How to transform a for loop into stream loop

Question

How to transform this code into a stream loop:

for(long l = 1L; l <= 250000; l++) {
     v = value.add(BigInteger.valueOf(myMethod.getInt()));
}

I need to get the 'v' as a unique BigInteger value.

I just need to run myMethod.getInt() 25000 times. myMethod.getInt() return a random int. — Roland
– Roland, Commented Oct 30, 2018 at 11:50

Michael · Accepted Answer · 2018-10-30 13:36:12Z

4

Fundamentally, it looks like your myMethod.getInt method is a generator. Therefore, the best way to do this, in my opinion, is to create an infinite stream from your generator.

IntStream.generate(myMethod::getInt)
    .mapToObj(BigInteger::valueOf)
    .limit(25000)
    .reduce(BigInteger.ZERO, BigInteger::add)

This is clearer because you don't have to specify a range - the range is not what you care about, the number of elements is (i.e. the size of the range). You also don't have to ignore the parameter when you're mapping.

edited Oct 30, 2018 at 13:36

answered Oct 30, 2018 at 13:28

Michael

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Ilya Zinkovich · Accepted Answer · 2018-10-31 13:38:22Z

3

BigInteger result = IntStream.range(0, 25000)
  .map(i -> myMethod.getInt())
  .mapToObj(BigInteger::valueOf)
  .reduce(BigInteger.valueOf(0), BigInteger::add)

Another answer with IntStream.generate(myMethod::getInt) and limit is more elegant :)

edited Oct 31, 2018 at 13:38

answered Oct 30, 2018 at 12:15

Ilya Zinkovich

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9 Comments

Roland Over a year ago

With this code I was not able to get the final value of the sum. Like int value = IntStream.range(0, 25000) .map(i -> myMethod.getInt()) .mapToObj(BigInteger::valueOf) .map(value::add) and I got two errors: Error:(16, 96) java: incompatible types: long cannot be converted to java.math.BigDecimal Error:(16, 79) java: incompatible types: no instance(s) of type variable(s) U exist so that java.util.stream.Stream<U> conforms to java.math.BigDecimal

Ilya Zinkovich Over a year ago

@Roland if value collects the sum you can go with .reduce(BigInteger.valueOf(0), BigInteger::add)

Holger Over a year ago

Instead of BigInteger.valueOf(0), you can use BigInteger.ZERO. Alternative solution are IntStream.generate(myMethod::getInt).limit(25000).mapToObj(BigInteger::valueOf) … or Stream.generate(() -> BigInteger.valueOf(myMethod.getInt())).limit(25000) ….

Holger Over a year ago

@Michael nope, I posted it as suggestion anyway, not as an answer. Besides that, finding such canonical solutions simultaneously, is not unusual.

Holger Over a year ago

@Roland a) wrong benchmark methodology b) whatever happens inside myMethod.getInt() (e.g. using a synchronized random generator). If all this method does, is returning a random int, how about

BigInteger result = ThreadLocalRandom.current().ints(25000, [min, max]) [.parallel()] .mapToObj(BigInteger::valueOf).reduce(BigInteger.ZERO, BigInteger::add);

? Though, to sum up 25000 int values, you don’t need BigInteger anyway, as even the long value range is sufficient.

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