Java: Putting items from List into HashMap efficiently [duplicate]

Question

I have a list here:

List<QueryStrings> queryStrings

and QueryStrings is just a simple class

public class QueryStrings {
    private Integer id;
    private String rel;
    private String impl;
}

I need to put the List into a HashMap where the id will be the key, I am doing it like this right now, looping the List one item at a time:

HashMap<Integer, QueryStrings> queryMap = new HashMap<>();
for (QueryStrings element : queryStrings) {
    Integer thisId = Integer.parseInt(element.getId());
    queryMap.put(thisId, element);
}

Is there a better way to this? I do not want to loop each items, is that possible?

Edit:

Sorry I should not parse the integer, so the code should look like:

HashMap<Integer, QueryStrings> queryMap = new HashMap<>();
for (QueryStrings element : queryStrings) {
    queryMap.put(element.getId(), element);
}

I believe something like this would work in Java8 Map<Integer, QueryStrings> queryMap = queryStrings.stream().collect(toMap(element.getId(), element)); — J. Knight
– J. Knight, Commented Jun 29, 2017 at 4:54
Why call parseInt()? id is already of type Integer, just use the ID directly. — markspace
– markspace, Commented Jun 29, 2017 at 4:54
Minus the extra call as noted in the current answer, looping is as "efficient" as any other method. Define "better way". — user2864740
– user2864740, Commented Jun 29, 2017 at 4:58
This is not worth worrying about. You can populate millions of items into a HashMap per second, as long as the hashcodes aren't degenerate. — user207421
– user207421, Commented Jun 29, 2017 at 5:57

Ryan Leach · Accepted Answer · 2017-06-29 05:19:52Z

7

Don't parse the int's if it's already an int!

HashMap<Integer, QueryStrings> queryMap = new HashMap<>();
for (QueryStrings element : queryStrings) {
    queryMap.put(element.getId(), element);
}

That said, if it's Java 8 you can have more concise syntax, although the performance differences are minimal.

list.stream().collect(Collectors.toMap(QueryStrings::getId, Function.identity()));

edited Jun 29, 2017 at 5:19

answered Jun 29, 2017 at 4:54

Ryan Leach

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2 Comments

J. Knight Over a year ago

His class does show it as an Integer though

Scary Wombat Over a year ago

did not catch that - definitely a better way to do is not to go from Integer -> String -> Integer :-)

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