Clash Royale CLAN TAG

Using nested apply functions instead of nested for loops

My objective here was to iterate across each column in a df and then for each column iterate down each row and perform a function. The specific function in this case replaces the NA values with the corresponding value in the final column, but the details of the function required are not relevant to the question here. I got the results I needed using two nested for loops like this:

df

NA

for (j in 1:ncol(df.i)) {
for (i in 1:nrow(df.i)) {
df.i[i,j] <- ifelse(is.na(df.i[i,j]), df.i[i,39], df.i[i,j])
}
}


However, I believe this should be possible using an apply(df.i, 1, function) nested within an apply(df.i, 2, function) But I'm not totally sure that is possible or how to do it. Does anyone know how to achieve the same thing with a nested use of the apply function?

apply(df.i, 1, function)

apply(df.i, 2, function)

apply

ifelse

df.i[,j] <- ifelse(is.na(df.i[,j]), df.i[,39], df.i[,j])

apply()

apply()

lapply()

1 Answer
1

Here are three ways to do what the inner instruction does.

First, a dataset example.

set.seed(5345) # Make the results reproducible
df.i <- matrix(1:400, ncol = 40)
is.na(df.i) <- sample(400, 50)


Now, the comment by @Dave2e: just one for loop, vectorize the inner most one.

for

df.i1 <- df.i # Work with a copy

for (j in 1:ncol(df.i1)) {
df.i1[,j] <- ifelse(is.na(df.i1[, j]), df.i1[, 39], df.i1[, j])
}


Then, fully vectorized, no loops at all.

df.i2 <- ifelse(is.na(df.i), df.i[, 39], df.i)


And your solution, as posted in the question.

for (j in 1:ncol(df.i)) {
for (i in 1:nrow(df.i)) {
df.i[i,j] <- ifelse(is.na(df.i[i,j]), df.i[i,39], df.i[i,j])
}
}


Compare the results.

identical(df.i, df.i1)
#[1] TRUE

identical(df.i, df.i2)
#[1] TRUE


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