Subsetting string vector by word count in one line

user_5905425 0
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I have a vector of strings

rownames
[1] "multifarmacias descuento" "multifarmacias"           "multifarmacias"

My goal is to subset rownames in one line by strings that only contain one word- the output would be

[1] "multifarmacias"           "multifarmacias"

I have tried the following but it throws an error:

rownames[which(sapply(strsplit(rownames, " "),length)) == 1]

Error in which(sapply(strsplit(rownames, " "), length)) : 
  argument to 'which' is not logical

Is there an elegant solution to subsetting a string vector by length of words the string?

stringr sapply subset r string
3 Answers

It would be easier with str_count

library(stringr)
rownames[str_count(rownames, "\\w+") == 1]
#[1] "multifarmacias" "multifarmacias"

If we use strsplit with lengths (from base R) would be more efficient

rownames[lengths(strsplit(rownames, "\\s+")) == 1]
#[1] "multifarmacias" "multifarmacias"

The error in OP's post is based on the wrong placement of ), It should be after the ==1 because which is applied directly on a vector of lengths i.e.

 which(c(2, 1, 1))

Error in which(c(2, 1, 1)) : argument to 'which' is not logical

data

rownames <- c("multifarmacias descuento", "multifarmacias", "multifarmacias")
user_37322710

A base R solution would be:

#Data
x <- c("multifarmacias descuento", "multifarmacias", "multifarmacias")

#Code
x[which(lapply(strsplit(x,split = ' '),length)==1)]

Output:

[1] "multifarmacias" "multifarmacias"
user_20808480

Here is another base R option using grep

grep("\\s",rownames,invert = TRUE,value = TRUE)

which gives

> grep("\\s",rownames,invert = TRUE,value = TRUE)
[1] "multifarmacias" "multifarmacias"
user_121587570
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