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Problem 2: Back and forth (100pts)

Problem

Implement backAndForth, a generator which takes in an iterator t and yields elements from t in multiple rounds.

In round \( i \), where \(i\) is a positive integer, backAndForth:

  1. yields consecutive \( i \) elements from t, if \( i \) is odd,
  2. skips consecutive \( i \) elements from t, if \( i \) is even.

实现 backAndForth，它是一个生成器，接受一个迭代器 t 并在多轮中产出 t 中的元素。

在第 \( i \) 轮中，其中 \(i\) 是一个正整数，backAndForth：

  1. 如果 \( i \) 是奇数，则产出 t 中连续的 \( i \) 个元素，
  2. 如果 \( i \) 是偶数，则跳过 t 中连续的 \( i \) 个元素。
def backAndForth(t):
    """Yields and skips elements from iterator t, back and forth.

    >>> list(backAndForth(iter([1, 2, 3, 4, 5, 6, 7, 8, 9])))
    [1, 4, 5, 6]
    >>> list(backAndForth(iter([1, 2, 2])))
    [1]
    >>> # generators allow us to represent infinite sequences!!!
    >>> def naturals():
    ...     i = 0
    ...     while True:
    ...         yield i
    ...         i += 1
    >>> m = backAndForth(naturals())
    >>> [next(m) for _ in range(9)]
    [0, 3, 4, 5, 10, 11, 12, 13, 14]
    """
    "*** YOUR CODE HERE ***"

Hints

  • 每次 next 后立刻 yield，避免遗漏。

Solutions

使用 try：

注意不要一次性使用 i 次 next，这可能遗漏最后一次输出。

def backAndForth(t):
    i = 1
    while True:
        try:
            for j in range(i):
                x = next(t)
                if i % 2: yield x
            i += 1
        except StopIteration: return