CODEWITHSUNDEEP
rdplyr
Jeremy K.
Jeremy K.

Reputation: 1792

create new columns with mutate_all

I have some data that looks like this:

  Year Revenue Cost Rent
1 2016    3000    4  100
2 2017    4000    5  100
3 2018    5000    6  100

df <- data.frame(stringsAsFactors=FALSE,
        Year = c(2016L, 2017L, 2018L),
     Revenue = c(3000L, 4000L, 5000L),
        Cost = c(4L, 5L, 6L),
        Rent = c(100L, 100L, 100L)
)

I'd like to divide everything say as a percentage of Rent:

library(dplyr)
df <- df %>% mutate_at(vars(Revenue:Rent), funs(. /Rent))

which works perfectly.

  Year Revenue Cost Rent
1 2016      30 0.04    1
2 2017      40 0.05    1
3 2018      50 0.06    1

The only thing: I've lost my original columns.

How can I do the mutate_all, so that I have new columns, say called Revenue_percentage_of_rent, Cost_percentage_of_rent ?

Upvotes: 6

Views: 801

Answers (2)

akrun
akrun

Reputation: 887951

The usage of funs would be deprecated in favor of list from dplyr_0.8.0 So, the option would be

library(dplyr)
df %>%
    mutate_at(vars(Revenue:Rent), list(percentage_of_rent = ~  ./Rent))
#  Year Revenue Cost Rent Revenue_percentage_of_rent Cost_percentage_of_rent Rent_percentage_of_rent
#1 2016    3000    4  100                         30                    0.04                       1
#2 2017    4000    5  100                         40                    0.05                       1
#3 2018    5000    6  100                         50                    0.06                       1

Upvotes: 5

Ronak Shah
Ronak Shah

Reputation: 389325

Name the column in the function in mutate_at

library(dplyr)
df %>% mutate_at(vars(Revenue:Rent), funs(percentage_of_rent = . /Rent))

You can do it with mutate_all but then it will also divide the Year column by Rent which I suppose you don't need.

A workaround to use mutate_all would be

df %>% select(-Year) %>% mutate_all(funs(percentage_of_rent = . /Rent))

but you loose Year column here.

Upvotes: 2

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