CODEWITHSUNDEEP
r
Kirk Fogg
Kirk Fogg

Reputation: 531

Using the ifelse statement in R

I would like to use the ifelse statement to create a new variable, say, z. However, one of the return values depends on the i-th column of a matrix. Here is a simple example

set.seed(1)
data <- data.frame(x = rnorm(10), y = rnorm(10), ind = rep(c(0, 1), 5))
m <- data.frame(matrix(rnorm(100), 10, 10))

z <- ifelse(data$ind == 1, data$x, sum(m[, i]))

I know the line with z won't run, but it illustrates what I would like to do. If a subject has the ind variable equal to 0, then I assign to z the sum of the 10 entries in m corresponding to subject i's column.

Could I do this with ifelse, or would I need a for loop? I'm trying to stay away from for loops, which is why I am trying ifelse in the first place.

Here is what z should look like:

z
 [1] -1.3367324  0.1836433  1.3413668  1.5952808  4.5120996 -0.8204684  1.2736029
 [8]  0.7383247  3.4748021 -0.3053884

Thanks!

Upvotes: 2

Views: 125

Answers (3)

akrun
akrun

Reputation: 886968

Or we can use arithmetic

colSums(m)*(data$ind==0) + (data$ind==1)*data$x
#     X1         X2         X3         X4         X5         X6         X7 
#-1.3367324  0.1836433  1.3413668  1.5952808  4.5120996 -0.8204684  1.2736029 
#        X8         X9        X10 
# 0.7383247  3.4748021 -0.3053884

Upvotes: 3

Jealie
Jealie

Reputation: 6267

Yes you can do it with ifelse and a one-liner, very close to what you wrote:

z <- ifelse(data$ind == 0, colSums(m), data$x)

Here is what R does when it executes this statement:

  1. it computes the boolean vector data$ind == 0, and stores into memory the two numeric vectors colSums(m) and data$x
  2. where (data$ind == 0) is True, it outputs colSums(m); where (data$ind == 0) is False, it outputs data$x

Upvotes: 4

jeremycg
jeremycg

Reputation: 24945

You can do it in a two-liner instead:

z <- data$x
z[data$ind == 0] <- colSums(m[,data$ind == 0])

 [1] -1.3367324  0.1836433  1.3413668  1.5952808  4.5120996 -0.8204684  1.2736029  0.7383247  3.4748021
[10] -0.3053884

more generally, you could use an apply function. This will in general be slower than a straight vectorised solution, like the above. Here's sapply:

sapply(1:nrow(data), function(x){ifelse(data$ind[x] == 1, data$x[x], sum(m[, x]))})

 [1] -1.3367324  0.1836433  1.3413668  1.5952808  4.5120996 -0.8204684  1.2736029  0.7383247  3.4748021
[10] -0.3053884

A benchmark:

microbenchmark::microbenchmark(
     sapply = sapply(1:nrow(data), function(x){ifelse(data$ind[x] == 1, data$x[x], sum(m[, x]))}), 
     vectorised = {z <- data$x;
                   z[data$ind == 0] <- colSums(m[,data$ind == 0])})
Unit: microseconds
       expr     min      lq     mean   median       uq     max neval cld
     sapply 391.297 408.193 423.6525 412.4170 423.7450 853.249   100   b
 vectorised 197.377 199.873 208.7701 202.5605 214.4645 284.545   100  a

Upvotes: 3

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