How do you map every combination of categorical variables in R?

Question

Using R, is there a way to take a data set and map out every possible combination of every categorical variable?

For example, let\'s say I had 10,000 rows of custom

Answer 1

That there is:

expand.grid(gender = c("male", "female"), tShirtSize = c("xs", "s","m","l","xl"))

Will return all the combinations in a dataframe. For the summary stats, try aggregate, e.g:

country = sample(c("america", "canadian"), 30, replace = TRUE)
gender = sample(c("male", "female"), 30, replace = TRUE)
x = abs(rnorm(30) * 1000)
aggregate(data.frame(x), by = list(country, gender), FUN = mean)

I run into errors if there are columns with strings in the dataframe, so I'd subset out the columns with numeric values.

Answer 2

Create some dummy data:

dataset <- data.frame(
    spend=10*runif(100),
    email=sample(c("yahoo","gmail","hotmail","aol"),100,replace=TRUE),
    browser=sample(c("Mozilla","IE","Chrome","Opera"),100,replace=TRUE),
    country=sample(c("USA","Canada","China","Australia",
      "Egypt","S.Korea","Brazil"),100,replace=TRUE))

Average the spend per combination:

with(dataset,by(spend,list(email,browser,country),mean))

Note the NAs for combinations without entries.

Or turn this into a three-dimensional array:

as.table(with(dataset,by(spend,list(email,browser,country),mean)))

Answer 3

Here's a method that utilizes dplyr

require(magrittr)
require(dplyr)    

set.seed(123)
dat = data.frame(email=sample(c("yahoo", "gmail"), 10000, replace=T),
                 browser=sample(c("mozilla", "ie"), 10000, replace=T),
                 country=sample(c("usa", "canada"), 10000, replace=T),
                 money=runif(10000))  
dat %>%
  group_by(email, browser, country) %>%
  summarize(mean = mean(money))
# email browser country      mean
# 1 gmail      ie  canada 0.5172424
# 2 gmail      ie     usa 0.4921908
# 3 gmail mozilla  canada 0.4934892
# 4 gmail mozilla     usa 0.4993923
# 5 yahoo      ie  canada 0.5013214
# 6 yahoo      ie     usa 0.5098280
# 7 yahoo mozilla  canada 0.4985357
# 8 yahoo mozilla     usa 0.4919743

EDIT: if you want to pass a list into group_by(), you'll need to use the not-non-standard evaluation counterpart, regroup(). For example,

mylist <- list("email", "browser", "country")
dat %>%
  regroup(mylist) %>%
  summarize(mean = mean(money))

also see dplyr: How to use group_by inside a function?

Answer 4

You can do this with aggregate:

set.seed(144)
dat = data.frame(email=sample(c("yahoo", "gmail"), 10000, replace=T),
                 browser=sample(c("mozilla", "ie"), 10000, replace=T),
                 country=sample(c("usa", "canada"), 10000, replace=T),
                 money=runif(10000))
aggregate(dat$money, by=list(browser=dat$browser, email=dat$email,
                             country=dat$country), mean)
#   browser email country         x
# 1      ie gmail  canada 0.4905588
# 2 mozilla gmail  canada 0.5064342
# 3      ie yahoo  canada 0.4894398
# 4 mozilla yahoo  canada 0.4959031
# 5      ie gmail     usa 0.5069363
# 6 mozilla gmail     usa 0.5088138
# 7      ie yahoo     usa 0.4957478
# 8 mozilla yahoo     usa 0.4993698

To get multiple columns like mean and count together, you can do:

res = aggregate(dat$money, by=list(browser=dat$browser, email=dat$email,
                                   country=dat$country),
                FUN=function(x) c(mean=mean(x), count=length(x)))
res
#   browser email country       x.mean      x.count
# 1      ie gmail  canada    0.4905588 1261.0000000
# 2 mozilla gmail  canada    0.5064342 1227.0000000
# 3      ie yahoo  canada    0.4894398 1267.0000000
# 4 mozilla yahoo  canada    0.4959031 1253.0000000
# 5      ie gmail     usa    0.5069363 1240.0000000
# 6 mozilla gmail     usa    0.5088138 1236.0000000
# 7      ie yahoo     usa    0.4957478 1213.0000000
# 8 mozilla yahoo     usa    0.4993698 1303.0000000