如果所有值(在该行或列中)为None,则从2D列表中删除行或列

Question

我有一个网格(6行,5列):

grid = [
        [None, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, None, None],
        ]

我增加了网格,它可能变成如下:

grid = [
        [{"some" : "thing"}, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, None, None],
        [None, None, None, {"something" : "else"}, None],
        [None, {"another" : "thing"}, None, None, None],
        [None, None, None, None, None],
        ]

我想删除包含所有内容 None的整个行和列.所以在前面的代码中,网格将转换为:

grid = [
        [{"some" : "thing"}, None, None],
        [None, None, {"something" : "else"}],
        [None, {"another" : "thing"}, None],
        ]

我删除了第1,2,5行(零索引)和第2列和第4列.

我现在删除行的方式:

for row in range(6):
    if grid[row] == [None, None, None, None, None]:
        del grid[row] 

我还没有一个像样的删除None列的方法.这样做有"pythonic"方式吗？

Answer 1

这不是最快的方式,但我认为这很容易理解:

def transpose(grid):
    return zip(*grid)

def removeBlankRows(grid):
    return [list(row) for row in grid if any(row)]

print removeBlankRows(transpose(removeBlankRows(transpose(grid))))

输出:

[[{'some': 'thing'}, None, None],
 [None, None, {'something': 'else'}],
 [None, {'another': 'thing'}, None]]

它是如何工作的:我zip用来编写一个转换行和列的函数.第二个函数removeBlankRows删除所有元素都为None的行(或在布尔上下文中计算为false的任何行).然后执行整个操作我转置网格,删除空白行(这是原始数据中的列),再次转置,然后删除空行.

如果仅删除None而不是其他评估为false的内容很重要,请将removeBlankRows函数更改为:

def removeBlankRows(grid):
    return [list(row) for row in grid if any(x is not None for x in row)]