我需要合并两个迭代器。我写了这个功能:合并两个已排序迭代器而不替换
def merge_no_repeat(iter1, iter2, key=None):
"""
a = iter([(2, 'a'), (4, 'a'), (6, 'a')])
b = iter([(1, 'b'), (2, 'b'), (3, 'b'), (4, 'b'), (5, 'b'), (6, 'b'), (7, 'b'), (8, 'b')])
key = lambda item: item[0]
fusion_no_repeat(a, b, key) ->
iter([(1, 'b'), (2, 'a'), (3, 'b'), (4, 'a'), (5, 'b'), (6, 'a'), (7, 'b'), (8, 'b')])
:param iter1: sorted iterator
:param iter2: sorted iterator
:param key: lambda get sorted key, default: lambda x: x
:return: merged iterator
"""
if key is None:
key = lambda x: x
element1 = next(iter1, None)
element2 = next(iter2, None)
while element1 is not None or element2 is not None:
if element1 is None:
yield element2
element2 = next(iter2, None)
elif element2 is None:
yield element1
element1 = next(iter1, None)
elif key(element1) > key(element2):
yield element2
element2 = next(iter2, None)
elif key(element1) == key(element2):
yield element1
element1 = next(iter1, None)
element2 = next(iter2, None)
elif key(element1) < key(element2):
yield element1
element1 = next(iter1, None)
这个函数有效。但我认为这太复杂了。是否有可能使用Python标准库使这个函数最简单?
['itertools.chain'](https://docs.python.org/2/库/ itertools.html#itertools.chain)? –
这个问题可能更适合http://codereview.stackexchange.com – jonrsharpe
@MathiasEttinger itertools.chain不会返回一个已排序的迭代器。 –