CODEWITHSUNDEEP
r
IloveR
IloveR

Reputation: 69

return to the position in matrix by

comN<-gtools::permutations(2,5,v=c(0,1),repeats=TRUE)

which(comN==c(0,0,0,0,1),arr.ind.col=T)

but it didn't give me the row number 2

I want to know c( 0,0,0,0,1) is on position comN [2,]

Upvotes: 2

Views: 32

Answers (2)

Ronak Shah
Ronak Shah

Reputation: 388907

We can use sweep to compare row-wise values of comN with the vector and use rowSums to count number of TRUE in each row

which(rowSums(sweep(comN, 2, c(0,0,0,0,1), `==`)) == ncol(comN))
#[1] 2

Instead of rowSums, we can also use apply since sweep already returns a matrix of logical vectors

which(apply(sweep(comN, 2,c(0,0,0,0,1), `==`), 1, all))

Upvotes: 0

Rui Barradas
Rui Barradas

Reputation: 76402

R's matrix operations are in column order, so the comparison in your which instruction compares the vector c(0,0,0,0,1) column by column, recycling it. Just see what happens in this simple example.

m <- matrix(1:8, ncol = 2)
m
#     [,1] [,2]
#[1,]    1    5
#[2,]    2    6
#[3,]    3    7
#[4,]    4    8

m == c(1, 5)
#      [,1]  [,2]
#[1,]  TRUE FALSE
#[2,] FALSE FALSE
#[3,] FALSE FALSE
#[4,] FALSE FALSE

The vector c(1, 5) is not compared with the first row, the comparisons are 

c(1, 2) == c(1, 5)

then c(1, 5) is recycled and compared with the next values in the first column. So when it reaches the second column the first comparison in that column is 

c(5, 6) == c(1, 5)
#[1] FALSE FALSE

returning c(FALSE, FALSE).

This is why I needed the transpose t() in my solutions below. A way could be

which(colSums(t(comN) == c(0,0,0,0,1)) == ncol(comN))
#[1] 2

or 

which(apply(t(comN) == c(0,0,0,0,1), 2, all))

The colSums solution is around 3 times faster.

Upvotes: 1

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