I have a class column in my data frame which carries the value "Bad" & "Good". I want to replace these string into 0 & 1 respectively.
I tried the following:
x = c("Bad", "Good", "Bad", "Bad", "Good")
factor(x)
factor(x, c(0, 1))
but, it converts the value in the dataset to NA
factor(x, c(0, 1))
[1] <NA> <NA> <NA> <NA> <NA>`
Levels: 0 1`
A convenient tidyverse approach would be to use dplyr's recode function.
df <- data.frame(x = c("Bad", "Good", "Bad", "Bad", "Good"))
df$x <- as.factor(df$x)
library(tidyverse)
df <- df %>%
mutate(x = recode(x,
"Bad" = "0",
"Good" = "1"))
That's assuming you want to keep it as a factor column. If you'd rather it be a numeric vector you'd simply add a second mutate call to convert it to numeric... EDIT: being careful to convert the values as numbers, not the underlying level codes.
df <- df %>%
mutate(x = recode(x,
"Bad" = "0",
"Good" = "1")) %>%
mutate(x = as.numeric(levels(x))[x])
You can use a named vector to map the text strings to numeric values.
df <- data.frame( x = c("Bad", "Good", "Bad", "Bad", "Good"), stringsAsFactors = FALSE)
lookup <- c("Bad" = 0, "Good" = 1)
df$new_x <- lookup[df$x]
Result
df
x new_x
1 Bad 0
2 Good 1
3 Bad 0
4 Bad 0
5 Good 1
# showing type info
str(df)
'data.frame': 5 obs. of 2 variables:
$ x : chr "Bad" "Good" "Bad" "Bad" ...
$ new_x: num 0 1 0 0 1
as.integer(x=="Good"). Yours doesnt work asfactorlooks for levels 0/ 1 which arent there. So be explicitfactor(x, levels=c("Bad", "Good"), labels=c(0, 1))