CODEWITHSUNDEEP
rfunctionmatrixapplysapply
Tim
Tim

Reputation: 99408

Apply a function to every row of a matrix or a data frame

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

For example, I would like to compute the density of a 2D standard Normal distribution on 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 the function to each row of out?

How to pass values for the other arguments besides the points to the function in the way you specify?

Upvotes: 164

Views: 324578

Answers (7)

rubengavidia0x
rubengavidia0x

Reputation: 590

A dplyr Approach using across, rowSums and rowMeans.

M <- matrix(1:9, nrow=3, byrow=TRUE)
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5    6
[3,]    7    8    9

M %>% as_tibble() %>%
  rowwise() %>% 
  mutate(sum = rowSums(across(where(is.numeric)))) %>% 
  mutate(mean = rowMeans(across(V1:V3))) %>%
  mutate(Max = max(V1:V3)) %>%
  mutate(Min = min(V1:V3)) %>%
  as.matrix()

     V1 V2 V3 sum mean Max Min
[1,]  1  2  3   6    2   3   1
[2,]  4  5  6  15    5   6   4
[3,]  7  8  9  24    8   9   7

Upvotes: 1

Pratham
Pratham

Reputation: 179

Apply does the job well, but is quite slow. Using sapply and vapply could be useful. dplyr's rowwise could also be useful Let's see an example of how to do row wise product of any data frame.

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

Note that assigning to variable before using vapply/sapply/ apply is good practice as it reduces time a lot. Let's see microbenchmark results

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))
)

Have a careful look at how t() is being used

Upvotes: 9

Viliam Simko
Viliam Simko

Reputation: 1841

Here is a short example of applying a function to each row of a matrix. (Here, the function applied normalizes every row to 1.)

Note: The result from the apply() had to 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

Upvotes: 17

DWAHL
DWAHL

Reputation: 156

Another approach if you want to use a varying portion of the dataset instead of a single value is to use rollapply(data, width, FUN, ...). Using a vector of widths allows you to apply a function on a varying window of the dataset. I've used this to build an adaptive filtering routine, though it isn't very efficient.

Upvotes: 2

aL3xa
aL3xa

Reputation: 36080

In case you want to apply common functions such as sum or mean, you should use rowSums or rowMeans since they're faster than apply(data, 1, sum) approach. Otherwise, stick with apply(data, 1, fun). You can pass additional arguments after FUN argument (as Dirk already suggested):

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

Upvotes: 18

IRTFM
IRTFM

Reputation: 263301

First step would be making the function object, then applying it. If you want a matrix object that has the same number of rows, you can predefine it and use the object[] form as illustrated (otherwise the returned value will be simplified to a 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 wanted to use other than your default parameters then the call should include named arguments after the function:

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

apply() can also be used on higher dimensional arrays and the MARGIN argument can be a vector as well as a single integer.

Upvotes: 5

Dirk is no longer here
Dirk is no longer here

Reputation: 368181

You simply 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 (silly) function to each row. You pass extra arguments to the function as fourth, fifth, ... arguments to apply().

Upvotes: 214

Related Questions