min max scaling/normalization in r for train and test data

user_5052482 0
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I am looking to create a function that takes in the training set and the testing set as its arguments, min-max scales/normalizes and returns the training set and uses those same values of minimum and range to min-max scale/normalize and return the test set.

So far this is the function I have come up with:

min_max_scaling <- function(train, test){

  min_vals <- sapply(train, min)
  range1 <- sapply(train, function(x) diff(range(x)))

  # scale the training data

  train_scaled <- data.frame(matrix(nrow = nrow(train), ncol = ncol(train)))

  for(i in seq_len(ncol(train))){
    column <- (train[,i] - min_vals[i])/range1[i]
    train_scaled[i] <- column
  }

  colnames(train_scaled) <- colnames(train)

  # scale the testing data using the min and range of the train data

  test_scaled <- data.frame(matrix(nrow = nrow(test), ncol = ncol(test)))

  for(i in seq_len(ncol(test))){
    column <- (test[,i] - min_vals[i])/range1[i]
    test_scaled[i] <- column
  }

  colnames(test_scaled) <- colnames(test)

  return(list(train = train_scaled, test = test_scaled))
}

The definition of min max scaling is similar to this question asked earlier on SO - Normalisation of a two column data using min and max values

My questions are:
1. Is there a way to vectorize the two for loops in the function? e.g. using sapply()
2. Are there any packages that allow us to do what we are looking to do here?

r
3 Answers

Here is the code for the min-max normalization. See this Wikipedia page for the formulae, and also other ways of performing feature scaling.

normalize <- function(x, na.rm = TRUE) {
    return((x- min(x)) /(max(x)-min(x)))
}

To get a vector, use apply instead of lapply.

as.data.frame(apply(df$name, normalize))

Update to address Holger's suggestion.

If you want to pass additional arguments to min() and max(), e.g., na.rm, then you can use:

normalize <- function(x, ...) {
    return((x - min(x, ...)) /(max(x, ...) - min(x, ...)))
}

x <- c(1, NA, 2, 3)

normalize(a)
# [1] NA NA NA NA

normalize(a, na.rm = TRUE)
# 0.0  NA 0.5 1.0

Just keep in mind, that whatever you pass to min() via the ellipsis ... you also implicitly pass to max(). In this case, this shouldn't be a big problem since both min() and max() share the same function signature.

user_88140470
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