How can I replace multiple strings in a vector using variables and not hard coding the pattern using-CodePudding

I have the following vector and data frame:

v <- c(1:20)
df <- data.frame(CHR=c(1:23), val=sample(100, 23, replace=TRUE))

and I want to create the following vector:

v2 <- c(as.character(v), "X", "Y", "Z")

In my real code the length of the vector and df will differ, so I am trying to create the v2 character vector using str_replace_all like so:

v2 <- str_replace_all(as.character(df$CHR), c(as.character(length(v)   1) = "X", as.character(length(v)   2) = "Y"), as.character(length(v)   3) = "Z"))

However this throws an error "unexpected '=' in "labels <- str_replace_all(as.character(df$CHR), c(as.character(length(v) 1) =""

I thought I was using str_replace_all correctly. If I simply run:

labels <- str_replace_all(as.character(df$CHR), c("21" = "X", "22" = "Y", "23" = "Z"))

I get the correct result but I don't want to hard code.

If my code evaluates to the same string pattern, why doesn't it work? please let me know if you can help.

CodePudding user response：

We may use setNames to create the named vector instead of passing an expression on the lhs of =

library(stringr)
str_replace_all(as.character(df$CHR), 
       setNames(c("X", "Y", "Z"), length(v)   1:3))

-output

[1] "1"  "2"  "3"  "4"  "5"  "6"  "7"  "8"  "9"  "10" "11" "12" "13" "14" "15" "16" "17" "18" "19" "20" "X" 
[22] "Y"  "Z"

If we want to use expression on the lhs, use := operator

library(dplyr)
dplyr::lst(!!as.character(length(v)   1) := "X", 
    !!as.character(length(v)   2) := "Y", !!as.character(length(v)   3) := "Z")
$`21`
[1] "X"

$`22`
[1] "Y"

$`23`
[1] "Z"

and then use it as

str_replace_all(as.character(df$CHR), 
   unlist(dplyr::lst(!!as.character(length(v)   1) := "X", 
   !!as.character(length(v)   2) := "Y",
   !!as.character(length(v)   3) := "Z")))
 [1] "1"  "2"  "3"  "4"  "5"  "6"  "7"  "8"  "9"  "10" "11" "12" "13" "14" "15" "16" "17" "18" "19" "20" "X"  "Y"  "Z"

CodePudding user response：

Update: without hard coding:

v2 <- c(intersect(v, df$CHR), LETTERS[(dim(df)[1] 1):26])

First answer: Maybe this one?

v2 <- c(intersect(v, df$CHR), LETTERS[24:26])

 [1] "1"  "2"  "3"  "4"  "5"  "6"  "7"  "8" 
 [9] "9"  "10" "11" "12" "13" "14" "15" "16"
[17] "17" "18" "19" "20" "X"  "Y"  "Z"