CODEWITHSUNDEEP
rdplyrrename
Lynn
Lynn

Reputation: 4408

Add space between a character in R

Background

I have a dataset, df. Whenever I try and rename the 'TanishaIsCool' column, I get an error: unexpected string constant. I wish to add spaces within my column name

TanishaIsCool   Hello
hi              hi

This is what I am doing:

df1 <- df %>% rename(Tanisha Is Cool = `TanishaIsCool` )

Desired output

 Tanisha Is Cool     Hello
 hi                  hi

dput

structure(list(TanishaIsCool = structure(1L, .Label = "hi", class = "factor"), 
Hello = structure(1L, .Label = "hi", class = "factor")), class = "data.frame", row.names = c(NA, 
-1L))

Upvotes: 2

Views: 2817

Answers (2)

Nadia
Nadia

Reputation: 61

Your attempt was nearly there, except missing the backquotes/backticks:

df1 %>% rename(`Tanisha Is Cool` = TanishaIsCool)

However, I believe you will find that most recommendations (and I agree completely after my own personal experience of struggling with one particular dataset...), state not to use spaces in your variable names, since you might find that when you have to reference these variables, you will have to always include the `` , which can get pretty cumbersome.

Just realised @thelatemail has answered exactly this in the comment!

Upvotes: 3

akrun
akrun

Reputation: 887991

We can use gsub to capture the lower case letter (([a-z])), then the upper case (([A-Z])) and in the replacement, use the backreference of the captured groups (\\1,\2`) and create space between them

colnames(df1) <- gsub("([a-z])([A-Z])", "\\1 \\2", colnames(df1))
df1
#  Tanisha Is Cool Hello
#1              hi    hi

With tidyverse and option is

library(dplyr)
library(stringr)
library(magrittr)
df1 %<>%
    rename_all(~ str_replace_all(., "([a-z])([A-Z])", "\\1 \\2"))

For selected columns, use rename_at

df1 %<>%
    rename_at(1, ~ str_replace_all(., "([a-z])([A-Z])", "\\1 \\2"))

Another option is regex lookaround

gsub("(?<=[a-z])(?=[A-Z])", " ", names(df1), perl = TRUE)
#[1] "Tanisha Is Cool" "Hello"

If we need to update only selected column names, then use an index, hre it is the first column

names(df1)[1] <- gsub("(?<=[a-z])(?=[A-Z])", " ", names(df1)[1], perl = TRUE)

Upvotes: 1

Related Questions