此问题涉及:Convert long state names embedded with other text to two-letter state abbreviations
下面的 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
我尝试按照以下方式获得矢量解决方案,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)?感谢您的帮助。
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"
最佳答案
我怀疑这可以矢量化。最好的情况是,您可以将 for
循环隐藏在 *apply
等效项下,或者使用 Reduce
,如下所示:
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)
它很奇特,但恕我直言,它不值得付出努力:它可能不比 for
循环快,而且更难理解,不是吗?在 R 中使用 for
有一点太多的耻辱。
关于r - 如何将此 'for' 循环转换为向量解,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25584046/