CODEWITHSUNDEEP
javafunctional-programmingjava-8functional-java
Nir Ben Yaacov
Nir Ben Yaacov

Reputation: 1182

Filtering on Java 8 List

I have a list of type List in functional java using List type provided by fj.data.List

import fj.data.List

List<Long> managedCustomers

I am trying to filter it using the following:

managedCustomers.filter(customerId -> customerId == 5424164219L)

I get this message

enter image description here

According to documentation, List has a filter method and this should work http://www.functionaljava.org/examples-java8.html

What am I missing?

Thanks

Upvotes: 4

Views: 2443

Answers (3)

azro
azro

Reputation: 54148

What you did seem a bit weird, Streams (to use filter) are commonly used like this (I don't know what you really want to do with the filtrate list, you can tell me in the comment tp get a more precise answer) : 

//Select and print
managedCustomers.stream().filter(customerId -> customerId == 5424164219L)
                         .forEach(System.out::println);

//Select and keep
ArrayList<> newList = managedCustomers.stream().filter(customerId -> customerId == 5424164219L)
                         .collect(Collectors.toList());

Upvotes: 4

Naman
Naman

Reputation: 31888

As already pointed out in the comment by @Alexis C 

managedCustomers.removeIf(customerId -> customerId != 5424164219L);

should get you the filtered list if the customerId equals 5424164219L.

Edit - The above code modifies the existing managedCustomers removing the other entries. And also the other way to do so is using the stream().filter() as - 

managedCustomers.stream().filter(mc -> mc == 5424164219L).forEach(//do some action thee after);

Edit 2 -

For the specific fj.List, you can use - 

managedCustomers.toStream().filter(mc -> mc == 5424164219L).forEach(// your action);

Upvotes: 5

Peter Lawrey
Peter Lawrey

Reputation: 533530

A lambda determines it's types by context. When you have a statement which doesn't compile, the javac sometimes gets confused and complains your lambda won't compile when the real reason is you have made some other mistake which is why it can't workout what the type of your lambda should be.

In this case, there is no List.filter(x) method, which is the only error you should see because unless you fix that your lambda is never going to make sense.

In this case, rather than use filter, you could use anyMatch as you already know there is only one possible value where customerId == 5424164219L

if (managedCustomers.stream().anyMatch(c -> c == 5424164219L) {
    // customerId 5424164219L found
}

Upvotes: 1

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