I'm trying to build a data frame that is entirely composed of 1s and 0s. It should be randomly built except for the fact that each column needs to add up to a specified value.
I would know how to do this if this was for just one data frame, but it needs to be built into a function, where in said function it will be done as an iterative process, up to 1000x.
An efficient approach would be to shuffle a vector with the appropriate number of 1s and 0s for each column. You could define the following function to generate a matrix with a specified number of rows and the number of 1s in each column:
build.mat <- function(nrow, csums) {
sapply(csums, function(x) sample(rep(c(0, 1), c(nrow-x, x))))
}
set.seed(144)
build.mat(5, 0:5)
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 0 0 0 0 1 1
# [2,] 0 0 0 1 0 1
# [3,] 0 0 0 0 1 1
# [4,] 0 1 1 1 1 1
# [5,] 0 0 1 1 1 1
To build a list, you might use lapply over the desired column sums for each matrix:
cslist <- list(1:3, c(4, 2))
set.seed(144)
lapply(cslist, build.mat, nrow=5)
# [[1]]
# [,1] [,2] [,3]
# [1,] 0 1 1
# [2,] 0 0 0
# [3,] 0 0 0
# [4,] 0 1 1
# [5,] 1 0 1
#
# [[2]]
# [,1] [,2]
# [1,] 0 0
# [2,] 1 0
# [3,] 1 1
# [4,] 1 0
# [5,] 1 1
nrow (since it's a function name). Also, the OP claims that this must be done many times, so if the number of rows is small and constant across these runs, it might make sense to precompute and store the vectors outside the function for quicker access: vecs <- sapply(setNames(0:n,0:n),function(x)rep(0:1, c(n-x, x))); apply(vecs[,as.character(csums)],2,sample)
nrow. I guess there might be some use cases were your proposal speeds things up, but for long, narrow data frames it would probably make the code much slower, since you would need to allocate a huge matrix vecs, most of which you would probably never use.If there are many more zeros than ones or vice versa, @akrun's approach may be faster:
build_01_mat <- function(n,n1s){
nc <- length(n1s)
zerofirst <- sum(n1s) < n*nc/2
tochange <- if (zerofirst) n1s else n-n1s
mat <- matrix(if (zerofirst) 0L else 1L,n,nc)
mat[cbind(
unlist(c(sapply((1:nc)[tochange>0],function(col)sample(1:n,tochange[col])))),
rep(1:nc,tochange)
)] <- if (zerofirst) 1L else 0L
mat
}
set.seed(1)
build_01_mat(5,c(1,3,0))
# [,1] [,2] [,3]
# [1,] 0 0 0
# [2,] 1 1 0
# [3,] 0 1 0
# [4,] 0 1 0
# [5,] 0 0 0
Some benchmarks:
require(rbenchmark)
# similar numbers of zeros and ones
benchmark(
permute=build.mat(1e7,1e7/2),
replace=build_01_mat(1e7,1e7/2),replications=10)[1:5]
# test replications elapsed relative user.self
# 1 permute 10 7.68 1.126 6.59
# 2 replace 10 6.82 1.000 6.27
# many more zeros than ones
benchmark(
permute=build.mat(1e6,rep(10,20)),
replace=build_01_mat(1e6,rep(10,20)),replications=10)[1:5]
# test replications elapsed relative user.self
# 1 permute 10 10.28 3.779 8.51
# 2 replace 10 2.72 1.000 2.23
# many more ones than zeros
benchmark(
permute=build.mat(1e6,1e6-rep(10,20)),
replace=build_01_mat(1e6,1e6-rep(10,20)),replications=10)[1:5]
# test replications elapsed relative user.self
# 1 permute 10 10.94 4.341 9.28
# 2 replace 10 2.52 1.000 2.09
forloop or runreplicateon it. You'll need to be much more specific, show what you've tried, and show sample input and desired output for this to be a good question.sample. For example, suppose, you want to create a vector of length 10 that sum to 5. i.e. 51s.v1 <- numeric(10); v1[sample(10, 5, replace=FALSE)] <- 1usingreplicateas @Gregor suggested, this can be looped. But, I am not sure whether the specified value is different for different columns. So, you may need to show some example to clear the confusion