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

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How do I apply a function to all rows of a data frame?

Use apply() function when you wanted to update every row in pandas DataFrame by calling a custom function. In order to apply a function to every row, you should use axis=1 param to apply().

How do I apply the same function to all rows and columns of a matrix in R?

One of the most famous and most used features of R is the *apply() family of functions, such as apply() , tapply() , and lapply() . Here, we'll look at apply() , which instructs R to call a user-specified function on each of the rows or each of the columns of a matrix.

How do I apply a function to each column in a DataFrame in R?

Apply any function to all R data frame You can set the MARGIN argument to c(1, 2) or, equivalently, to 1:2 to apply the function to each value of the data frame. If you set MARGIN = c(2, 1) instead of c(1, 2) the output will be the same matrix but transposed. The output is of class “matrix” instead of “data.

What is apply () in R?

The apply() collection is a part of R essential package. This family of functions helps us to apply a certain function to a certain data frame, list, or vector and return the result as a list or vector depending on the function we use.


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().


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

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

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.


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


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.