Apply a function to each row of the matrix or to a data frame

Question

Suppose I have an n by 2 matrix and a function that takes a 2-vector as one of its arguments. I would like to apply a function to each row of the matrix and get an n-vector. How to do it in R?

For example, I would like to calculate the density of a two-dimensional standard normal distribution over three points:

bivariate.density(x = c(0, 0), mu = c(0, 0), sigma = c(1, 1), rho = 0){ exp(-1/(2*(1-rho^2))*(x[1]^2/sigma[1]^2+x[2]^2/sigma[2]^2-2*rho*x[1]*x[2]/(sigma[1]*sigma[2]))) * 1/(2*pi*sigma[1]*sigma[2]*sqrt(1-rho^2)) } out <- rbind(c(1, 2), c(3, 4), c(5, 6)) 

How to apply a function to each out line?

How do I pass values ​​for arguments other than dots to a function, as you specify?

Answer 1

You just use the apply() function:

 R> M <- matrix(1:6, nrow=3, byrow=TRUE) R> M [,1] [,2] [1,] 1 2 [2,] 3 4 [3,] 5 6 R> apply(M, 1, function(x) 2*x[1]+x[2]) [1] 4 10 16 R> 

This takes a matrix and applies a (dumb) function to each row. You pass additional arguments to the function as the fourth, fifth, ... arguments in apply() .

Answer 2

If you want to apply general functions like sum or mean, you should use rowSums or rowMeans , since they are faster than apply(data, 1, sum) . Otherwise, stick to apply(data, 1, fun) . You can pass additional arguments after the FUN argument (as Dirk already pointed out):

 set.seed(1) m <- matrix(round(runif(20, 1, 5)), ncol=4) diag(m) <- NA m [,1] [,2] [,3] [,4] [1,] NA 5 2 3 [2,] 2 NA 2 4 [3,] 3 4 NA 5 [4,] 5 4 3 NA [5,] 2 1 4 4 

Then you can do something like this:

 apply(m, 1, quantile, probs=c(.25,.5, .75), na.rm=TRUE) [,1] [,2] [,3] [,4] [,5] 25% 2.5 2 3.5 3.5 1.75 50% 3.0 2 4.0 4.0 3.00 75% 4.0 3 4.5 4.5 4.00 

Answer 3

Here is a brief example of applying a function to each row of a matrix. (Here the function used normalizes each line to 1.)

Note. The result of apply() must be transposed using t() to get the same layout as the input matrix A

 A <- matrix(c( 0, 1, 1, 2, 0, 0, 1, 3, 0, 0, 1, 3 ), nrow = 3, byrow = TRUE) t(apply(A, 1, function(x) x / sum(x) )) 

Result:

  [,1] [,2] [,3] [,4] [1,] 0 0.25 0.25 0.50 [2,] 0 0.00 0.25 0.75 [3,] 0 0.00 0.25 0.75 

Answer 4

The first step is to create a function object and then use it. If you want the matrix object to have the same number of rows, you can predefine it and use the object [] form, as shown (otherwise the return value will be simplified for the vector):

 bvnormdens <- function(x=c(0,0),mu=c(0,0), sigma=c(1,1), rho=0){ exp(-1/(2*(1-rho^2))*(x[1]^2/sigma[1]^2+ x[2]^2/sigma[2]^2- 2*rho*x[1]*x[2]/(sigma[1]*sigma[2]))) * 1/(2*pi*sigma[1]*sigma[2]*sqrt(1-rho^2)) } out=rbind(c(1,2),c(3,4),c(5,6)); bvout<-matrix(NA, ncol=1, nrow=3) bvout[] <-apply(out, 1, bvnormdens) bvout [,1] [1,] 1.306423e-02 [2,] 5.931153e-07 [3,] 9.033134e-15 

If you want to use parameters other than the default parameters, then the call must include named arguments after the function:

 bvout[] <-apply(out, 1, FUN=bvnormdens, mu=c(-1,1), rho=0.6) 

apply () can also be used for arrays with a higher dimension, and the argument MARGIN can be either a vector or a single integer.

Answer 5

Another approach, if you want to use the variable part of the data set instead of a single value, is to use rollapply(data, width, FUN, ...) . Using a width vector allows you to apply a function in a changing dataset window. I used this to create an adaptive filtering procedure, although this is not very efficient.

Answer 6

To apply the work is good, but rather slowly. Using sapply and vapply may be helpful. dplyr rowwise can also be useful. Let's look at an example of how to make a series of wise products of any data frame.

 a = data.frame(t(iris[1:10,1:3])) vapply(a, prod, 0) sapply(a, prod) 

Note that assigning a variable before using vapply / sapply / apply is good practice, as it greatly reduces time. See the results of microdetection.

 a = data.frame(t(iris[1:10,1:3])) b = iris[1:10,1:3] microbenchmark::microbenchmark( apply(b, 1 , prod), vapply(a, prod, 0), sapply(a, prod) , apply(iris[1:10,1:3], 1 , prod), vapply(data.frame(t(iris[1:10,1:3])), prod, 0), sapply(data.frame(t(iris[1:10,1:3])), prod) , b %>% rowwise() %>% summarise(p = prod(Sepal.Length,Sepal.Width,Petal.Length)) ) 

Use caution when using t ()