Question

from collections.abc import Sequence

class Map(Sequence):
    """ Represents a map for a floor as a matrix """

    def __init__(self, matrix):
        """ Takes a map as a matrix """
        self.matrix = matrix
        self.height = len(matrix)
        self.width = len(matrix[0])
        super().__init__()

    def __getitem__(self, item):
        """ Needed by Sequence """
        return self.matrix[item]

    def __len__(self):
        """ Needed by Sequence """
        return len(self.matrix)

    def search(self, entity):
        """ Returns a generator of tuples that contain the x and y for every element in the map that matches 'entity' """
        for row in range(self.height):
            for column in range(self.width):
                if matrix[row][column] == entity:
                    yield (row, column)


# Examples

gmap = Map([[0, 0, 0],
           [0, 1, 0],
           [0, 0, 0]])

for entity in gmap:
    print(entity)

如何实施__iter__以便

for entity in gmap:
    print(entity)

产生0 0 0 0 1 0 0 0 0而不是

[0, 0, 0]
[0, 1, 0]
[0, 0, 0]

这样可以避免我需要子类Sequence，并且会使search()的代码更整洁

此外，他们还应该使用其他任何魔术方法吗？（除__str__之外，我在迭代工作后就这样做了）

Answer 1

您可以像这样实施__iter__()：

from itertools import chain

def __iter__(self):
    return chain.from_iterable(self.matrix)

itertools.chain.from_iterable()采用可迭代的迭代次数并将它们组合在一起。它创建了一个生成器，因此不会使用额外的内存。

迭代矩阵类的Python

1 个答案: