我们可以使用rowCumsums创建一个逻辑向量，然后进行赋值

library(matrixStats)
df[rowCumsums(rowCumsums(df == 1)) > 1] <- NA

• 输出

> df
  T1 T2 T3 T4 T5
1  0  1 NA NA NA
2  0  0  1 NA NA
3  0  0  0  0  0
4  0  0  0 NA NA
5  0  0  0  1 NA

使用apply从底数R得到which.max的第一个1
first1 = which.max(x == 1)
生成一个可排序的整数，并将不在序列中的值划分为子集。mySeq <-ifelse(first1 == 1, length(x), first1)
子集值为NA
x[-seq(mySeq)] <- NA
返回值

df[] <- t(apply(df, 1, \(x) {
  first1 = which.max(x == 1)
  mySeq <-ifelse(first1 == 1, length(x), first1)
  x[-seq(mySeq)] <- NA
  x
  }))

T1 T2 T3 T4 T5
1  0  1 NA NA NA
2  0  0  1 NA NA
3  0  0  0  0  0
4  0  0  0 NA NA
5  0  0  0  1 NA

下面是第一次透视到长格式的替代方法：

library(dplyr)
library(tidyr)

df %>% 
  pivot_longer(everything()) %>% 
  group_by(group =as.integer(gl(n(),max(parse_number(name)),n()))) %>% 
  mutate(value1 = ifelse(1:n() > which(value == 1)[1], NA, value),
         y = sum(value, na.rm = TRUE)) %>% 
  mutate(value = ifelse(y == 0, value, value1), .keep="unused") %>% 
  pivot_wider(names_from = name, values_from = value)

group    T1    T2    T3    T4    T5
  <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     0     1    NA    NA    NA
2     2     0     0     1    NA    NA
3     3     0     0     0     0     0
4     4     0     0     0    NA    NA
5     5     0     0     0     1    NA