如何减少列表中的列表

Question

我的直觉告诉我，reduce(来自functools import reduce)会提供我正在寻找的东西，但我无法在这里使用它。

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

你要找的是:

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)]))

如果你愿意使用numpy，它就像这样简单:

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

乌利希期刊指南演示:

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]])