Using the features in Java 8, what is the most concise way of transforming all the values of a list? [duplicate]

Question

Using the new features of Java 8, what is the most concise way of transforming all the values of a List<String>?

Given this:

List<String> words = Arrays.asList("Apple", "Ananas", "Mango", "Banana", "Beer");

I am currently doing this:

for (int n = 0; n < words.size(); n++) {
    words.set(n, words.get(n).toUpperCase());
}

How can the new Lambdas, Collections and Streams API in Java 8 help:

1. transform the values in-place (without creating a new list)

2. transform the values into a new result list.

Answer 1

This is what I came up with:

Given the list:

List<String> keywords = Arrays.asList("Apple", "Ananas", "Mango", "Banana", "Beer");

(1) Transforming them in place

Maybe I am missing it, there does not seem to be a 'apply' or 'compute' method that takes a lambda for List. So, this is the same as with old Java. I can not think of a more concise or efficient way with Java 8.

for (int n = 0; n < keywords.size(); n++) {
    keywords.set(n, keywords.get(n).toUpperCase());
}

Although there is this way which is no better than the for(..) loop:

IntStream.range(0,keywords.size())
    .forEach( i -> keywords.set(i, keywords.get(i).toUpperCase()));

(2) Transform and create new list

List<String> changed = keywords.stream()
    .map( it -> it.toUpperCase() ).collect(Collectors.toList());