CODEWITHSUNDEEP
rarguments
Jose
Jose

Reputation: 25

how to assign each element of a list as arguments to a function in a loop in R?

I'm new to R. I'd like to get a number of statistics on the numeric columns (say, column C) of a data frame (dt) based on the combination of factor columns (say, columns A and B). First, I want the results by grouping both columns A and B, and then the same operations by A alone and by B alone. I've written a code that looks like the one below. I have a list of the factor combinations that I'd like to test (groupList) and then for each iteration of the loop I feed an element of that list as the argument to "by". However, as surely you can see, it doesn't work. R doesn't recognize the elements of the list as arguments to the function "by". Any ideas on how to make this work? Any pointer or suggestion is welcome and appreciated.

groupList <- list(".(A, B)", "A", "B")

for(i in 1:length(groupList)){
  output <- dt[,list(mean=mean(C),
                     sd=sd(C),
                     min=min(C),
                     median=median(C),
                     max=max(C)),
               by = groupList[i]]

  Here insert code to save each output
}

Upvotes: 1

Views: 735

Answers (3)

thelatemail
thelatemail

Reputation: 93813

Your groupList can be restructured as a list of character vectors. Then you can either use lapply or the existing for loop with an added eval() to interpret the by= input properly:

set.seed(1)
dt <- data.table(A=rep(1:2,each=5), B=rep(1:5,each=2), C=1:10)

groupList <- list(c("A", "B"), c("A"), c("B"))

lapply(
  groupList,
  function(x) {
    dt[, .(mean=mean(C), sd=sd(C)), by=x]
  }
)

out <- vector("list", 3)
for(i in 1:length(groupList)){
  out[[i]] <- dt[, .(mean=mean(C), sd=sd(C)), by=eval(groupList[[i]]) ]
}

str(out)
#List of 3
# $ :Classes ‘data.table’ and 'data.frame':      6 obs. of  4 variables:
#  ..$ A   : int [1:6] 1 1 1 2 2 2
#  ..$ B   : int [1:6] 1 2 3 3 4 5
#  ..$ mean: num [1:6] 1.5 3.5 5 6 7.5 9.5
#  ..$ sd  : num [1:6] 0.707 0.707 NA NA 0.707 ...
#  ..- attr(*, ".internal.selfref")=<externalptr> 
# $ :Classes ‘data.table’ and 'data.frame':      2 obs. of  3 variables:
#  ..$ A   : int [1:2] 1 2
#  ..$ mean: num [1:2] 3 8
#  ..$ sd  : num [1:2] 1.58 1.58
#  ..- attr(*, ".internal.selfref")=<externalptr> 
# $ :Classes ‘data.table’ and 'data.frame':      5 obs. of  3 variables:
#  ..$ B   : int [1:5] 1 2 3 4 5
#  ..$ mean: num [1:5] 1.5 3.5 5.5 7.5 9.5
#  ..$ sd  : num [1:5] 0.707 0.707 0.707 0.707 0.707

Upvotes: 0

hpesoj626
hpesoj626

Reputation: 3619

For demonstration, I used the mtcars data set. Here is one way with the dplyr package.

library(dplyr)

# create a vector of functions that you need
describe <- c("mean", "sd", "min", "median", "max")

# group by the variable gear
mtcars %>%
  group_by(gear) %>%
  summarise_at(vars(mpg), describe) 

# group by the variable carb
mtcars %>%
  group_by(carb) %>%
  summarise_at(vars(mpg), describe) 

# group by both gear and carb
mtcars %>%
  group_by(gear, carb) %>%
  summarise_at(vars(mpg), describe)

Upvotes: 0

Myosotis
Myosotis

Reputation: 53

I guess aggregate function can solve your problem. Let us say you have a dataframe df contains three columns A,B,C,given as:

df<-data.frame(A=rep(letters[1:3],3),B=rep(letters[4:6],each=3),C=1:9)

If you want calculate mean of C by factor A, try:

aggregate(formula=C~A,data=df,FUN=mean)

by factor B, try:

aggregate(formula=C~B,data=df,FUN=mean)

by factor A and B, try:

aggregate(formula=C~A+B,data=df,FUN=mean)

Upvotes: 2

Related Questions