下面是一个使用Tidyverse语法的解决方案：

library(tidyverse)

df <- tribble(
  ~Person_id, ~Days, ~Outcome,
  50L,   94L,       1L,
  234L,  102L,       1L,
  555L,   50L,       0L
)

df |> 
  mutate(
    Days = if_else(Outcome == 0L, 100, Days),
    Outcome = if_else(Days > 100, 0, Outcome),
    Days = if_else(Days > 100, 100, Days)
  )
#> # A tibble: 3 × 3
#>   Person_id  Days Outcome
#>       <int> <dbl>   <dbl>
#> 1        50    94       1
#> 2       234   100       0
#> 3       555   100       0

创建于2023-03-23带有reprex v2.0.2

简单的事情怎么样，分两步来做？

data <- structure(list(Person_id = c(50L, 234L, 555L), 
                       Days = c(934L, 102L, 50L), 
                       Outcome = c(1L, 1L, 0L)), 
                  class = "data.frame", row.names = c(NA, -3L))

data$Days[data$Days > 100] <- 100
data$Outcome[data$Days >= 100] <- 0
print(data)

使用case_when的方法

library(dplyr)

dfold %>% 
  mutate(Days = case_when(Days > 100 ~ 100, 
                          Days < 100 & Outocme == 0 ~ 100, 
                          Days < 100 & Outocme == 1 ~ Days))
  Person_id Days Outocme
1        50   94       1
2       234  100       1
3       555  100       0