Create a list concatenating a list element and character string in R

Question

I am trying to create a list of requests called requests. I have a list of hundreds of products, but I simplify for this example code.

products <- list()
products[[1]] <- c("77295T", "67619Q", "9528QR")
products[[2]] <- c("B7295T", "676193")
requests <- vector("list",length(products[[1]]))
length(requests)
i <- 1
for (i in 1:(length(products[[1]])))  {
    requests[[i]] <- cat('"',noquote(products[[1]][[i]]),'~XPORT"', sep = '')  
   }

results in:

"77295T~XPORT""67619Q~XPORT""9528QR~XPORT"

The end result I am looking for is a request list of length 3 with the following three elements:

requests[[1]] 
"77295T~XPORT"

requests[[2]]
"67619Q~XPORT"

requests[[3]]
"9528QR~XPORT"

Background: Eventually I want requests to be a nested list (a list of lists); in my dataset products is a list, so that is why I provide products[[2]], even though my question relates to just iterating over products[[1]]. Replacingproducts[[1]]` with an equivalent products1 also creates the same output. That code is here:

products1 <- c("77295T", "67619Q", "9528QR")
i <- 1
for (i in 1:(length(products1))) {
  requests[[i]] <- cat('"',noquote(products1[[i]]),'~XPORT"', sep = '')  
  }

This more simple code provides the same results:

"77295T~XPORT""67619Q~XPORT""9528QR~XPORT"

I do realize that the for loop I called may not be necessary. I am trying to gradually learn how to "vectorize" my R code. Any suggestion for further vector-elegance in this code would be appreciated.

Probably requests <- as.list(paste(products[[1]], "XPORT", sep="~")) is the simplest answer. — lmo
– lmo, Commented Jun 1, 2017 at 14:25
lapply(products[[1]], function(x){ paste0(x, "~XPORT")}) should also work — KenHBS
– KenHBS, Commented Jun 1, 2017 at 14:27
@lmo you mean, this code without the loop, right? That does the trick. — mrlcpa
– mrlcpa, Commented Jun 1, 2017 at 14:38
Yes. A loop is not needed with that code. In this instance, paste does the work under the hood. — lmo
– lmo, Commented Jun 1, 2017 at 14:40

KenHBS · Accepted Answer · 2017-06-01 15:31:07Z

3

I guess what you are looking for is this:

requests <- lapply(products[[1]], function(x){paste0(x, "~XPORT")})
requests
# [[1]]
# [1] "77295T~XPORT"
#
# [[2]]
# [1] "67619Q~XPORT"
#
# [[3]]
# [1] "9528QR~XPORT"

And if you wish to do this to all vectors in products, I would use a loop:

request <- list()
for(i in 1:length(products)){
   request[[i]] <- lapply(products[[i]], function(x){paste0(x, "~XPORT")})
}

I'm sure this is just one of many ways to accomplish this.

Edit

An alternative to the above without any loops, would be to use sapply instead of for

sapply(products, function(y){
  lapply(y, function(x) paste0(x, "~XPORT"))
})

edited Jun 1, 2017 at 15:31

answered Jun 1, 2017 at 14:33

KenHBS

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3 Comments

mrlcpa Over a year ago

Both parts of this code works. Thank you. I am wondering if there is a way to avoid the second loop. I tried req4 <- lapply(prods, function(x){paste0(x, "~XPORT")}), but that doesn't generate a nested list, just a list of two, where the first looks like this: "77295T~XPORT" "67619Q~XPORT" "9528QR~XPORT"

KenHBS Over a year ago

@mrlcpa I just added an alternative that works without using a loop

mrlcpa Over a year ago

The alternative without loops works, thank you. When I replace sapply with lapply, the resulting all.equal(request1,request2) brings back TRUE. So I guess in this case they function the same.

S.Bhattarai · Accepted Answer · 2017-06-01 15:28:40Z

0

As suggested by Ken ,to get the nested list without using the loop, as.list() might be helpful on this case.

products <- list()
products[[1]] <- c("77295T", "67619Q", "9528QR")
products[[2]] <- c("B7295T", "676193")
requests <- lapply(products, function(x){ as.list(paste0(x, "~XPORT"))})
requests

[[1]]
[[1]][[1]]
[1] "77295T~XPORT"

[[1]][[2]]
[1] "67619Q~XPORT"

[[1]][[3]]
[1] "9528QR~XPORT"


[[2]]
[[2]][[1]]
[1] "B7295T~XPORT"

[[2]][[2]]
[1] "676193~XPORT"

answered Jun 1, 2017 at 15:28

S.Bhattarai

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