Java filter with lambda

Question

I'm trying to make a filter with lambda. There is a list of string, which contains values, that should be sorted out and there's the other list, which contains different values (that's the one, that should be filtered). So what I've been trying to do is:

stringList = stringList.stream()
       .filter(e ->toBeSortedOutList.forEach(outSorted->!e.startsWith(outSorted)))
       .collect(Collectors.toList());

Unfortunately, I am allways getting the error, that boolean can't be converted to void. So what I wanted to know is, why that I am getting this error and if it's possible to filter by iterating trough a list and using its values for the filter? Thanks for your answers!

What exactly do you want to do? Could you provide a sample input / output? — Tunaki
– Tunaki, Commented Jan 11, 2016 at 9:46
Could you provide an example of such lists and the output you're trying to get for them? — Mureinik
– Mureinik, Commented Jan 11, 2016 at 9:47
Generally, you should try to find out, what a method does, before using it. There is no reason to assume that a void method, like forEach, can be used as a predicate. I hope, you've already heard about that thing called API documentation — Holger
– Holger, Commented Jan 11, 2016 at 9:53

Tagir Valeev · Accepted Answer · 2016-01-11 09:47:05Z

8

Use noneMatch:

stringList = stringList.stream()
   .filter(e ->toBeSortedOutList.stream().noneMatch(outSorted->e.startsWith(outSorted)))
   .collect(Collectors.toList());

In general avoid using forEach unless you need an explicit side-effect (like output to System.out). Usually there are better alternatives to solve your problem.

By the way you can also modify your list in-place using removeIf:

stringList.removeIf(e -> toBeSortedOutList.stream()
                         .anyMatch(outSorted->e.startsWith(outSorted)));

answered Jan 11, 2016 at 9:47

Tagir Valeev

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2 Comments

Misha Over a year ago

outSorted->e.startsWith(outSorted)) can also be expressed as a method reference e::startsWith

Tagir Valeev Over a year ago

@Misha, agreed, I would use method reference in my code. However this might look cryptic for the beginners.

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