Blockwise sum of matrix elements

2019-01-26 12:11发布

I want to go from something like this:

1> a = matrix(c(1,4,2,5,2,5,2,1,4,4,3,2,1,6,7,4),4)
1> a
     [,1] [,2] [,3] [,4]
[1,]    1    2    4    1
[2,]    4    5    4    6
[3,]    2    2    3    7
[4,]    5    1    2    4

To something like this:

     [,1] [,2]
[1,]   12   15
[2,]   10   16

...without using for-loops, plyr, or otherwise without looping. Possible? I'm trying to shrink a geographic lat/long dataset from 5 arc-minutes to half-degree, and I've got an ascii grid. A little function where I specify blocksize would be great. I've got hundreds of such files, so things that allow me to do it quickly without parallelization/supercomputers would be much appreciated.

3条回答
Summer. ? 凉城
2楼-- · 2019-01-26 12:53

I guess that might help you, but still it uses sapply which can be considered as loop-ish tool.

a <- matrix(c(1,4,2,5,2,5,2,1,4,4,3,2,1,6,7,4),4)
block.step <- 2
res <- sapply(seq(1, nrow(a), by=block.step), function(x) 
    sapply(seq(1, nrow(a), by=block.step), function(y) 
        sum(a[x:(x+block.step-1), y:(y+block.step-1)])
    )
)
res

Is it anyhow helpful ?

查看更多
够拽才男人
3楼-- · 2019-01-26 12:56
tapply(a, list((row(a) + 1L) %/% 2L, (col(a) + 1L) %/% 2L), sum)
#    1  2
# 1 12 15
# 2 10 16

I used 1L and 2L instead of 1 and 2 so indices remain integers (as opposed to numerics) and it should run faster that way.

查看更多
啃猪蹄的小仙女
4楼-- · 2019-01-26 13:00

You can use matrix multiplication for this.

# Computation matrix:

mat <- function(n, r) {
  suppressWarnings(matrix(c(rep(1, r), rep(0, n)), n, n/r))
}

Square-matrix example, uses a matrix and its transpose on each side of a:

# Reduce a 4x4 matrix by a factor of 2:

x <- mat(4, 2)
x
##      [,1] [,2]
## [1,]    1    0
## [2,]    1    0
## [3,]    0    1
## [4,]    0    1

t(x) %*% a %*% x
##      [,1] [,2]
## [1,]   12   15
## [2,]   10   16

Non-square example:

b <- matrix(1:24, 4 ,6)
t(mat(4, 2)) %*% b %*% mat(6, 2)
##      [,1] [,2] [,3]
## [1,]   14   46   78
## [2,]   22   54   86
查看更多
登录 后发表回答