CODEWITHSUNDEEP
r
Zweifler
Zweifler

Reputation: 451

Filling a matrix in R

I am trying to fill some rows of a (500,2) matrix with the row vector (1,0) using this code, last line is to verify the result:

data<-matrix(ncol=2,nrow=500)
data[41:150,]<-matrix(c(1,0),nrow=1,ncol=2,byrow=TRUE)
data[41:45,]

But the result is

> data[41:45,]
     [,1] [,2]
[1,]    1    1
[2,]    0    0
[3,]    1    1
[4,]    0    0
[5,]    1    1

instead of

> data[41:45,]
     [,1] [,2]
[1,]    1    0
[2,]    1    0
[3,]    1    0
[4,]    1    0
[5,]    1    0

(1) What am I doing wrong?

(2) Why aren't the row indices in the result 41, 42, 43, 44 and 45?

Upvotes: 2

Views: 15451

Answers (3)

Andre Elrico
Andre Elrico

Reputation: 11480

You could use rep.

data[41:150,] <- rep(1:0, each=150-41+1)

#> data[41:45,]
#     [,1] [,2]
#[1,]    1    0
#[2,]    1    0
#[3,]    1    0
#[4,]    1    0
#[5,]    1    0

I think MichaelChirico approach is the cleanest/savest to use.

Upvotes: 2

lebatsnok
lebatsnok

Reputation: 6449

You're trying to fill a part of the matrix, so the block you're trying to drop in there should be of the right size:

 data[41:150,]<-matrix(c(1,0),nrow=110,ncol=2,byrow=TRUE)
 # nrow = 110, instead of 1 !!!!

Otherwise your piece-to-be-added will be reverted to vector and added columnwise. Try, for example, this:

 data[41:150,] <- matrix(c(1,2,3,4,5), nrow=5, ncol=2, byrow=TRUE)
 data[41:45,]
     [,1] [,2]
[1,]    1    1
[2,]    3    3
[3,]    5    5
[4,]    2    2
[5,]    4    4

Can one complain? Yes, and now. No, because R behaves as documented (matrices are vectors with dimension attributes, and recycling works on vectors). Yes, because although recycling can be convenient, it may create false expectations.

Why aren't row indices 41,42,43,... ? I don't know, that's just the way matrices and vectors behave.

> (1:10)[5:6]
[1] 5 6

(Notice there's [1] in the output, not [5].)

Data frames behave differently, so you would see the original line numbers for slices:

 as.data.frame(data)[45:50,]

Upvotes: 4

MichaelChirico
MichaelChirico

Reputation: 34703

It will be cleaner to just do this column-wise:

data[41:150, 1L] = 1
data[41:150, 2L] = 0

You could also accomplish this in one line with matrix indexing like so:

data[cbind(rep(41:150, each = 2L), 1:2)] = 1:0

Upvotes: 3

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