Mutating multiple columns dynamically while conditioning on specific rows

Question

I know there are several similar questions around here, but none of them seems to address the precise issue I\'m having.

set.seed(4)
df = data.frame(
  Key = c(\         


        

Answer 1

With this dplyr command,

df %>% mutate_at(.vars = vars(cols), .funs = function(x) ifelse(df$Key == "A", 0, x))

You are actually evaluating the statement df$Key == "A", n times, where n=the number of columns you have.

One work around is to pre-define the rows you want to change:

idx = which(DF$Key=="A")
DF %>% mutate_at(.vars = vars(cols), .funs = function(x){x[idx]=0;x})

A cleaner and better way, correctly pointed out by @IceCreamToucan (see comments below), is to use the function replace, while passing it the extra parameters:

DF %>% mutate_at(.vars = vars(cols), replace, DF$Key == 'A', 0)

We can put all these approaches to test, and I think dplyr and data.table are comparable.

#simulate data
set.seed(100)
Key = sample(LETTERS[1:3],1000000,replace=TRUE)
DF = as.data.frame(data.frame(Key,matrix(runif(1000000*10),nrow=1000000,ncol=10)))
DT = as.data.table(DF)

cols = grep("[35789]", names(DF), value = TRUE)

#long method
system.time(DF %>% mutate_at(.vars = vars(cols), .funs = function(x) ifelse(DF$Key == "A", 0, x)))
user  system elapsed 
  0.121   0.035   0.156 

#old base R way
system.time(DF[idx,cols] <- 0)
   user  system elapsed 
  0.085   0.021   0.106 

#dplyr
# define function
func = function(){
       idx = which(DF$Key=="A")
       DF %>% mutate_at(.vars = vars(cols), .funs = function(x){x[idx]=0;x})
}
system.time(func())
user  system elapsed 
  0.020   0.006   0.026

#data.table
system.time(DT[Key=="A", (cols) := 0])
   user  system elapsed 
  0.012   0.001   0.013 
#replace with dplyr
system.time(DF %>% mutate_at(.vars = vars(cols), replace, DF$Key == 'A', 0))
user  system elapsed 
  0.007   0.001   0.008