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Removing Odd Numbers

 
Greenhorn
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I am currently writing a program that lists numbers the user enters and totals the sums of all, even and odd using an array. When calculating the sum of the even numbers the odd numbers in the list should be removed. Here is the code:



I am able to add the even numbers but I am having trouble remove the odd numbers. Any tips?
 
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What problem you are facing?
 
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You don't have to remove the numbers, all you have to do is use IntStream.filter() and IntStream.sum() twice.

If you want to do this in one pass, you can also use one loop over the numbers array, and keep two accumulator variables for the odd and even numbers.
 
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Can you partition the stream into two and add odd and even in one pass?

Duplicating discussion in our Java8 forum.
 
Stephan van Hulst
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Campbell Ritchie wrote:Can you partition the stream into two and add odd and even in one pass?



Not easily. Since you can only call one terminal operation per stream, you'd have to extend IntStream with a reduceIf(IntPredicate, int, IntBinaryOperator) method, that returns a ConditionallyReducingIntStream, which has an elseReduce(int, IntBinaryOperator) method, which returns an IntStream of reduced values on which you can perform the terminal operation.

For a concrete example, check this post: https://coderanch.com/t/631548/java/java/Lambda-FizzBuzz#3028752

In short, this is a pain.
 
Stephan van Hulst
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It would look like this:
 
Campbell Ritchie
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But you do need your extended class with the else method.
 
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What's wrong with using Collectors.partitioningBy(predicate, reducer)?
 
Campbell Ritchie
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Otherwise you would end up with some horror like this?I presume you have come across the bitwise and operator as an alternative to the remainder operator? Once you have tried Stream operations with int[] arrays, you will start cursing the very existence of arrays of primitives. I am assuming that numbers is a List<Integer>, in which case how do you get Integer::parseInt to compile? I am allowing the compiler to do its own unnboxing rather than trying to do it explicitly with map methods.

Can you use Collectors#groupingBy to create a Map<Integer, Integer> with remainders and totals?
 
Stephan van Hulst
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The problem with using a collector is that you still have to pass the elements twice, which is what Campbell's initial question was about (I think). Once to partition, and once to process.

Campbell, I'm not sure you code compiles, because your lambda closes over non-final variables it attempts to assign values to.

As for primitive arrays and primitive streams:
 
Campbell Ritchie
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You are right; the sums are non‑final, so that won't work.
 
Campbell Ritchie
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You can create a local Sum class which is mutable so its instances can be final. You can get that to compile, surely. You too can get -99% for code style
 
Stephan van Hulst
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Ew :P
 
Campbell Ritchie
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Stephan van Hulst wrote:Ew :P

Is that worse than my -99%?
 
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Piet Souris wrote:What's wrong with using Collectors.partitioningBy(predicate, reducer)?


I think this is the best advice so far. Untested:
 
Stephan van Hulst
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Forgot all about summingInt()!

Piet was right.
 
Campbell Ritchie
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Partitioning rather than grouping? Of course. Well done PS
 
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