如何减少列表中的列表

Question

我的直觉说降低（从功能上的导入降低）会提供我想要的东西，但我无法在这里使用它来缠绕我的头。

def test_reduce_lists_by_summing_them():
    """
    Sum each item from the first list with the same positional item from the subsequent list. 
    The result is the next first list and appended to the returned result list.
    """
    input_ = [[1, 0, 0], [0, 1, 0], [0, 0, 2], [3, 0, 0]]
    expected_output = [[1, 1, 0], [1, 1, 2], [4, 1, 2]]

    def f(x, y):
        """Return the sum of two positional items"""
        return x + y

    output = []
    for group in zip(input_[0:], input_[1:]):
        # Obvisously not working since the first list is not generated but just taken grp[0]
        output.append(list(map(f, group[0], group[1])))
    assert output == expected_output

Answer 1

您正在寻找的是累积：

from itertools import accumulate

# perhaps, convert it to a list explicitly
accumulate(input_, lambda x, y: list(map(sum, zip(x, y))))

# a more explicit equivalent:
list(accumulate(input_, lambda x, y: [x_+y_ for x_, y_ in zip(x, y)]))

如果您'愿意使用numpy，它将很简单：

np.array(input_).cumsum(axis=0)

upd demo：

In [36]: list(accumulate(input_, lambda x, y: list(map(sum, zip(x, y)))))
Out[36]: [[1, 0, 0], [1, 1, 0], [1, 1, 2], [4, 1, 2]]

In [37]: list(accumulate(input_, lambda x, y: [x_+y_ for x_, y_ in zip(x, y)]))
Out[37]: [[1, 0, 0], [1, 1, 0], [1, 1, 2], [4, 1, 2]]

In [38]: np.array(input_).cumsum(axis=0)
Out[38]: 
array([[1, 0, 0],
       [1, 1, 0],
       [1, 1, 2],
       [4, 1, 2]])