CODEWITHSUNDEEP
rmatrixdimension
Jolanda Kossakowski
Jolanda Kossakowski

Reputation: 461

R: changing dimensions of matrix

I have the following dataset:

day <- c(rep(17,4), rep(18,2))
beep <- c(74.50, 77.50, 89.50, 75.25, 58.25, 81.25)
m <- cbind(day, beep)
m
     day  beep
[1,]  17 74.50
[2,]  17 77.50
[3,]  17 89.50
[4,]  17 75.25
[5,]  18 58.25
[6,]  18 81.25

what I want is to turn this dataset into a matrix with the amount of days (in this case 2) as the amount of columns. This is how it would like:

      [,1]  [,2]
[1,] 74.50 58.25
[2,] 77.50 81.25
[3,] 89.50    NA
[4,] 75.25    NA

Since this person had 4 beeps on day 1, and 2 beeps on day 2, necessarily 2 NAs must be within the matrix. I'd love to know how I could turn the above dataset in this, without manually adjusting it like I did now to make the example.

Upvotes: 0

Views: 13189

Answers (3)

user13177385
user13177385

Reputation: 1

This might be simpler than others assuming that the dim of the matrix is known.

n <- m[, "beep"]
length(n) <- 8 # Since n is a vector with 6 elements, this will make it 8 element- 
               # vector with two NAs attached.
dim(n) <- c(4, 2)  # change the dimension of the vector to a 4 by 2 matrix
print(n)

      [,1]  [,2]
[1,] 74.50 58.25
[2,] 77.50 81.25
[3,] 89.50    NA
[4,] 75.25    NA

Upvotes: 0

Maciej
Maciej

Reputation: 3303

You could also use reshape function from stats package, but you need to transform your matrix to data.frame and form data.frame to matrix. But I think that it is more flexible way, because you create id for variable that you want to be responsible for columns (in your case day).

m.df<-as.data.frame(m) ## convert to data.frame
m.df$id<-ave(m.df$day,m.df$day,FUN=seq_along) ### create index with ave function (more solutions: http://stackoverflow.com/questions/8997638/numbering-by-groups )
m2<-reshape(m.df,idvar='id',timevar='day',direction='wide') ## reshape data timevar is responsible for columns and with direction you tell how data set should be expaned.
as.matrix(m2) ### convert back to matrix

Upvotes: 1

Roland
Roland

Reputation: 132989

I agree with @flodel's comment, but here is a way:

m2 <- unstack(m, beep~day)
nrow <- max(sapply(m2, length))
m2 <- sapply(m2, function(x) {
  length(x) <- nrow
  x
})

#        17    18
#[1,] 74.50 58.25
#[2,] 77.50 81.25
#[3,] 89.50    NA
#[4,] 75.25    NA

Upvotes: 1

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