Question

我试图创建一个功能，告诉用户他们所在房间的出口。每个子列表都是一个房间。所有非零数字都是我的二维列表中每个子列表的出口。所有零都不是退出。我一直得到输出＆＃34;当我的地图的第一个值是$this->Auth->user('username');时，连续4次没有退出到0号房间＆＃34; 它显然有2和3哪个有出口。我想知道我做错了什么？

[0, 2, 3, 0]

Answer 1

删除elif条件并使用else，因为没有其他可能的答案。除此之外

map=[[0, 2, 3, 0], [0, 0, 4, 1], [1, 4, 0, 0], [2, 0, 0, 3], [0, 2, 3, 0], [0, 0, 4, 1], [1, 4, 0, 0], [2, 0, 0, 3], [0, 2, 3, 0], [0, 0, 4, 1], [1, 4, 0, 0], [2, 0, 0, 3], [0, 2, 3, 0], [0, 0, 4, 1], [1, 4, 0, 0], [2, 0, 0, 3]]

class Player: 
    def __init__(self, room_number = 0): 
        self.room_number = room_number

    def exit_count(self):
        for n in range(0,4):
            # print n
            # print map[self.room_number][n]
            if map[self.room_number][n] == 0:
                print("There is no exit to room {0}\n".format(self.room_number))
            else:
                print("There is an exit to room {0}\n".format(self.room_number))

p = Player(room_number=4)

p.exit_count()

奏

There is no exit to room 4

There is an exit to room 4

There is an exit to room 4

There is no exit to room 4

[Finished in 0.0s]

Answer 2

只需使用以下

即可

def findexits(rooms):
    for room in rooms:
        yield [door for door in room if door]

exits = findexits(map) # this is a generator
# list(exits) => [[2,3], [..],...]
# If you wish to calculate the number of exits, employ the generator
# before consuming
exit_counts = [len(exit_) for exit_ in exits]
# this consumes it.

您可以通过各种方式修改findexits。

对于具有2维列表的循环

2 个答案: