lag method

Iterable<Iterable<T>> lag(int offset, { required T defaultValue, })

Returns an iterable that projects the elements in this iterable upon other elements in this iterable offset backwards by a given value.

Every element in this iterable is paired with the element appearing in the iterable offset prior to it. Each pairing is then returned as an element in the resulting iterable.

For the first offset elements in the iterable, the element is paired with the value passed to defaultValue.

If the value of offset is zero, the iterable is unchanged. If it is less than zero, it is converted to the positive equivalent.

Example:

void main() {
  final list = <Object>['a', 'b', 'c', 'd'];
  final result = list.lag(2, defaultValue: 'e');

  // Result: [['a', 'e'], ['b', 'e'], ['c', 'a'], ['d', 'b']]
}

Implementation

Iterable<Iterable<T>> lag(
  int offset, {
  required T defaultValue,
}) sync* {
  final queue = Queue<T>();
  offset = offset < 0 ? -offset : offset;
  var index = 0;

  for (var o in this) {
    queue.add(o);
    if (index >= offset) {
      yield [o, queue.removeFirst()];
    } else {
      yield [o, defaultValue];
    }
    index++;
  }
}