若元素为x，前一个元素为na，则替换向量中的元素，单位为R foo

这里有一种方法

result <- c("a","a",NA,"whatever","a","a","b","b")

将其包装到函数中以便于使用：

> foo
[1] "a"        "a"        NA         "whatever" "a"       
[6] "a"        "b"        "b"

无论什么一个简单的解决方案：

> foo <- c("a","a",NA,"b","a","a","b","b")
> whatever(foo)
[1] "a"        "a"        NA         "whatever" "a"       
[6] "a"        "b"        "b"

一个完全复杂的解决方案，使用
嵌入
只是为了好玩：

library(zoo)
foo[-1][rollapply(foo, 2, identical, c(NA, "b"))] <- "whatever"

foo[which](
申请(
嵌入（foo，2），
1.
函数（x）x[1]=“b”和is.na（x[2]）
)
) + 1
]福
[1] “a”“a”“NA”“无论什么”“a”“a”“b”“b”
我正在与“上一个元素”的情况作斗争。我找不到满足这个条件的方法。哇，简单又接近我所尝试的，但是我不太明白-1在这里的用法。foo[-1]删除列表的第一个元素，这是如何工作的？如果我可以问…@Chargaff，你永远不需要替换第一个元素，因为它本来就没有前一个元素。@Chargaff评估
foo
的第一个元素没有意义，因为根据定义，它没有前一个元素。因此，
foo[-1]
贯穿始终（头（）从另一端截断），而不是
foo
@Chargaff我在答案中添加了另一个解决方案。
> foo
[1] "a"        "a"        NA         "whatever" "a"       
[6] "a"        "b"        "b"

whatever <- function(x) {
  nas <- which(is.na(x))
  bs <- which(x == "b")
  x[bs[match(nas, bs - 1)]] <- "whatever"
  x
}

> foo <- c("a","a",NA,"b","a","a","b","b")
> whatever(foo)
[1] "a"        "a"        NA         "whatever" "a"       
[6] "a"        "b"        "b"

foo[-1][foo[-1] == "b" & is.na(head(foo, -1))] <- "whatever"

library(zoo)
foo[-1][rollapply(foo, 2, identical, c(NA, "b"))] <- "whatever"

foo[which(
          apply(
                embed(foo,2),
                1,
                function(x) x[1]=="b" & is.na(x[2])
               )
         ) + 1
    ] <- "whatever"

> foo
[1] "a" "a" NA "whatever" "a" "a" "b" "b"