Cartesian product data frame

Question

I have three or more independent variables represented as R vectors, like so:

A <- c(1,2,3)
B <- factor(c('x','y'))
C <- c(0.1,0.5)

and I want to take the Cartesian product of all of them and put the result into a data frame, like this:

A B C
1 x 0.1
1 x 0.5
1 y 0.1
1 y 0.5
2 x 0.1
2 x 0.5
2 y 0.1
2 y 0.5
3 x 0.1
3 x 0.5
3 y 0.1
3 y 0.5

I can do this by manually writing out calls to rep:

d <- data.frame(A = rep(A, times=length(B)*length(C)),
                B = rep(B, times=length(A), each=length(C)),
                C = rep(C, each=length(A)*length(B))

but there must be a more elegant way to do it, yes? product in itertools does part of the job, but I can't find any way to absorb the output of an iterator and put it into a data frame. Any suggestions?

p.s. The next step in this calculation looks like

d$D <- f(d$A, d$B, d$C)

so if you know a way to do both steps at once, that would also be helpful.

Answer 1

You can use expand.grid(A, B, C)

---

EDIT: an alternative to using do.call to achieve the second part, is the function mdply from the package plyr:

library(plyr)

d = expand.grid(x = A, y = B, z = C)
d = mdply(d, f)

To illustrate its usage using a trivial function 'paste', you can try

d = mdply(d, 'paste', sep = '+');

Answer 2

There's a function manipulating dataframe, which is helpful in this case.

It can produce various join(in SQL terminology), while Cartesian product is a special case.

You have to convert the varibles to data frames first, because it take data frame as parameters.

so something like this will do:

A.B=merge(data.frame(A=A), data.frame(B=B),by=NULL);
A.B.C=merge(A.B, data.frame(C=C),by=NULL);

The only thing to care about is that rows are not sorted as you depicted. You may sort them manually as you wish.

merge(x, y, by = intersect(names(x), names(y)),
      by.x = by, by.y = by, all = FALSE, all.x = all, all.y = all,
      sort = TRUE, suffixes = c(".x",".y"),
      incomparables = NULL, ...)

"If by or both by.x and by.y are of length 0 (a length zero vector or NULL), the result, r, is the Cartesian product of x and y"

see this url for detail: http://stat.ethz.ch/R-manual/R-patched/library/base/html/merge.html

Answer 3

With library tidyr one can use tidyr::crossing (order will be as in OP):

library(tidyr)
crossing(A,B,C)
# A tibble: 12 x 3
#        A B         C
#    <dbl> <fct> <dbl>
#  1     1 x       0.1
#  2     1 x       0.5
#  3     1 y       0.1
#  4     1 y       0.5
#  5     2 x       0.1
#  6     2 x       0.5
#  7     2 y       0.1
#  8     2 y       0.5
#  9     3 x       0.1
# 10     3 x       0.5
# 11     3 y       0.1
# 12     3 y       0.5 

The next step would be to use tidyverse and especially the purrr::pmap* family:

library(tidyverse)
crossing(A,B,C) %>% mutate(D = pmap_chr(.,paste,sep="_"))
# A tibble: 12 x 4
#        A B         C D      
#    <dbl> <fct> <dbl> <chr>  
#  1     1 x       0.1 1_1_0.1
#  2     1 x       0.5 1_1_0.5
#  3     1 y       0.1 1_2_0.1
#  4     1 y       0.5 1_2_0.5
#  5     2 x       0.1 2_1_0.1
#  6     2 x       0.5 2_1_0.5
#  7     2 y       0.1 2_2_0.1
#  8     2 y       0.5 2_2_0.5
#  9     3 x       0.1 3_1_0.1
# 10     3 x       0.5 3_1_0.5
# 11     3 y       0.1 3_2_0.1
# 12     3 y       0.5 3_2_0.5

Answer 4

Consider using the wonderful data.table library for expressiveness and speed. It handles many plyr use-cases (relational group by), along with transform, subset and relational join using a fairly simple uniform syntax.

library(data.table)
d <- CJ(x=A, y=B, z=C)  # Cross join
d[, w:=f(x,y,z)]  # Mutates the data.table

or in one line

d <- CJ(x=A, y=B, z=C)[, w:=f(x,y,z)]

Answer 5

Here's a way to do both, using Ramnath's suggestion of expand.grid:

f <- function(x,y,z) paste(x,y,z,sep="+")
d <- expand.grid(x=A, y=B, z=C)
d$D <- do.call(f, d)

Note that do.call works on d "as-is" because a data.frame is a list. But do.call expects the column names of d to match the argument names of f.

Answer 6

Using cross join in sqldf:

library(sqldf)

A <- data.frame(c1 = c(1,2,3))
B <- data.frame(c2 = factor(c('x','y')))
C <- data.frame(c3 = c(0.1,0.5))

result <- sqldf('SELECT * FROM (A CROSS JOIN B) CROSS JOIN C') 

Answer 7

I can never remember that standard function expand.grid. So here's another version.

crossproduct <- function(...,FUN='data.frame') {
  args <- list(...)
  n1 <- names(args)
  n2 <- sapply(match.call()[1+1:length(args)], as.character)
  nn <- if (is.null(n1)) n2 else ifelse(n1!='',n1,n2)
  dims <- sapply(args,length)
  dimtot <- prod(dims)
  reps <- rev(cumprod(c(1,rev(dims))))[-1]
  cols <- lapply(1:length(dims), function(j)
                 args[[j]][1+((1:dimtot-1) %/% reps[j]) %% dims[j]])
  names(cols) <- nn
  do.call(match.fun(FUN),cols)
}

A <- c(1,2,3)
B <- factor(c('x','y'))
C <- c(.1,.5)

crossproduct(A,B,C)

crossproduct(A,B,C, FUN=function(...) paste(...,sep='_'))