Question

如何在我的程序中的addCoordinate函数的嵌套for循环中应用嵌套列表解析，而不影响输出。我已经得到它以输出中实现的格式打印迷宫，但我希望使我的整个程序更紧凑并提高其时间复杂度。

((* -> *) -> * -> *) ->
((* -> *) -> * -> *) ->
 (* -> *) -> * -> *

我的计划的输出是：

from itertools import product
class Maze:
    mazeboard, x1, y1, list = [], 0, 0, []

    def __init__(self):
        """
        Constructor - You may modify this, but please do not add any extra parameters
        """

    def addCoordinate(self, x, y, blockType):
        #print self.x1
        if self.x1 < x :
            self.x1 = x
        if self.y1 < y:
            self.y1 = y
        if  len(self.mazeboard) <= self.x1 or  len(self.mazeboard) <= self.y1:
            modified_board = [[1 for a in range(self.x1 + 1)] for b in range(self.y1 + 1)]
            for a in range(len(self.mazeboard)):
                for b in range(len(self.mazeboard[a])):
                    modified_board[a][b] = self.mazeboard[a][b]
            self.mazeboard = modified_board
        self.mazeboard[y][x] = blockType

    def printMaze(self):
        for a in range(self.x1 + 1):
            for b in range(self.y1 + 1):
                if self.mazeboard[a][b] == 0:
                    print " ",
                else:
                    print "*",
            print ""


def mazeTest():
    myMaze = Maze()
    myMaze.addCoordinate(1, 0, 0)
    myMaze.addCoordinate(1, 1, 0)
    myMaze.addCoordinate(7, 1, 0)
    myMaze.addCoordinate(1, 2, 0)
    myMaze.addCoordinate(2, 2, 0)
    myMaze.addCoordinate(3, 2, 0)
    myMaze.addCoordinate(4, 2, 0)
    myMaze.addCoordinate(6, 2, 0)
    myMaze.addCoordinate(7, 2, 0)
    myMaze.addCoordinate(4, 3, 0)
    myMaze.addCoordinate(7, 3, 0)
    myMaze.addCoordinate(4, 4, 0)
    myMaze.addCoordinate(7, 4, 0)
    myMaze.addCoordinate(3, 5, 0)
    myMaze.addCoordinate(4, 5, 0)
    myMaze.addCoordinate(7, 5, 0)
    myMaze.addCoordinate(1, 6, 0)
    myMaze.addCoordinate(2, 6, 0)
    myMaze.addCoordinate(3, 6, 0)
    myMaze.addCoordinate(4, 6, 0)
    myMaze.addCoordinate(6, 6, 0)
    myMaze.addCoordinate(7, 6, 0)
    myMaze.addCoordinate(5, 7, 0)
    myMaze.printMaze()
mazeTest()

Answer 1

您可以通过几种不同的方式改进：

创建坐标列表，而不是多次调用addCoordinate()。
通过将迷宫中的字典转换为*和字符，然后将它们全部合并为一个可打印的字符串，来改善您的迷宫打印。
使用product将旧版mazeboard复制到更新后的版本中。

例如：

from itertools import product

class Maze:
    mazeboard, x1, y1 = [], 0, 0

    def __init__(self):
        """
        Constructor - You may modify this, but please do not add any extra parameters
        """

    def addCoordinate(self, x, y, blockType):
        if self.x1 < x :
            self.x1 = x
        if self.y1 < y:
            self.y1 = y
        if  len(self.mazeboard) <= self.x1 or len(self.mazeboard) <= self.y1:
            modified_board = [[1 for a in range(self.x1 + 1)] for b in range(self.y1 + 1)]
            if len(self.mazeboard):
                for a, b in product(range(len(self.mazeboard)), range(len(self.mazeboard[0]))):
                    modified_board[a][b] = self.mazeboard[a][b]            

            self.mazeboard = modified_board
        self.mazeboard[y][x] = blockType

    def printMaze(self):
        print '\n'.join(' '.join(map({1:'*', 0:' '}.get, row)) for row in self.mazeboard)


def mazeTest():
    myMaze = Maze()

    coords = [
        (1, 0, 0), (1, 1, 0), (7, 1, 0), (1, 2, 0), (2, 2, 0), 
        (3, 2, 0), (4, 2, 0), (6, 2, 0), (7, 2, 0), (4, 3, 0), (7, 3, 0), 
        (4, 4, 0), (7, 4, 0), (3, 5, 0), (4, 5, 0), (7, 5, 0), (1, 6, 0), 
        (2, 6, 0), (3, 6, 0), (4, 6, 0), (6, 6, 0), (7, 6, 0), (5, 7, 0)]

    for coord in coords:
        myMaze.addCoordinate(*coord)

    myMaze.printMaze()

mazeTest()

在函数中应用嵌套列表推导

1 个答案: