如何使用R自动执行两个矩阵的每一列之间的运算？

Question

I have general function written for MAPE (mean absolute percentage error)

mape <- function(y, yhat)
     mean(abs((y - yhat)/y))

I want to calculate MAPE between each column of two different matrices. Suppose they are in following format

y = matrix(c(11, 12, 12, 12, 14, 16, 23, 21, 28),byrow=TRUE,ncol=3)

and predicted as

yp = matrix(c(12, 13, 14, 12, 15, 17, 24, 22, 28),byrow=TRUE,ncol=3)

This can be manually done for each column as mape(y[,1],yp[,1])

How do I automate such process (any other operation also - not only MAPE) of performing operations between each columns of large dimension matrix using R? can FOR loops be avoided using apply/sapply?

Answer 1

Update:

Of course mape can be vectorised:

mapeVec <- function(y, yhat)
  colMeans(abs((y-yhat)/y))

f3 <- function() { mapeVec(y, yp) } 

Unit: milliseconds
 expr       min        lq      mean    median        uq      max neval cld
 f1() 33.677431 34.121107 35.494355 34.441823 35.078125 46.16782   100  b 
 f2() 33.558224 33.970123 35.609414 34.239525 34.881354 49.99195   100  b 
 f3()  8.344952  8.525146  9.218695  8.568763  8.709681 17.82791   100 a  

identical(f1(), f3())  # TRUE

Old part:

sapply with a sequence seq(nrow(y)) should do the trick:

mape <- function(y, yhat)
  mean(abs((y - yhat)/y))

y <- matrix(c(11, 12, 12, 12, 14, 16, 23, 21, 28), nrow = 3, ncol = 3)

yp = matrix(c(12, 13, 14, 12, 15, 17, 24, 22, 28), nrow = 3, ncol = 3)

sapply(seq(nrow(y)), function(id) { mape(y[,id], yp[,id]) })

Microbenchmark:

library(microbenchmark)

mape <- function(y, yhat)
  mean(abs((y - yhat)/y))

y <- matrix(rnorm(1000000), nrow = 1000, ncol = 1000)

yp = matrix(rnorm(1000000), nrow = 1000, ncol = 1000)

f1 <- function() { sapply(seq(nrow(y)), function(id) { mape(y[,id], yp[,id]) }) }

f2 <- function() { 
  a <- vector(mode = "numeric", length = nrow(y))
  for(id in seq(nrow(y))) { 
    a[id] <- mape(y[,id], yp[,id]) 
  }
  a
}

microbenchmark(
  f1(),
  f2()
)

Results:

Unit: milliseconds
 expr      min       lq     mean   median       uq      max neval cld
 f1() 33.28310 34.15209 36.57389 35.42845 36.20803 48.11936   100   a
 f2() 34.14755 34.78859 37.65782 36.33395 37.06874 64.10664   100   a

Personal comment:

f1 ( sapply() approach) looks much more compact and "clean".