 91, 90),
                 diff3.1 = c(140, 92, 91),
                 date1 = ymd(c('2014-10-12', '2014-06-01', '2015-02-10')),
                 date2 = as.Date(c('2015-10-17', '2014-06-06', '2015-04-20')),
                 date3 = as.Date(c('', '2014-08-13', '2015-12-01')),
                 m1 = c('4','4','10'),
                 m2 = c('3', '0', ' 1'),
                 m3 = c('10', '7', '5'))

df_new <- df %>%
  mutate(
    m4 = case_when(
      diff3.1 >= diff2.1 & diff3.1 < 91 ~ pmax(m1,m2,m3),
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 > 90 ~ m1,
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 < 91 ~ pmax(m1,m2)
    ),
    date4 = case_when(
      diff3.1 >= diff2.1 & diff3.1 < 91 ~ case_when(
        pmax(m1,m2,m3) == m1 ~ date1,
        pmax(m1,m2,m3) == m2 ~ date2,
        pmax(m1,m2,m3) == m3 ~ date3
      ),
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 > 90 ~ date1,
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 < 91 ~ case_when(
        pmax(m1,m2) == m1 ~ date1,
        pmax(m1,m2) == m2 ~ date2
      )
    )
  )

运行后得到的date4完全符合你给出的预期结果。

方案2：基于已有m4匹配date4 #

如果已经生成了m4，不想修改原有生成逻辑，可以用行向匹配实现：

df_new <- df %>%
  mutate(
    m4 = case_when(
      diff3.1 >= diff2.1 & diff3.1 < 91 ~ pmax(m1,m2,m3),
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 > 90 ~ m1,
      diff3.1 > diff2.1 & diff3.1 > 90 & diff2.1 < 91 ~ pmax(m1,m2)
    )
  ) %>%
  rowwise() %>%
  mutate(
    date4 = c_across(date1:date3)[which(c_across(m1:m3) == m4)[1]]
  ) %>%
  ungroup()

注意事项 #

如果存在同一行多个m列的值和m4相等的情况，方案2会取第一个匹配到的日期，这种场景优先使用方案1保证逻辑准确性。

内容的提问来源于stack exchange，提问作者Youknowme