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

Question

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?

Dirk Eddelbuettel · Accepted Answer · 2010-11-21T04:14:27.557

197

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().

edited Nov 21 '10 at 04:14

answered Nov 21 '10 at 04:05

Dirk Eddelbuettel

347,098
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Thanks! What if the rows of the matrix is not the first arg of the function? How to specify which arg of the function each row of the matrix is assigned to? – Tim Nov 21 '10 at 04:10
Read the help for `apply()` -- it sweeps by row (when the second arg is 1, else by column), and the current row (or col) is always the first argument. That is how things are defined. – Dirk Eddelbuettel Nov 21 '10 at 04:15
1

@Tim : if you use an internal R function and the row is not the first arg, do as Dirk did and make your own custom function where row **is** the first arg. – Joris Meys Nov 22 '10 at 12:58
3

The plyr package provides a wide range of these apply kinds of functions. It also provides more functionality, including parallel processing. – Paul Hiemstra Nov 30 '11 at 13:29
Can you explain what `1` means in `apply(M, 1...)`? – cryptic0 Nov 08 '17 at 18:24
7

@cryptic0 this answer is late, but for googlers, the second argument in apply is the `MARGIN` argument. Here it means apply the function to the rows (the first dimension in `dim(M)`). If it were 2, it would apply the function to the columns. – De Novo Mar 05 '18 at 08:24

score 17 · Answer 2 · answered Nov 04 '14 at 12:37

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

aL3xa · Answer 3 · 2010-11-21T18:14:25.397

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

score 7 · Answer 4 · answered May 29 '17 at 15:32

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

It might be more fair to compare the apply family if you used `b — DaSpeeg, Dec 13 '18 at 16:59

score 6 · Answer 5 · answered Nov 21 '10 at 15:01

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.

score 2 · Answer 6 · edited Nov 30 '11 at 06:34

2

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.

edited Nov 30 '11 at 06:34

joran

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answered Sep 21 '11 at 16:29

DWAHL

156
6

score 0 · Answer 7 · answered Mar 29 '22 at 23:28

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

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

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