r - Convert total records per species into record x species matrix -

imagine have 9 sampling records 3 species distributed such:

sp1 sp2 sp3 3 1 5 

what want obtain records x species matrix, , fill 1s , 0s such:

sp1 sp2 sp3  1 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 

the number of columns matches number of species , number of rows number of records. note each row represents unique record 1 species.

another option create row/column index, use sparsematrix library(matrix) create sparse matrix, can converted matrix of 0 , 1s as.matrix.

it not clear whether initial dataset matrix or not. assuming matrix 3 column , 1 row, column index replicating sequence of columns elements of 'm1'. should work if vector. data.frame, have use rep(seq_along(df1), unlist(df1)). then, create sparsematrix, specifying row index sequence of 'ci' , column index ('ci') , value 'x' 1.

library(matrix) ci <- rep(seq_along(m1), m1) m2 <- as.matrix(sparsematrix(seq_along(ci), ci, x=1)) colnames(m2) <- colnames(m1) m2 #      sp1 sp2 sp3 # [1,]   1   0   0 # [2,]   1   0   0 # [3,]   1   0   0 # [4,]   0   1   0 # [5,]   0   0   1 # [6,]   0   0   1 # [7,]   0   0   1 # [8,]   0   0   1 # [9,]   0   0   1 

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a base r approach to create matrix of 0 , replace elements corresponds row/column index 1.

m2 <- matrix(0, nrow=length(ci), ncol=ncol(m1),                    dimnames=list(null, colnames(m1)))  m2[cbind(seq_along(ci), ci)] <- 1 m2 #      sp1 sp2 sp3 # [1,]   1   0   0 # [2,]   1   0   0 # [3,]   1   0   0 # [4,]   0   1   0 # [5,]   0   0   1 # [6,]   0   0   1 # [7,]   0   0   1 # [8,]   0   0   1 # [9,]   0   0   1 

data

m1 <- structure(c(3l, 1l, 5l), .dim = c(1l, 3l), .dimnames = list(null,  c("sp1", "sp2", "sp3")))