一种方法（用NA代替.以保留数据类型）：match找到第一个NA位置，replace将该位置- 1（上一个）中的值替换为NA。apply(data, 1, \(x) ...)对每行应用该函数。最后，t转置结果（因为默认情况下，apply会将结果强制转换为列。

data = data.frame(id = 1:6, X1 = c(5L, 4L, 7L, 8L, 1L, 5L), X2 = c(5L, 
NA, 1L, 2L, 5L, 7L), X3 = c(6L, NA, 3L, 4L, NA, 7L), X4 = c(NA, 
NA, 5L, 2L, NA, 7L), X_final = c(6L, 4L, 5L, 2L, 5L, 7L), X_final_location = c("X3", 
"X1", "X4", "X4", "X2", "X4"))

data[,2:5] <- t(apply(data[,2:5], 1 , function(x) replace(x, match(NA, x) - 1, NA)))

data
#>   id X1 X2 X3 X4 X_final X_final_location
#> 1  1  5  5 NA NA       6               X3
#> 2  2 NA NA NA NA       4               X1
#> 3  3  7  1  3  5       5               X4
#> 4  4  8  2  4  2       2               X4
#> 5  5  1 NA NA NA       5               X2
#> 6  6  5  7  7  7       7               X4

使用拆分的另一种方法（按行分组）：

split(data, row.names(data)) <- 
   lapply(split(data, row.names(data)), \(x) replace(x, x$X_final_location, "."))

另一种方法，因为您已经有了$X_final_location中的位置。正如问题注解中提到的，如果回归分析的目标是保留数值，则首选NA值。

data_orig <- data.frame(
    id = c(1, 2, 3, 4, 5, 6),
    X1 = c(5, 4, 7, 8, 1, 5),
    X2 = c(5, NA, 1, 2, 5, 7),
    X3 = c(6, NA, 3, 4, NA, 7),
    X4 = c(NA, NA, 5, 2, NA, 7),
    X_final = c(6, 4, 5, 2, 5, 7),
    X_final_location = c("X3", "X1", "X4", "X4", "X2", "X4")
)

data_new <- data_orig
for (i in seq_len(nrow(data_new))) {
    data_new[i, data_new$X_final_location[i]] <- NA
}

data_new

#   id X1 X2 X3 X4 X_final X_final_location
# 1  1  5  5 NA NA       6               X3
# 2  2 NA NA NA NA       4               X1
# 3  3  7  1  3 NA       5               X4
# 4  4  8  2  4 NA       2               X4
# 5  5  1 NA NA NA       5               X2
# 6  6  5  7  7 NA       7               X4