Create variables based on regular expressions with a loop in r - r

Question

I need help to create variables based on regular expressions.
This is my dataframe:
df <- data.frame(a=c("blue", "red", "yellow", "yellow", "yellow", "yellow", "red"), b=c("apple", "orange", "peach", "lemon", "pineapple", "tomato", NA))
Basically, what I want to do is this, but in one step:
regx_1 <- as.numeric(grep("^[a-z]{5}$", df$b))
regx_2 <- as.numeric(grep("^[a-z]{6,}$", df$b))
df$fruit_1 <- NA
df$fruit_1[regx_1 + 1] <- as.character(df$b[regx_1])
df$fruit_2 <- NA
df$fruit_2[regx_2 + 1] <- as.character(df$b[regx_2])
Here is my try:
regex1 <- "^[a-z]{5}$"
regex2 <- "^[a-z]{6,}$"
regex <- c(regex1, regex1)
make_non_matches_NA <- function(vec, pattern){
df[[newvariable]] <- NA
df[[newvariable]][as.numeric(grep(pattern, vec)) + 1] <- as.character(vec[as.numeric(grep(pattern, vec))])
return(newvariable)
}
df[c("fruit1", "fruit2")] <- lapply(regex, make_non_matches_NA, vec = df$b)
EDIT: Why is my approach wrong? (Please note that the actual problem is bigger, so I have to stick to an approach, where a repetition of a pattern should be avoided)
Any help is much appreciated!

Answer 1

Having numbered items in a your workspace is a good sign that they really belong
to a list, so they are formally linked and we can work with them much more easily. So let's do that first.
regex <- c("^[a-z]{5}$", "^[a-z]{6,}$")
Our core functionality is to copy a source vector, but remove elements that don't match, and leave NA in their place, so we'll make a function for that, and we'll name it explicitly so we understand intuitively what it's doing (and as will our colleagues next reader on SO ;) ) :
make_non_matches_NA <- function(vec, pattern){
# logical indices of matches
matches_lgl <- grepl(pattern, vec)
# the elements which don't match should be NA
vec[!matches_lgl] <- NA
# resulting vector should be returned
vec
}
Let's test this with first pattern
make_non_matches_NA(df$b, regex[[1]])
#> [1] apple <NA> peach lemon <NA> <NA>
#> Levels: apple lemon orange peach pineapple tomato
So far so good! now let's test it with all regex, we avoid for loops when we can generally in R because we have clearer tools like lapply(). Here I want to apply this function to all regex expressions :
lapply(regex, make_non_matches_NA, vec = df$b)
#> [[1]]
#> [1] apple <NA> peach lemon <NA> <NA>
#> Levels: apple lemon orange peach pineapple tomato
#>
#> [[2]]
#> [1] <NA> orange <NA> <NA> pineapple tomato
#> Levels: apple lemon orange peach pineapple tomato
Great, it works!
But I want this in my data.frame, not as a separate list, so I will assign this result to the relevant names in my df directly
df[c("fruit1", "fruit2")] <- lapply(regex, make_non_matches_NA, vec = df$b)
# then print my updated df
df
#> a b fruit1 fruit2
#> 1 1 apple apple <NA>
#> 2 2 orange <NA> orange
#> 3 3 peach peach <NA>
#> 4 4 lemon lemon <NA>
#> 5 5 pineapple <NA> pineapple
#> 6 6 tomato <NA> tomato
tada!

Answer 2

I don't if this qualifies as "at one step" but you could try mutate from the dplyr package:
df <- data.frame(a=c(1:6), b=c("apple", "orange", "peach", "lemon", "pineapple", "tomato"),
stringsAsFactors = FALSE)
Note that I set stringsAsFactors = FALSE inside data.frames.
dplyr::mutate(df, fruit_1 = if_else(grepl("^[a-z]{5}$", b), b, NA_character_),
fruit_2 = if_else(grepl("^[a-z]{6}$", b), b, NA_character_))
a b fruit_1 fruit_2
1 1 apple apple <NA>
2 2 orange <NA> orange
3 3 peach peach <NA>
4 4 lemon lemon <NA>
5 5 pineapple <NA> <NA>
6 6 tomato <NA> tomato