python-3.x - 如何用鼠标绘制并保存为 1 和 0-6ren

python-3.x - 如何用鼠标绘制并保存为 1 和 0

转载作者：行者123 更新时间：2023-12-04 11:00:18

我已经在我的主游戏中实现了自定义关卡代码。但我需要一个关卡编辑器，以便更轻松地制作自定义关卡。我使用按钮来覆盖屏幕，当点击时，它会放置一个块(下面打印是一个测试，如果你将代码与我的主要问题结合起来，它会对我有很大帮助。奇怪的是，它不会打印和滞后，所以我需要代码来绘制和导出为我需要的格式(代码中的格式)。

import sys
pygame.init()
screen = pygame.display.set_mode((1280, 720))
white = (255,255,255)
grey = (128,128,128)
class button():
    def __init__(self, color, x,y,width,height, text=''):
        self.color = color
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.text = text

    def draw(self,win, outline=None):
        #Call this method to draw the button on the screen

        pygame.draw.rect(win, self.color, (self.x,self.y,self.width,self.height),0)

        if self.text != '':
            font = pygame.font.SysFont('comicsans', 60)
            text = font.render(self.text, 1, (0,0,0))
            screen.blit(text, (round(self.x) + (round(self.width/2) - round(text.get_width()/2)), round(self.y) + (round(self.height/2) - round(text.get_height()/2))))

    def isOver(self, pos):
        #Pos is the mouse position or a tuple of (x,y) coordinates
        if pos[0] > self.x and pos[0] < self.x + self.width:
            if pos[1] > self.y and pos[1] < self.y + self.height:
                return True

        return False

WIDTH = 64
HEIGHT = 36
TILESIZE = 20
clock = pygame.time.Clock()
#the format i need it in
tm = [
      [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
      [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
    ]

buttonD = button((0,255,0), 640, 360, 20, 20, ' ')
while True:
    pygame.display.update()
    screen.fill(white)  
    for row in range(HEIGHT):
        for column in range(WIDTH):
            if tm[row][column] == 1:
                pygame.draw.rect(screen, grey, [column*TILESIZE, row*TILESIZE, TILESIZE, TILESIZE])
                wallRect = pygame.Rect(column*TILESIZE, row*TILESIZE, TILESIZE, TILESIZE)   
            if tm[row][column] == 0:
                buttonD = button((255,255,255), column*TILESIZE, row*TILESIZE, TILESIZE, TILESIZE, ' ')
                buttonD.draw(screen, (0,0,0))
    for event in pygame.event.get():
        pos = pygame.mouse.get_pos()
        if event.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
        if event.type == pygame.MOUSEBUTTONDOWN:
            if buttonD.isOver(pos):
                print('Hi')
clock.tick(60)

我希望它打印嗨，但它不打印和卡住

最佳答案

为什么需要按钮？

计算鼠标按下时单元格的索引:

column = event.pos[0] // TILESIZE
row = event.pos[1] // TILESIZE

并切换单元格的状态

tm[row][column] = 1 if tm[row][column] == 0 else 0

例如。:

while True:

    for event in pygame.event.get():
        pos = pygame.mouse.get_pos()
        if event.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
        if event.type == pygame.MOUSEBUTTONDOWN:
            column = event.pos[0] // TILESIZE
            row = event.pos[1] // TILESIZE
            if row < len(tm) and column < len(tm[row]):
                tm[row][column] = 1 if tm[row][column] == 0 else 0

    screen.fill(white)  
    for row in range(HEIGHT):
        for column in range(WIDTH):
            c = grey if tm[row][column] == 1 else (255,255,255)
            pygame.draw.rect(screen, c, [column*TILESIZE, row*TILESIZE, TILESIZE, TILESIZE])

    pygame.display.update()
    clock.tick(60)

此外，您可以突出显示“鼠标所在的单元格并绘制包含单元格行和列的文本:

class button():
    def __init__(self, color, x,y,width,height, text=''):
        self.color = color
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.font = font = pygame.font.SysFont('comicsans', 20)

    def draw(self, win, x, y):
        pygame.draw.rect(win, self.color, (self.x,self.y,self.width,self.height),0)
        text = self.font.render(str(x) + " / " + str(y), 1, (0,0,0))
        screen.blit(text, (round(self.x) + (round(self.width/2) - round(text.get_width()/2)), round(self.y) + (round(self.height/2) - round(text.get_height()/2))))

buttonD = button((0,255,0), 640, 360, 20, 20, ' ')
while True:

    for event in pygame.event.get():
        pos = pygame.mouse.get_pos()
        if event.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
        if event.type == pygame.MOUSEBUTTONDOWN:
            column, row = event.pos[0] // TILESIZE, event.pos[1] // TILESIZE
            if row < len(tm) and column < len(tm[row]):
                tm[row][column] = 1 if tm[row][column] == 0 else 0

    mouse_pos = pygame.mouse.get_pos()
    mcol, mrow = mouse_pos[0] // TILESIZE, mouse_pos[1] // TILESIZE
    buttonD.x, buttonD.y = mcol*TILESIZE, mrow*TILESIZE

    screen.fill(white)  
    for row in range(HEIGHT):
        for column in range(WIDTH):
            c = grey if tm[row][column] == 1 else (255,255,255)
            pygame.draw.rect(screen, c, [column*TILESIZE, row*TILESIZE, TILESIZE, TILESIZE])

    buttonD.draw(screen, mcol, mrow)    

    pygame.display.update()
    clock.tick(60)

关于python-3.x - 如何用鼠标绘制并保存为 1 和 0，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/58847079/

文章推荐： ruby-on-rails - 删除附件时跳过 ActiveStorage 文件清除

文章推荐： sql-server - SQL Server 异常处理

文章推荐： scala - 无法在 Akka Stream 中使用 GraphStage 类运行 SourceShape

python - Python 中的集群或合并集群以减少组数 (Python)
我正在处理一组标记为 160 个组的 173k 点。我想通过合并最接近的(到 9 或 10 个组)来减少组/集群的数量。我搜索过 sklearn 或类似的库，但没有成功。我猜它只是通过 knn 聚类
python - python 列表的子集基于同一列表的元素组，pythonically
我有一个扁平数字列表，这些数字逻辑上以 3 为一组，其中每个三元组是 (number, __ignored, flag[0 or 1])，例如: [7,56,1, 8,0,0, 2,0,0, 6,1,
python - 激活 Python 虚拟环境并在另一个 Python 脚本中调用 Python 脚本
我正在使用 pipenv 来管理我的包。我想编写一个 python 脚本来调用另一个使用不同虚拟环境(VE)的 python 脚本。如何运行使用 VE1 的 python 脚本 1 并调用另一个 p
python - 在焕然一新的 Python 环境中以编程方式从 Python 内部执行 Python 文件
假设我有一个文件 script.py 位于 path = "foo/bar/script.py"。我正在寻找一种在 Python 中通过函数 execute_script() 从我的主要 Python
python - 从 python 脚本但在 python 脚本之外运行 python 脚本
这听起来像是谜语或笑话，但实际上我还没有找到这个问题的答案。问题到底是什么？我想运行 2 个脚本。在第一个脚本中，我调用另一个脚本，但我希望它们继续并行，而不是在两个单独的线程中。主要是我不希望第
python - 使用不同的 python 从 python 运行 python 脚本
我有一个带有 python 2.5.5 的软件。我想发送一个命令，该命令将在 python 2.7.5 中启动一个脚本，然后继续执行该脚本。我试过用 #!python2.7.5 和http://re
python - 为什么从 Python 命令行调用 Python 时 Python 无法找到并运行我的脚本？
我在 python 命令行(使用 python 2.7)中，并尝试运行 Python 脚本。我的操作系统是 Windows 7。我已将我的目录设置为包含我所有脚本的文件夹，使用: os.chdir("
python - 使用动态版本的 Python 执行嵌入的 Python 代码时出现致命的 Python 错误
剧透:部分解决(见最后)。以下是使用 Python 嵌入的代码示例: #include int main(int argc, char** argv) { Py_SetPythonHome
python - python 中识别 python 数组或列表中最大累积差异的最快方法是什么？
假设我有以下列表，对应于及时的股票价格: prices = [1, 3, 7, 10, 9, 8, 5, 3, 6, 8, 12, 9, 6, 10, 13, 8, 4, 11] 我想确定以下总体上最
python - (Python) 通过单选按钮 python 更新背景
所以我试图在选择某个单选按钮时更改此框架的背景。我的框架位于一个类中，并且单选按钮的功能位于该类之外。 (这样我就可以在所有其他框架上调用它们。) 问题是每当我选择单选按钮时都会出现以下错误: co
python - python 中的字符串与正则表达式比较在 python 中失败
我正在尝试将字符串与 python 中的正则表达式进行比较，如下所示， #!/usr/bin/env python3 import re str1 = "Expecting property name
python - python 如何加载Boost.Python 库？
考虑以下原型(prototype) Boost.Python 模块，该模块从单独的 C++ 头文件中引入类“D”。 /* file: a/b.cpp */ BOOST_PYTHON_MODULE(c)
python - python 检查模块 python 的问题
如何编写一个程序来“识别函数调用的行号？” python 检查模块提供了定位行号的选项，但是， def di(): return inspect.currentframe().f_back.f_l
python - 系统 python 与用户 python
我已经使用 macports 安装了 Python 2.7，并且由于我的 $PATH 变量，这就是我输入 $ python 时得到的变量。然而，virtualenv 默认使用 Python 2.6，除
python - [Python] : Python re. 长字符串行的搜索速度优化
我只想问如何加快 python 上的 re.search 速度。我有一个很长的字符串行，长度为 176861(即带有一些符号的字母数字字符)，我使用此函数测试了该行以进行研究: def getExe
python - 编辑字符串 python 正则表达式 python
list1= [u'%app%%General%%Council%', u'%people%', u'%people%%Regional%%Council%%Mandate%', u'%ppp%%Ge
python - Python 映射中的副作用(Python "do" block )
这个问题在这里已经有了答案: Is it Pythonic to use list comprehensions for just side effects? (7 个答案) 关闭 4 个月前。告
python - 使用其值逻辑组合两个 python 列表 - Python
我想用 Python 将两个列表组合成一个列表，方法如下: a = [1,1,1,2,2,2,3,3,3,3] b= ["Sun", "is", "bright", "June","and" ,"Ju
python - Boost.Python python 链接错误
我正在运行带有最新 Boost 发行版 (1.55.0) 的 Mac OS X 10.8.4 (Darwin 12.4.0)。我正在按照说明 here构建包含在我的发行版中的教程 Boost-Pyth
python - 在 Python 中仅使用内置库制作一个基本的网络抓取工具 - Python
学习 Python，我正在尝试制作一个没有任何第 3 方库的网络抓取工具，这样过程对我来说并没有简化，而且我知道我在做什么。我浏览了一些在线资源，但所有这些都让我对某些事情感到困惑。 html 看起来

行者123

个人简介

我是一名优秀的程序员,十分优秀！

作者热门文章

滴滴打车优惠券免费领取

全站热门文章

首页

博学

6Ren·AI

商城

python-3.x - 如何用鼠标绘制并保存为 1 和 0