Apply a function to dataframe subsetted by all possible combinations of categorical variables

Question

An example dataframe with categorical variables catA, catB, and catC. Obs is some observed value.

catA <- rep(factor(c(\"a\",\"b\",\"c\")), length.out=10         


        

Answer 1

Using only vectorized functions and base R

# Find all possible subsets of your data
combVars <- c("catA", "catB", "catC")
subsets <- lapply(0:length(combVars), combn, x = combVars, simplify = FALSE)
subsets <- do.call(c, subsets)
# Calculate means by each subset
meanValues <- lapply(subsets, function(x) aggregate(dat[["obs"]], by = dat[x], FUN = mean))
# Pull them all into one dataframe
Reduce(function(x,y) merge(x,y,all=TRUE), meanValues)

Answer 2

This isn't the cleanest solution, but I think it gets close to what you want.

getAllSubs <- function(df, lookup, fun) {

  out <- lapply(1:nrow(lookup), function(i) {

    df_new <- df

    if(length(na.omit(unlist(lookup[i,]))) > 0) {

      for(j in colnames(lookup)[which(!is.na(unlist(lookup[i,])))]) {
        df_new <- df_new[df_new[,j] == lookup[i,j],]
      }  
    } 
    fun(df_new)  
  })

  if(mean(sapply(out, length) ==1) == 1) {
    out <- unlist(out)
  } else {
    out <- do.call("rbind", out)
  }

  final <- cbind(lookup, out)
  final[is.na(final)] <- NA
  final
}

As it is currently written you have to construct the lookup table beforehand, but you could just as easily move that construction into the function itself. I added a few lines at the end to make sure it could accomodate outputs of different lengths and so NaNs were turned into NAs, just because that seemed to create a cleaner output. As it is currently written, it applies the function to the entire original data frame in cases where all columns are NA.

dat_out <- getAllSubs(dat, allsubs, function(x) mean(x$obs, na.rm = TRUE))

head(dat_out,20)

   catA catB catC      out
1  <NA> <NA> <NA> 47.25446
2     a <NA> <NA> 51.54226
3     b <NA> <NA> 46.45352
4     c <NA> <NA> 43.63767
5  <NA>    1 <NA> 47.23872
6     a    1 <NA> 66.59281
7     b    1 <NA> 32.03513
8     c    1 <NA> 40.66896
9  <NA>    2 <NA> 45.16588
10    a    2 <NA> 50.59323
11    b    2 <NA> 51.02013
12    c    2 <NA> 33.15251
13 <NA>    3 <NA> 51.67809
14    a    3 <NA> 48.13645
15    b    3 <NA> 57.92084
16    c    3 <NA> 49.27710
17 <NA>    4 <NA> 44.93515
18    a    4 <NA> 40.36266
19    b    4 <NA> 44.26717
20    c    4 <NA> 50.74718

Answer 3

ans <- with(dat, tapply(obs, list(catA, catB, catC), mean))
ans <- data.frame(expand.grid(dimnames(ans)), results=c(ans))
names(ans)[1:3] <- names(dat)[1:3]

str(ans)
# 'data.frame':  36 obs. of  4 variables:
#  $ catA   : Factor w/ 3 levels "a","b","c": 1 2 3 1 2 3 1 2 3 1 ...
#  $ catB   : Factor w/ 4 levels "1","2","3","4": 1 1 1 2 2 2 3 3 3 4 ...
#  $ catC   : Factor w/ 3 levels "d","e","f": 1 1 1 1 1 1 1 1 1 1 ...
#  $ results: num  69.7 NA NA 55.3 NA ...

Answer 4

An alternative approach, one function to get all combinations of variables and another to apply a function over all subsets. The combinations function was stolen from another post...

## return all combinations of vector up to maximum length n
multicombn <- function(dat, n) {
    unlist(lapply(1:n, function(x) combn(dat, x, simplify=F)), recursive=F)
}

For allsubs, vars is of form c("catA","catB","catC"), out.name = "mean". func needs to be written in form that ddply would take,

func=function(x) mean(x$obs, na.rm=TRUE)

library(plyr)
allsubs <- function(indat, vars, func=NULL, out.name=NULL) {
    results <- data.frame()
    nvars <- rev(multicombn(vars,length(vars)))
    for(i in 1:length(nvars)) {
        results <-
            rbind.fill(results, ddply(indat, unlist(nvars[i]), func))
    }
    if(!missing(out.name)) names(results)[length(vars)+1] <- out.name
    results
}

One difference between this answer and shwaund's, this does not return rows for empty subsets, so no NAs in results column.

allsubs(dat, c("catA","catB","catc"), func, out.name="mean")
> head(allsubs(dat, vars, func, out.name = "mean"),20)
   catA catB catC     mean
1     a    1    d 56.65909
2     a    2    d 54.98116
3     a    3    d 37.52655
4     a    4    d 58.29034
5     b    1    e 52.88945
6     b    2    e 50.43122
7     b    3    e 52.57115
8     b    4    e 59.45348
9     c    1    f 52.41637
10    c    2    f 34.58122
11    c    3    f 46.80256
12    c    4    f 51.58668
13 <NA>    1    d 56.65909
14 <NA>    1    e 52.88945
15 <NA>    1    f 52.41637
16 <NA>    2    d 54.98116
17 <NA>    2    e 50.43122
18 <NA>    2    f 34.58122
19 <NA>    3    d 37.52655
20 <NA>    3    e 52.57115