CODEWITHSUNDEEP
rfor-loopdata.tablelapply
Jose
Jose

Reputation: 25

How to use a named variable within a function

Assume the following dummy data frame: 

dt <- data.table(A=c("a", "a", "a", "b", "b", "b", "c", "c", "c", "d", "d", "d"), 
             B=c("e", "e", "e", "e", "e", "e", "f", "f", "f", "f", "f", "f"), 
             C=1:12, 
             D=13:24)

I'd like to calculate some stadistics (say, mean and standard deviation) per each numeric column ("C" and "D") and each time grouped by the factor columns c("A"), c("B"), and c("A", "B). In the actual data frame, I have about 40 numeric columns, 10 factor columns that group in different combinations and a large list of statistics I'd like to calculate. Based on the answer (by @thelatemail) I got from a previous question, I know I can use the code below to deal with factor groupings (by=) using a list:

groupList <- list(c("A", "B"), c("A"), c("B"))
out <- vector("list", 3)
out <- lapply(
  groupList,
  function(x) {
    dt[, .(mean=mean(C), sd=sd(C)), by=x]
  }
)

Now I'd like to go a step further and create a variable containing a list of the names of numeric columns in the data frame and use the name of that variable within the function above. I came out with the following code but unfortunately, it doesn't work. My idea is to use a loop to extract a value from measureList at each turn and place that value within the mean, sd functions. Any ideas? The loop is how I tend to think of these things but I'll be glad to get rid of it if it makes the code faster or more efficient (particularly because one of the factor columns I have has 90 levels). I'd appreciate any pointer to solve this problem! Thanks.

factorList <- list(c("A"), c("B"), c("A", "B"))
measureList <- list(c("C"), c("D"))

out <- vector("list", 2)
for(i in 1:length(measureList)){
  out[[i]] <-lapply(
    factorList,
    function(x) {
      dt[, .(mean=mean(eval(measureList[[i]])), 
             sd=sd(eval(measureList[[i]]))),
         by = x]
    }
  )
}

Upvotes: 1

Views: 109

Answers (3)

Onyambu
Onyambu

Reputation: 79338

You can use outer with a vectorized function or use Map as shown below: 

m = function(x,y)dt[, .(mean=mean(get(y)), sd=sd(get(y))), by=x]

c(outer(factorList,measureList,Vectorize(m)))

or

Map(m,rep(factorList,each=length(measureList)),measureList)

EDIT:

TO HAVE THE NAMES:

m = function(x,y)setNames(dt[, .(mean(get(y)),sd(get(y))), by=x],
                          c(head(names(dt),length(x)),paste(c("mean","sd"),y,sep="_")))

c(outer(factorList,measureList,Vectorize(m)))

Upvotes: 1

Jaap
Jaap

Reputation: 83275

Another possibility is to use the new groupingsets function from :

groupingsets(dt
             , j = lapply(.SD, function(x) list(mean(x), sd(x)))
             , by = c('A','B')
             , sets = factorList)[, type := c('mean','sd')][]

which gives:

      A    B        C        D type
 1:    a <NA>        2       14 mean
 2:    a <NA>        1        1   sd
 3:    b <NA>        5       17 mean
 4:    b <NA>        1        1   sd
 5:    c <NA>        8       20 mean
 6:    c <NA>        1        1   sd
 7:    d <NA>       11       23 mean
 8:    d <NA>        1        1   sd
 9: <NA>    e      3.5     15.5 mean
10: <NA>    e 1.870829 1.870829   sd
11: <NA>    f      9.5     21.5 mean
12: <NA>    f 1.870829 1.870829   sd
13:    a    e        2       14 mean
14:    a    e        1        1   sd
15:    b    e        5       17 mean
16:    b    e        1        1   sd
17:    c    f        8       20 mean
18:    c    f        1        1   sd
19:    d    f       11       23 mean
20:    d    f        1        1   sd

Upvotes: 2

zack
zack

Reputation: 5415

This uses dplyr and purrr, but I think it works. 

library(dplyr)
library(purrr)

combos <- expand.grid(factorList, measureList)
map2(combos[, 1],
     combos[, 2],
     ~ dt %>% group_by_at(.x) %>% summarize_at(.y, funs(mean, sd)))

Upvotes: 1

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