How to convert this 'for' loop to a vector solution

This question is related to: Convert long state names embedded with other text to two-letter state abbreviations

Following for loop code works well.

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"

Data:

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

I tried following to have a vector solution, but they do not work:

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))

How can I convert it to a vector solution (using apply etc, without using any special packages)? Thanks for your help.

Edit: output of akrun's solution:

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"

Answer1:

I doubt this can be vectorized. At best you can hide the for loop under an *apply equivalent, or using Reduce like here:

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)

It's fancy and all but it is IMHO not worth the effort: it is probably not faster than a for loop and it is much harder to understand, isn't it? There's a little too much stigma around using for in R.

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