Transform List into Map using only two keys and odd or even list indexes as values – Java 8 Stream

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I would like to transform list into map using as key values only two string values. Then as values just list of strings containing elements from odd or even index positions from input list. Here is old fashion code:

Map<String, List<String>> map = new HashMap<>();

List<String> list = Arrays.asList("one", "two", "three", "four");

map.put("evenIndex", new ArrayList<>());
map.put("oddIndex", new ArrayList<>());
for (int i = 0; i < list.size(); i++) {
    if(i % 2 == 0)

How to transform this code into Java 8 using streams to get this result?

{evenIndex=[one, three], oddIndex=[two, four]}

My current messy attempt require modifying elements of list, but definitly must be better option.

List<String> listModified = Arrays.asList("++one", "two", "++three", "four");

map =
                               str -> str.startsWith("++") ? "evenIndex" : "oddIndex"));

Or maybe someone help me with this incorrect solution?

IntStream.range(0, list.size())
         .collect(Collectors.groupingBy( i -> i % 2 == 0 ? "even" : "odd",
                  Collectors.toMap( (i -> i ) , i -> list.get(i) ) )));

which return this:

{even={0=one, 2=three}, odd={1=two, 3=four}}


You were on the right track with streaming over the indexes:

import static;
import static;
import static;

                i -> i % 2 == 0 ? "even" : "odd", 
                mapping(list::get, toList())

If you are ok with having your map be indexed by a boolean you can use partitioningBy:

IntStream.range(0, list.size())
                i -> i % 2 == 0, 
                mapping(list::get, toList())

Source: stackoverflow