计算一个元组的可数的行和列总数

Question

这需要在计算结果时yield。不应在任何时候存储所有数据。这应该支持大于内存的数据流。你知道吗

For each row add an int at the beginning that is the total of the row.

Once the entire input has been processed add a final row with the totals of every column in the input. This should include the initial total column and you should treat columns that are missing on a given row as zeros.

The row totals are the first column instead of the last (as is more common) because it makes rows of different length easier to handle.

例如：

    def func([(1,2,3), (4,5)]) = [(6,1,2,3),(9,4,5),(15,5,7,3)]

Answer 1

希望你能从中学到一些东西：

from itertools import izip_longest

def func(rows):
    totals = []
    for row in rows:
        row = (sum(row),) + row
        totals = [sum(col) for col in izip_longest(totals, row, fillvalue=0)]
        yield row
    yield tuple(totals)

>>> list(func([(1,2,3), (4,5)]))
[(6, 1, 2, 3), (9, 4, 5), [15, 5, 7, 3]]

此代码遍历所有行，生成一个元组，元组由求和列和原始列组成。你知道吗

^{}将当前行中的项与totals中相应的项配对，以保持每列的运行总数。^选择{}是因为它可以处理不同长度的行，并且可以为缺少的项提供填充值（本例中为0）。你知道吗