Slicing

Taking a window of the sequence by position or by a predicate. Take and TakeWhile consume only what they emit, so either one bounds an infinite source; TakeLast and DropLast cannot, since they have to reach the end first.

Take

func (s Seq[T]) Take(n int) Seq[T]

Take yields at most the first n elements, consuming exactly as many as it yields. Panics if n is negative.

// Consumes exactly what it emits, so it bounds an infinite source.
fmt.Println(catena.Generate(1, func(n int) int { return n + 1 }).
  Take(3).
  Collect())

[1 2 3]

TakeWhile

func (s Seq[T]) TakeWhile(pred func(T) bool) Seq[T]

TakeWhile yields elements until pred first returns false.

// Stops at the first element that fails — unlike Filter, which would
// keep testing the rest.
fmt.Println(catena.Of(1, 2, 9, 3).
  TakeWhile(func(n int) bool { return n < 5 }).
  Collect())

[1 2]

TakeLast

func (s Seq[T]) TakeLast(n int) Seq[T]

TakeLast yields the final n elements. Panics if n is negative.

Caution

Buffers n elements and fully drains the source before emitting — hangs on infinite input

fmt.Println(catena.Range(1, 8, 1).TakeLast(3).Collect())

[5 6 7]

Drop

func (s Seq[T]) Drop(n int) Seq[T]

Drop skips the first n elements. Panics if n is negative.

fmt.Println(catena.Of("a", "b", "c", "d").Drop(2).Collect())

[c d]

DropWhile

func (s Seq[T]) DropWhile(pred func(T) bool) Seq[T]

DropWhile skips elements until pred first returns false, then yields the rest.

// Drops only the leading run; once the predicate fails, everything
// after is kept.
fmt.Println(catena.Of(0, 0, 3, 0, 5).
  DropWhile(func(n int) bool { return n == 0 }).
  Collect())

[3 0 5]

DropLast

func (s Seq[T]) DropLast(n int) Seq[T]

DropLast yields all but the final n elements, emitting with an n-element lag. Panics if n is negative.

Caution

Buffers n elements

fmt.Println(catena.Range(1, 6, 1).DropLast(2).Collect())

[1 2 3]

Step

func (s Seq[T]) Step(n int) Seq[T]

Step yields the first element and every nth element after it. Panics if n <= 0.

// The first element always survives, then every nth after it.
fmt.Println(catena.Range(0, 10, 1).Step(3).Collect())

[0 3 6 9]