Question

此问题与：Convert long state names embedded with other text to two-letter state abbreviations

有关

以下for循环代码效果很好。

for(r in 1:nrow(states.list)) {
    states = sub(states.list[r,1], states.list[r,2], states)
}

states
[1] "Plano NJ"      "NC"            "xyz"           "AL 02138"      "TX"            "Town IA 99999"

数据：

states <- c("Plano New Jersey", "NC", "xyz", "Alabama 02138", "Texas", "Town Iowa 99999")

states.list = structure(list(state.name = structure(c(4L, 1L, 5L, 2L, 3L), .Label = c("Alabama", 
"Iowa", "Minnesota", "New Jersey", "Texas"), class = "factor"), 
    state.abb = structure(c(4L, 1L, 5L, 2L, 3L), .Label = c("AL", 
    "IA", "MN", "NJ", "TX"), class = "factor")), .Names = c("state.name", 
"state.abb"), class = "data.frame", row.names = c(NA, -5L))

states.list
  state.name state.abb
1 New Jersey        NJ
2    Alabama        AL
3      Texas        TX
4       Iowa        IA
5  Minnesota        MN

我尝试过使用矢量解决方案，但它们不起作用：

apply(states.list, 1, function(x) {
    sapply(states, function(y) {
        sub(   x[1], x[2],   y
        )
        })
})

sapply(states, function(x) sub(states.list[,1], states.list[,2], x))

apply(states.list, 1, function(x) sub(x[1],x[2], states))

如何将其转换为矢量解决方案（使用apply等，而不使用任何特殊包）？谢谢你的帮助。

编辑： akrun解决方案的输出：

sapply ( seq_len(nrow(states.list)), function(i) {
+ sub(states.list[i,1], states.list[i,2], states[i])
+ })
[1] "Plano NJ"      "NC"            "xyz"           "Alabama 02138" "Texas"

Answer 1

我怀疑这可以被矢量化。最多可以将for循环隐藏在*apply等效项下，或者使用Reduce，例如：

ARGS <- split(states.list, seq_len(nrow(states.list)))
FUN  <- function(x, y) gsub(as.character(y$state.name),
                            as.character(y$state.abb), x)
Reduce(FUN, ARGS, states)

这很奇特，除了它是恕我直言不值得的努力：它可能不比for循环更快，它更难理解，不是吗？在R中使用for时有点太多的耻辱。

如何转换这个＆＃39; for＆＃39;循环到矢量解决方案

1 个答案: