CODEWITHSUNDEEP
rconditional-statements
PedroA
PedroA

Reputation: 1925

R - work on data frame rows based on condition

I'm trying to understand how can I work on the rows of a data frame based on a condition. Having a data frame like this

> d<-data.frame(x=c(0,1,2,3), y=c(1,1,1,0))
> d
  x y
1 0 1
2 1 1
3 2 1
4 3 0

how can I add +1 to all rows that contain a value of zero? (note that zeros can be found in any column), so that the result would look like this:

  x y
1 1 2
2 1 1
3 2 1
4 4 1

The following code seems to do part of the job, but is just printing the rows where the action was taken, the number of times it was taken (2)...

> for(i in 1:nrow(d)){
+     d[d[i,]==0,]<-d[i,]+1
+ }
> d
  x y
1 1 2
2 4 1
3 1 2
4 4 1

I'm sure there is a simple solution for this, maybe an apply function?, but I'm not getting there.

Thanks.

Upvotes: 4

Views: 2267

Answers (1)

Julius Vainora
Julius Vainora

Reputation: 48191

Some possibilities:

# 1
idx <- which(d == 0, arr.ind = TRUE)[, 1]
d[idx, ] <- d[idx, ] + 1
# 2
t(apply(d, 1, function(x) x + any(x == 0)))
# 3
d + apply(d == 0, 1, max)

The usage of which for vectors, e.g. which(1:3 > 2), is quite common, whereas it is used less for matrices: by specifying arr.ind = TRUE what we get is array indices, i.e. coordinates of every 0:

which(d == 0, arr.ind = TRUE)
     row col
[1,]   1   1
[2,]   4   2

Since we are interested only in rows where zeros occur, I take the first column of which(d == 0, arr.ind = TRUE) and add 1 to all the elements in these rows by d[idx, ] <- d[idx, ] + 1.

Regarding the second approach, apply(d, 1, function(x) x) would be simply going row by row and returning the same row without any modifications. By any(x == 0) we check whether there are any zeros in a particular row and get TRUE or FALSE. However, by writing x + any(x == 0) we transform TRUE or FALSE to 1 or 0, respectively, as required.

Now the third approach. d == 0 is a logical matrix, and we use apply to go over its rows. Then when applying max to a particular row, we again transform TRUE, FALSE to 1, 0 and find a maximal element. This element is 1 if and only if there are any zeros in that row. Hence, apply(d == 0, 1, max) returns a vector of zeros and ones. The final point is that when we write A + b, where A is a matrix and b is a vector, the addition is column-wise. In this way, by writing d + apply(d == 0, 1, max) we add apply(d == 0, 1, max) to every column of d, as needed.

Upvotes: 3

Related Questions