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For example, I have a dataset named "animals". animals = read_csv("animals.csv")

It contains a column called "Species", where the value of Species can be "a","b","c". See below.

Weight Height Species <dbl> <dbl> <chr> 1 12.3 15.7 a 2 8.1 17.9 a 3 11.3 14.4 b 4 12.1 18 b 5 6.8 18.1 a 6 16.3 18.3 a 7 19.6 19.2 c 8 22.2 19 c

Now I want to rename the values of a,b,c. a is "tiger", b is "lion", c is "elephant" so it becomes below. And I need to use factor function to rename them, how can I do it please?

Weight Height Species <dbl> <dbl> <chr> 1 12.3 15.7 tiger 2 8.1 17.9 tiger 3 11.3 14.4 lion 4 12.1 18 lion 5 6.8 18.1 tiger 6 16.3 18.3 tiger 7 19.6 19.2 elephant 8 22.2 19 elephant
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3 Answers 3

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Here is a base R option using factor

transform( df, Species = c('tiger', 'lion', 'elephant')[as.integer(factor(Species,levels = c('a', 'b', 'c')))] )

which gives

 Weight Height Species 1 12.3 15.7 tiger 2 8.1 17.9 tiger 3 11.3 14.4 lion 4 12.1 18.0 lion 5 6.8 18.1 tiger 6 16.3 18.3 tiger 7 19.6 19.2 elephant 8 22.2 19.0 elephant
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If Species isn't already a factor, you can turn it into one by animals$Species = as.factor(animals$Species)

Then, you can change the levels by doing:

levels(animals$Species) = c("tiger","lion","elephant")
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2 Comments

Thanks! This is what I wanted. One more question from me. I tested using below which works as well, is there any difference between as.factor and factor? animals$Species = factor(animals$Species) levels(animals$Species) = c("tiger","lion","elephant")
Apparently as.factor woks best in some cases. If you're curious, check: stackoverflow.com/questions/39279238/….
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Here is an option with left_join

keydat <- data.frame(Species = c('a', 'b', 'c'), value = c('tiger', 'lion', 'elephant')) library(dplyr) df1 <- df1 %>% left_join(keydat) %>% mutate(Species = value, value = NULL)

-output

df1 # Weight Height Species #1 12.3 15.7 tiger #2 8.1 17.9 tiger #3 11.3 14.4 lion #4 12.1 18.0 lion #5 6.8 18.1 tiger #6 16.3 18.3 tiger #7 19.6 19.2 elephant #8 22.2 19.0 elephant

Or using a named vector to match and replace

nm1 <- setNames(c('tiger', 'lion', 'elephant'), c('a', 'b', 'c')) df1$Species <- nm1[df1$Species]

Or an option with fct_recode in case if the column is factor (should also work with character class)

library(forcats) df1$Species <- fct_recode(factor(df1$Species), !!!setNames(names(nm1), nm1) )

data

df1 <- structure(list(Weight = c(12.3, 8.1, 11.3, 12.1, 6.8, 16.3, 19.6, 22.2), Height = c(15.7, 17.9, 14.4, 18, 18.1, 18.3, 19.2, 19), Species = c("a", "a", "b", "b", "a", "a", "c", "c")), class = "data.frame", row.names = c("1", "2", "3", "4", "5", "6", "7", "8"))
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1 Comment

Brilliant using setNames, +1!

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