CODEWITHSUNDEEP
rloopsmatrixadditionrows
Bobbie Cattani
Bobbie Cattani

Reputation: 1

How to add sum rows of a table or matrix in R?

I have a matrix in R, that looks like the following: 

     [,1] [,2] [,3] [,4]
[1,]    0   1     1    1
[2,]    2   12   22   32
[3,]    3   13   23   33
[4,]    4   14   24   34
[5,]    5   15   25   35
[6,]    6   16   26   36
[7,]    7   17   27   37
[8,]    8   18   28   38
[9,]    9   19   29   39
[10,]   10   20   30   40

I want to add every two rows are together, so that [1,] and [2,], create a new vector (say A). I need to do this repeatedly (so then [3,] and [4,] etc). I need to do this in a loop or some other way if that is possible (my actual data has 49 columns and 1000 rows). If possible, I would like it to create a new matrix with all the new data in, so from my original data I would have 500 rows, or here 5 rows?

I have tried the following code:

 dips = matrix()
 X <- seq(1, by = 2, len = 1000)
 for(i in X)
 {
 dips[i] = population[i,] + population[i+1,]
}

I keep getting a warning, "number of items to replace is not a multiple of replacement length". And the data produced doesn't contain all 49 columns.

Sorry if this is a basic question or has already been solved, I am a beginner. Thank you in advance!

Upvotes: 0

Views: 1286

Answers (2)

Qwfqwf
Qwfqwf

Reputation: 518

In the future, make sure to provide a minimal reproducible example of your data to make it easier for us to follow! You can show how you created dummy data (see my example below) or use something like dput() to produce a text version of the data.

There are few things that I notice:

  1. You need to make your output matrix the appropriate size if you're going to be adding by reference like that. So something like matrix(nrow = 500, ncol = 49)
  2. If you have 1000 rows and you want to add every other together, you actually only want to do the addition 500 times. Your X has length of 1000, so it will try to find rows that are higher than the largest row index in population. Take a look at tail(X) and you will see that it goes up to 1999.
  3. Finally, you are traversing a vector of every other number (1, 3, 5, etc.) and also using that as your row index, so the matrix that you are trying to populate will have blank rows every other, which I don't think is what you want. (Correct me if I'm wrong).

Here is a small example with some dummy data that I made. The notable changes compared to your example are that I:

  1. Made the output matrix have the numbers of rows and columns that I desired (same number of columns as population and half as many rows)
  2. Made X be half as long as nrow(population)
  3. Use 1:length(X) instead of 1:X in my for loop

Example:

population <- matrix(1:400, byrow = F, ncol = 4)

sum <- matrix(nrow = nrow(population)/2, ncol = 4)
X <- seq(1, by = 2, len = nrow(population)/2)
for (i in 1:length(X)){
  sum[i,] <- population[ X[i] ] + population[ X[i]+1 ]
}

Upvotes: 0

JasonAizkalns
JasonAizkalns

Reputation: 20473

This is probably a duplicate, but I couldn't locate a great match quickly, so here goes...you can use rowsum:

my_matrix <- structure(c(0L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 12L, 
                         13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 1L, 22L, 23L, 24L, 25L, 
                         26L, 27L, 28L, 29L, 30L, 1L, 32L, 33L, 34L, 35L, 36L, 37L, 38L, 
                         39L, 40L), .Dim = c(10L, 4L), .Dimnames = list(NULL, NULL))

rowsum(my_matrix, as.integer(gl(nrow(my_matrix), 2, nrow(my_matrix))))
#   [,1] [,2] [,3] [,4]
# 1    2   13   23   33
# 2    7   27   47   67
# 3   11   31   51   71
# 4   15   35   55   75
# 5   19   39   59   79

Upvotes: 4

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