Reverse Sort a stream

up vote
7
down vote

favorite

I want to reverse sort a stream such as below but getting compile time error as "The method sorted() in the type IntStream is not applicable for the arguments (( o1, o2) -> {})". Can anyone correct this

IntStream.range(1, 100)

         .filter(x -> x%2 != 0)

         .sorted((o1,o2) -> -o1.compareTo(o2))

         .forEach(System.out::println);

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

I'm curious as to why you would use streams for something like this. It would seem to only add a bunch of overhead.
– JimmyJames
7 mins ago

add a comment |

up vote
7
down vote

favorite

IntStream.range(1, 100)

         .filter(x -> x%2 != 0)

         .sorted((o1,o2) -> -o1.compareTo(o2))

         .forEach(System.out::println);

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

I'm curious as to why you would use streams for something like this. It would seem to only add a bunch of overhead.
– JimmyJames
7 mins ago

add a comment |

up vote
7
down vote

favorite

IntStream.range(1, 100)

         .filter(x -> x%2 != 0)

         .sorted((o1,o2) -> -o1.compareTo(o2))

         .forEach(System.out::println);

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

IntStream.range(1, 100)

         .filter(x -> x%2 != 0)

         .sorted((o1,o2) -> -o1.compareTo(o2))

         .forEach(System.out::println);

java sorting java-8 java-stream

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

edited 1 hour ago

Nicholas K

4,85641031

edited 1 hour ago

Nicholas K

4,85641031

edited 1 hour ago

Nicholas K

4,85641031

asked 2 hours ago

Sundresh

612

asked 2 hours ago

Sundresh

612

asked 2 hours ago

Sundresh

612

I'm curious as to why you would use streams for something like this. It would seem to only add a bunch of overhead.
– JimmyJames
7 mins ago

add a comment |

I'm curious as to why you would use streams for something like this. It would seem to only add a bunch of overhead.
– JimmyJames
7 mins ago

I'm curious as to why you would use streams for something like this. It would seem to only add a bunch of overhead.
– JimmyJames
7 mins ago

add a comment |

3 Answers
3

active

oldest

votes

up vote
8
down vote

since you're working with an IntStream it only has one overload of the sorted method and it's the natural order (which makes sense).

instead, box the stream from IntStream to Stream<Integer> then it should suffice:

 IntStream.range(1, 100)

          .filter(x -> x % 2 != 0)

          .boxed()  // <--- boxed to Stream<Integer>

          .sorted((o1,o2) -> -o1.compareTo(o2)) // now we can call compareTo on Integer

          .forEach(System.out::println);

A better approach would be:

IntStream.range(1, 100)

         .filter(x -> x % 2 != 0)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

why?

There's already a built-in comparator to perform reverse order as shown above.

if you want to guarantee that the elements are to be seen in the sorted order when printing then utilise forEachOrdered (big shout out to @Holger for always reminding me)

btw, in JDK9 you can simplify this to:

IntStream.iterate(1, i  -> i <= 99, i -> i + 2)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

With this approach, we can avoid the filter intermediate operation and increment in 2's.

and finally you could simplify it even further with:

IntStream.iterate(99, i  -> i > 0 , i -> i - 2)

         .forEachOrdered(System.out::println);

With this approach, we can avoid filter, boxed,sorted et al.

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

add a comment |

up vote
4
down vote

If that is your real code, then it may be more efficient to use IntStream.iterate and generate numbers from 99 to 0:

IntStream.iterate(99, i -> i - 1)

    .limit(100)

    .filter(x -> x % 2 != 0)

    .forEachOrdered(System.out::println);

answered 2 hours ago

ernest_k

18.8k41838

add a comment |

up vote
2
down vote

How about simple util such as :

private IntStream reverseSort(int from, int to) {

    return IntStream.range(from, to)

                    .filter(x -> x % 2 != 0)

                    .sorted()

                    .map(i -> to - i + from - 1);

}

Credits: Stuart Marks for the reverse util.

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
8
down vote

since you're working with an IntStream it only has one overload of the sorted method and it's the natural order (which makes sense).

instead, box the stream from IntStream to Stream<Integer> then it should suffice:

 IntStream.range(1, 100)

          .filter(x -> x % 2 != 0)

          .boxed()  // <--- boxed to Stream<Integer>

          .sorted((o1,o2) -> -o1.compareTo(o2)) // now we can call compareTo on Integer

          .forEach(System.out::println);

A better approach would be:

IntStream.range(1, 100)

         .filter(x -> x % 2 != 0)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

why?

There's already a built-in comparator to perform reverse order as shown above.

if you want to guarantee that the elements are to be seen in the sorted order when printing then utilise forEachOrdered (big shout out to @Holger for always reminding me)

btw, in JDK9 you can simplify this to:

IntStream.iterate(1, i  -> i <= 99, i -> i + 2)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

With this approach, we can avoid the filter intermediate operation and increment in 2's.

and finally you could simplify it even further with:

IntStream.iterate(99, i  -> i > 0 , i -> i - 2)

         .forEachOrdered(System.out::println);

With this approach, we can avoid filter, boxed,sorted et al.

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

add a comment |

up vote
8
down vote

since you're working with an IntStream it only has one overload of the sorted method and it's the natural order (which makes sense).

instead, box the stream from IntStream to Stream<Integer> then it should suffice:

 IntStream.range(1, 100)

          .filter(x -> x % 2 != 0)

          .boxed()  // <--- boxed to Stream<Integer>

          .sorted((o1,o2) -> -o1.compareTo(o2)) // now we can call compareTo on Integer

          .forEach(System.out::println);

A better approach would be:

IntStream.range(1, 100)

         .filter(x -> x % 2 != 0)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

why?

There's already a built-in comparator to perform reverse order as shown above.

if you want to guarantee that the elements are to be seen in the sorted order when printing then utilise forEachOrdered (big shout out to @Holger for always reminding me)

btw, in JDK9 you can simplify this to:

IntStream.iterate(1, i  -> i <= 99, i -> i + 2)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

With this approach, we can avoid the filter intermediate operation and increment in 2's.

and finally you could simplify it even further with:

IntStream.iterate(99, i  -> i > 0 , i -> i - 2)

         .forEachOrdered(System.out::println);

With this approach, we can avoid filter, boxed,sorted et al.

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

add a comment |

up vote
8
down vote

since you're working with an IntStream it only has one overload of the sorted method and it's the natural order (which makes sense).

instead, box the stream from IntStream to Stream<Integer> then it should suffice:

 IntStream.range(1, 100)

          .filter(x -> x % 2 != 0)

          .boxed()  // <--- boxed to Stream<Integer>

          .sorted((o1,o2) -> -o1.compareTo(o2)) // now we can call compareTo on Integer

          .forEach(System.out::println);

A better approach would be:

IntStream.range(1, 100)

         .filter(x -> x % 2 != 0)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

why?

There's already a built-in comparator to perform reverse order as shown above.

if you want to guarantee that the elements are to be seen in the sorted order when printing then utilise forEachOrdered (big shout out to @Holger for always reminding me)

btw, in JDK9 you can simplify this to:

IntStream.iterate(1, i  -> i <= 99, i -> i + 2)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

With this approach, we can avoid the filter intermediate operation and increment in 2's.

and finally you could simplify it even further with:

IntStream.iterate(99, i  -> i > 0 , i -> i - 2)

         .forEachOrdered(System.out::println);

With this approach, we can avoid filter, boxed,sorted et al.

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

since you're working with an IntStream it only has one overload of the sorted method and it's the natural order (which makes sense).

instead, box the stream from IntStream to Stream<Integer> then it should suffice:

 IntStream.range(1, 100)

          .filter(x -> x % 2 != 0)

          .boxed()  // <--- boxed to Stream<Integer>

          .sorted((o1,o2) -> -o1.compareTo(o2)) // now we can call compareTo on Integer

          .forEach(System.out::println);

A better approach would be:

IntStream.range(1, 100)

         .filter(x -> x % 2 != 0)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

why?

There's already a built-in comparator to perform reverse order as shown above.

if you want to guarantee that the elements are to be seen in the sorted order when printing then utilise forEachOrdered (big shout out to @Holger for always reminding me)

btw, in JDK9 you can simplify this to:

IntStream.iterate(1, i  -> i <= 99, i -> i + 2)

         .boxed()

         .sorted(Comparator.reverseOrder())

         .forEachOrdered(System.out::println);

With this approach, we can avoid the filter intermediate operation and increment in 2's.

and finally you could simplify it even further with:

IntStream.iterate(99, i  -> i > 0 , i -> i - 2)

         .forEachOrdered(System.out::println);

With this approach, we can avoid filter, boxed,sorted et al.

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

edited 1 hour ago

answered 2 hours ago

Aomine

36k62960

answered 2 hours ago

Aomine

36k62960

answered 2 hours ago

Aomine

36k62960

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

add a comment |

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

When we are at usual reminders, never use a comparator function like (o1,o2) -> -o1.compareTo(o2). It is perfectly legal for a compareTo implementation to return Integer.MIN_VALUE, in which case negation will fail and produce inconsistent results. A valid reverse comparator would be (o1,o2) -> o2.compareTo(o1), or just Comparator.reverseOrder() as you have shown, which does already the job right. By the way, I wouldn’t use iterate but rather IntStream.range(0, 50).map(i -> 99-i*2) which is more efficient in a lot of cases.
– Holger
7 mins ago

add a comment |

up vote
4
down vote

If that is your real code, then it may be more efficient to use IntStream.iterate and generate numbers from 99 to 0:

IntStream.iterate(99, i -> i - 1)

    .limit(100)

    .filter(x -> x % 2 != 0)

    .forEachOrdered(System.out::println);

answered 2 hours ago

ernest_k

18.8k41838

add a comment |

up vote
4
down vote

If that is your real code, then it may be more efficient to use IntStream.iterate and generate numbers from 99 to 0:

IntStream.iterate(99, i -> i - 1)

    .limit(100)

    .filter(x -> x % 2 != 0)

    .forEachOrdered(System.out::println);

answered 2 hours ago

ernest_k

18.8k41838

add a comment |

up vote
4
down vote

If that is your real code, then it may be more efficient to use IntStream.iterate and generate numbers from 99 to 0:

IntStream.iterate(99, i -> i - 1)

    .limit(100)

    .filter(x -> x % 2 != 0)

    .forEachOrdered(System.out::println);

answered 2 hours ago

ernest_k

18.8k41838

If that is your real code, then it may be more efficient to use IntStream.iterate and generate numbers from 99 to 0:

IntStream.iterate(99, i -> i - 1)

    .limit(100)

    .filter(x -> x % 2 != 0)

    .forEachOrdered(System.out::println);

answered 2 hours ago

ernest_k

18.8k41838

answered 2 hours ago

ernest_k

18.8k41838

answered 2 hours ago

ernest_k

18.8k41838

answered 2 hours ago

ernest_k

18.8k41838

add a comment |

up vote
2
down vote

How about simple util such as :

private IntStream reverseSort(int from, int to) {

    return IntStream.range(from, to)

                    .filter(x -> x % 2 != 0)

                    .sorted()

                    .map(i -> to - i + from - 1);

}

Credits: Stuart Marks for the reverse util.

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

add a comment |

up vote
2
down vote

How about simple util such as :

private IntStream reverseSort(int from, int to) {

    return IntStream.range(from, to)

                    .filter(x -> x % 2 != 0)

                    .sorted()

                    .map(i -> to - i + from - 1);

}

Credits: Stuart Marks for the reverse util.

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

add a comment |

up vote
2
down vote

How about simple util such as :

private IntStream reverseSort(int from, int to) {

    return IntStream.range(from, to)

                    .filter(x -> x % 2 != 0)

                    .sorted()

                    .map(i -> to - i + from - 1);

}

Credits: Stuart Marks for the reverse util.

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

How about simple util such as :

private IntStream reverseSort(int from, int to) {

    return IntStream.range(from, to)

                    .filter(x -> x % 2 != 0)

                    .sorted()

                    .map(i -> to - i + from - 1);

}

Credits: Stuart Marks for the reverse util.

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

edited 1 hour ago

answered 1 hour ago

nullpointer

38.3k1073146

answered 1 hour ago

nullpointer

38.3k1073146

answered 1 hour ago

nullpointer

38.3k1073146

add a comment |

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