单击网格的单元格时是否可以更新 pygame 绘图属性？

Question

我正在接近 PyGame，但我遇到了一个问题：我正在尝试绘制一个由单个矩形图形组成的网格，我希望当在所述矩形上释放鼠标左键时，这些矩形中的每一个都被填充或清空。我用一个元组矩阵表示的网格，每个元组代表每个矩形，格式如下（列，行）以符合pygame像素的坐标，然后我创建另一个元组矩阵以从鼠标位置元组（colpixel，rowpixel）转换为矩形元组，我还创建了一个字典，将矩形元组作为键与值可以是 1 或 0 配对，该值表示矩形是否被填充，这是我在释放鼠标按钮时更改的值。这是代码：

import pygame

pygame.init()

SCREEN_WIDTH = 40
SCREEN_HEIGHT = 20

DIM = 20

ROWS = int(SCREEN_HEIGHT / DIM)
COLUMNS = int(SCREEN_WIDTH / DIM)


screen = pygame.display.set_mode((SCREEN_WIDTH, SCREEN_HEIGHT))

pygame.display.set_caption('Game of Life')


# filling rect_matrix with tuple, each one representing a rectangle of the grid (columns, rows).
# filling rect_map by pairing the tuples from rect_matrix with a value that can be 0 or 1, default 1.
rect_matrix = []
rect_map = {}

for i in range(ROWS):
    temp = []
    for j in range(COLUMNS):
        temp.append(pygame.Rect((j * DIM, i * DIM), (DIM, DIM)))
        rect_map[(j, i)] = 1
    rect_matrix.append(temp)


# filling mouse_to_rectangle, sort of a decoder to convert from groups of tuples
# (representing mouse coordinates in pixels) to a single tuple (representing single rectangles)
mouse_to_rectangle = []

ii = 0
for i in range(SCREEN_WIDTH):
    jj = 0
    temp = []
    for j in range(SCREEN_HEIGHT):
        temp.append((ii, jj))
        if ((j + 1) % DIM == 0):
            jj += 1
    if ((i + 1) % DIM == 0):
        ii += 1
    mouse_to_rectangle.append(temp)


is_running = True

while is_running:

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            is_running = False
            pygame.quit()
            exit()
        if event.type == pygame.MOUSEBUTTONUP:

            if rect_map[mouse_to_rectangle[pygame.mouse.get_pos()[0]][pygame.mouse.get_pos()[1]]] == 1:
                rect_map[mouse_to_rectangle[pygame.mouse.get_pos()[0]][pygame.mouse.get_pos()[1]]] = 0
            else:
                rect_map[mouse_to_rectangle[pygame.mouse.get_pos()[0]][pygame.mouse.get_pos()[1]]] = 1



    for i, ii in zip(rect_matrix, range(len(rect_matrix))):
        for j, jj in zip(i, range(len(i))):
            pygame.draw.rect(screen, (100, 100, 100), j, rect_map[(jj, ii)])

    pygame.display.update()

我设法填充（rect_map 中的相对值变为 1）一个矩形，当指向它时松开左键，但是当我尝试再次单击现在已填充的矩形时，它不会为空。我用 1 或 0 表示“是否填充”属性，因为当

width

中的

pygame.draw.rect

参数设置为 0 时，矩形被填充，如果使用

>0

值，则矩形的边框将一样粗作为以像素为单位的指定值。

Answer 1

您错过了在绘制网格之前清除显示。因此，所有绘制的矩形都永远留在那里：

screen.fill((0, 0, 0))

不管怎样，你已经把算法复杂化了。可以通过将鼠标位置的分量除以

DIM

（

//

（地板除法）运算符）来计算研磨中的位置：

mousepos = pygame.mouse.get_pos()
gridpos = mousepos[0] // DIM, mousepos[1] // DIM

最后，网格状态可以轻松更改，

mouse_to_rectangle

根本不需要：

if rect_map[gridpos] == 1:
    rect_map[gridpos] = 0
else:
    rect_map[gridpos] = 1

看例子：

import pygame

pygame.init()

SCREEN_WIDTH = 200
SCREEN_HEIGHT = 200

DIM = 20

ROWS = int(SCREEN_HEIGHT / DIM)
COLUMNS = int(SCREEN_WIDTH / DIM)

screen = pygame.display.set_mode((SCREEN_WIDTH, SCREEN_HEIGHT))
pygame.display.set_caption('Game of Life')

# filling rect_matrix with tuple, each one representing a rectangle of the grid (columns, rows).
# filling rect_map by pairing the tuples from rect_matrix with a value that can be 0 or 1, default 1.
rect_matrix = []
rect_map = {}

for i in range(ROWS):
    temp = []
    for j in range(COLUMNS):
        temp.append(pygame.Rect((j * DIM, i * DIM), (DIM, DIM)))
        rect_map[(j, i)] = 1
    rect_matrix.append(temp)

is_running = True

while is_running:

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            is_running = False
            pygame.quit()
            exit()
        if event.type == pygame.MOUSEBUTTONUP:

            mousepos = pygame.mouse.get_pos()
            gridpos = mousepos[0] // DIM, mousepos[1] // DIM
            if rect_map[gridpos] == 1:
                rect_map[gridpos] = 0
            else:
                rect_map[gridpos] = 1

    screen.fill((0, 0, 0))

    for i, ii in zip(rect_matrix, range(len(rect_matrix))):
        for j, jj in zip(i, range(len(i))):
            pygame.draw.rect(screen, (100, 100, 100), j, rect_map[(jj, ii)])

    pygame.display.update()