'Using dplyr to conditionally replace values in a column

I have an example data set with a column that reads somewhat like this:

Candy
Sanitizer
Candy
Water
Cake
Candy
Ice Cream
Gum
Candy
Coffee

What I'd like to do is replace it into just two factors - "Candy" and "Non-Candy". I can do this with Python/Pandas, but can't seem to figure out a dplyr based solution. Thank you!

r dplyr

Solution 1:^[1]

Assuming your data frame is dat and your column is var:

dat = dat %>% mutate(candy.flag = factor(ifelse(var == "Candy", "Candy", "Non-Candy")))

Solution 2:^[2]

In dplyr and tidyr

dat %>% 
    mutate(var = replace(var, var != "Candy", "Not Candy"))

Significantly faster than the ifelse approaches. Code to create the initial dataframe can be as below:

library(dplyr)
dat <- as_data_frame(c("Candy","Sanitizer","Candy","Water","Cake","Candy","Ice Cream","Gum","Candy","Coffee"))
colnames(dat) <- "var"

Solution 3:^[3]

Another solution with dplyr using case_when:

dat %>%
    mutate(var = case_when(var == 'Candy' ~ 'Candy',
                           TRUE ~ 'Non-Candy'))

The syntax for case_when is condition ~ value to replace. Documentation here.

Probably less efficient than the solution using replace, but an advantage is that multiple replacements could be performed in a single command while still being nicely readable, i.e. replacing to produce three levels:

dat %>%
    mutate(var = case_when(var == 'Candy' ~ 'Candy',
                           var == 'Water' ~ 'Water',
                           TRUE ~ 'Neither-Water-Nor-Candy'))

Solution 4:^[4]

No need for dplyr. Assuming var is stored as a factor already:

non_c <- setdiff(levels(dat$var), "Candy")

levels(dat$var) <- list(Candy = "Candy", "Non-Candy" = non_c)

See ?levels.

This is much more efficient than the ifelse approach, which is bound to be slow:

library(microbenchmark)
set.seed(01239)
smp <- data.frame(sample(dat$var, 1e6, TRUE))
names(smp) <- "var"

times <- 
  replicate(50, 
            {cop <- smp
            s <- get_nanotime()
            levs <- setdiff(levels(cop$var), "Candy")
            levels(cop$var) <- list(Candy = "Candy", "Non-Candy" = levs)
            d1 <- get_nanotime() - s
            cop <- smp
            s <- get_nanotime()
            cop = cop %>%
              mutate(candy.flag = factor(ifelse(var == "Candy", 
                                                "Candy", "Non-Candy")))
            d2 <- get_nanotime() - s
            cop <- smp
            s <- get_nanotime()
            cop$var <- 
              factor(cop$var == "Candy", labels = c("Non-Candy", "Candy"))
            d3 <- get_nanotime() - s
            c(levels = d1, dplyr = d2, direct = d3)})

(x <- apply(times, 1, median))[2]/x[1]
#    dplyr   direct 
# 8.894303 4.962791

That is, this is 9 times faster.

Solution 5:^[5]

I didn't benchmark this, but at least in some cases with more than one condition, a combination of mutate and a list seems to provide an easy solution:

# assuming that all sweet things fall in one category

dat <- data.frame(var = c("Candy", "Sanitizer", "Candy", "Water", "Cake", "Candy", "Ice Cream", "Gum", "Candy", "Coffee"))

conditions <- list("Candy" = TRUE, "Sanitizer" = FALSE, "Water" = FALSE, 
"Cake" = TRUE, "Ice Cream" = TRUE, "Gum" = TRUE, "Coffee" = FALSE)

dat %>% mutate(sweet = conditions[var])

Solution 6:^[6]

When you only need two values, a simple ifelse() is prettiet, I think.

Furthermore, embedded ifelses can simulate the same situation as the case_when solution proposed by PhJ (I do like his readability, though)!

dat %>%
    mutate(
        var = ifelse(var == "Candy", "Candy", "Non-Candy")
    )

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1
Solution 2
Solution 3	PhJ
Solution 4	Community
Solution 5	ulrich
Solution 6	ZKA

'Using dplyr to conditionally replace values in a column

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Solution 4:[4]

Solution 5:[5]

Solution 6:[6]