In a pipe chain, I want to use a named vector to create a new column which matches the names of the vector with the string of a column:
library(tidyverse)
df <- data.frame(my_label = c("car", "house", "Bike", "ca"),
xx = c(1, 2, 3, 5))
# my_label xx
# 1 car 1
# 2 house 2
# 3 Bike 3
# 4 ca 5
named_vars <- c(
"car" = "Nice car",
"ca" = "Cat",
"house" = "Large house")
# car ca house
# "Nice car" "Cat" "Large house"
The following code works if the named vector contains all the strings within the column, which it doesn't in this case so it returns an NA (if it is missing I want to keep the original (Bike in this example):
df %>%
mutate(new = named_vars[my_label])
# my_label xx new
# 1 car 1 Nice car
# 2 house 2 Large house
# 3 Bike 3 <NA>
# 4 ca 5 Cat
As a workaround, this produces the output I want:
df %>%
mutate(new = ifelse(my_label %in% names(named_vars), named_vars[my_label], my_label))
# my_label xx new
# 1 car 1 Nice car
# 2 house 2 Large house
# 3 Bike 3 Bike
# 4 ca 5 Cat
I am wondering is there a shorter way to write this?
I tried stringr::str_replace_all but it combines ca and car to give incorrect output (Nice Catr instead of Nice car):
library(stringr)
df %>%
mutate(new = str_replace_all(my_label, named_vars))
# my_label xx new
# 1 car 1 Nice Catr
# 2 house 2 Large house
# 3 Bike 3 Bike
# 4 ca 5 Cat
Any suggestions? thanks
Use coalesce:
df %>%
mutate(new = coalesce(named_vars[my_label], my_label))
# my_label xx new
# 1 car 1 Nice car
# 2 house 2 Large house
# 3 Bike 3 Bike
# 4 ca 5 Cat
These two statements are functionality equivalent here:
coalesce(named_vars[my_label], my_label)
if_else(is.na(named_vars[my_label]), my_label, named_vars[my_label])