Using the features in Java 8, what is the most concise way of transforming all the values of a list?
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());
The Coordinator
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Updated on November 08, 2020Comments
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The Coordinator over 3 years
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:
transform the values in-place (without creating a new list)
transform the values into a new result list.