为什么past.builtins地图行为错误？

这是他们实施map的错误。这是他们的代码：

def oldmap(func, \*terables):
    """
    map(function, sequence[, sequence, ...]) -> list
    Return a list of the results of applying the function to the
    items of the argument sequence(s).  If more than one sequence is
    given, the function is called with an argument list consisting of
    the corresponding item of each sequence, substituting None for
    missing values when not all sequences have the same length.  If
    the function is None, return a list of the items of the sequence
    (or a list of tuples if more than one sequence).
    Test cases:
    >>> oldmap(None, 'hello world')
    ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
    >>> oldmap(None, range(4))
    [0, 1, 2, 3]
    More test cases are in past.tests.test_builtins.
    """
    zipped = itertools.zip_longest(\*terables)
    l = list(zipped)
    if len(l) == 0:
        return []
    if func is None:
        result = l
    else:
        result = list(starmap(func, l))

    # Inspect to see whether it's a simple sequence of tuples
    try:
        if max([len(item) for item in result]) == 1:
            return list(chain.from_iterable(result))
        # return list(flatmap(func, result))
    except TypeError as e:
        # Simple objects like ints have no len()
        pass
    return result

错误的地方是：

# Inspect to see whether it's a simple sequence of tuples

在其实现中，如果callable返回一个len值为1的对象列表，则将这些对象“解压缩”并返回一个扁平列表。我不确定这是哪里来的，因为据我所知，Python 2甚至对于元组都没有这样做：

# Python 2
print(map(lambda x: (x,), [1, 2]))
# [(1,), (2,)]

如果您想继续，那么在库代码存储库中似乎存在一个未解决的问题。

回答来源：stackoverflow