如何使pygame konami代码更简单？

Question

我目前正在开发一个pygame游戏，我的朋友问我是否可以实现一个konami代码。 我认为这是一个伟大的想法，所以我实施了它

我做这件事的方式肯定会奏效，因为我不必处理任何错误。 然而代码比预期的要大一点，所以我想知道如何使它变小并且仍然工作，因为我不知道有什么好的方法使它变小

守则：

konami_code = ['', '', '', '', '', '', '', '', '', '']

key = pygame.key.get_pressed()
for event in pygame.event.get():
    if event.type == pygame.KEYUP:
        if key[pygame.K_UP]:
            if konami_code[0] == '':
                konami_code[0] = 'UP'

            elif konami_code[0] == 'UP' \
                    and konami_code[1] == '':
                konami_code[1] = 'UP'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']

        if key[pygame.K_DOWN]:
            if konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == '':
                konami_code[2] = 'DOWN'

            elif konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == '':
                konami_code[3] = 'DOWN'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']

        if key[pygame.K_LEFT]:
            if konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == '':
                konami_code[4] = 'LEFT'

            elif konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == 'LEFT' \
                    and konami_code[5] == 'RIGHT' \
                    and konami_code[6] == '':
                konami_code[6] = 'LEFT'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']

        if key[pygame.K_RIGHT]:
           if konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == 'LEFT' \
                    and konami_code[5] == '':
                konami_code[5] = 'RIGHT'

            elif konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == 'LEFT' \
                    and konami_code[5] == 'RIGHT' \
                    and konami_code[6] == 'LEFT' \
                    and konami_code[7] == '':
                konami_code[7] = 'RIGHT'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']

        if key[pygame.K_b]:
            if konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == 'LEFT' \
                    and konami_code[5] == 'RIGHT' \
                    and konami_code[6] == 'LEFT' \
                    and konami_code[7] == 'RIGHT' \
                    and konami_code[8] == '':
                konami_code[8] = 'B'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']

        if key[pygame.K_a]:
            if konami_code[0] == 'UP' \
                    and konami_code[1] == 'UP' \
                    and konami_code[2] == 'DOWN' \
                    and konami_code[3] == 'DOWN' \
                    and konami_code[4] == 'LEFT' \
                    and konami_code[5] == 'RIGHT' \
                    and konami_code[6] == 'LEFT' \
                    and konami_code[7] == 'RIGHT' \
                    and konami_code[8] == 'B' \
                    and konami_code[9] == '':
                konami_code[9] = 'A'

            else:
                konami_code = ['', '', '', '', '', '', '', '', '', '']



    if konami_code[0] == 'UP' \
        and konami_code[1] == 'UP' \
        and konami_code[2] == 'DOWN' \
        and konami_code[3] == 'DOWN' \
        and konami_code[4] == 'LEFT' \
        and konami_code[5] == 'RIGHT' \
        and konami_code[6] == 'LEFT' \
        and konami_code[7] == 'RIGHT' \
        and konami_code[8] == 'B' \
        and konami_code[9] == 'A':
    print('code completed')

Answer 1

您可以将konami代码存储为值列表。然后将用户输入与该列表进行比较：

import pygame

CODE = [pygame.K_UP, pygame.K_UP, pygame.K_DOWN, pygame.K_DOWN, pygame.K_LEFT, pygame.K_RIGHT, pygame.K_LEFT, pygame.K_RIGHT, pygame.K_b, pygame.K_a]

pygame.init()
screen = pygame.display.set_mode([500, 500])
code = []
index = 0
running = True
while running:
    key = pygame.key.get_pressed()
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False
        elif event.type == pygame.KEYDOWN:
            if event.key == CODE[index]:
                code.append(event.key)
                index += 1
                if code == CODE:
                    index = 0
                    print('Bingo!')
            else:
                code = []
                index = 0

pygame.display.quit()
pygame.quit()

Answer 2

我认为在这里问这个问题并不合适（也许代码审查是一个更好的地方），但我对您的代码做了一些改进：

import pygame

pygame.init()
pygame.font.init()
display = pygame.display.set_mode((500, 300))
font = pygame.font.SysFont('arial', 20)

sequence = ['up', 'up', 'down', 'right', 'left', 'z', 'x', 'z', 'up', 'down']
text_render = font.render(' '.join(sequence), False, (255, 255, 255))
input_sequence = []

clock = pygame.time.Clock()

while True:
    clock.tick(60)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.quit()

        if event.type == pygame.KEYDOWN:
            if event.key == pygame.K_UP:
                input_sequence.append('up')

            if event.key == pygame.K_DOWN:
                input_sequence.append('down')

            if event.key == pygame.K_RIGHT:
                input_sequence.append('right')

            if event.key == pygame.K_LEFT:
                input_sequence.append('left')

            if event.key == pygame.K_z:
                input_sequence.append('z')

            if event.key == pygame.K_x:
                input_sequence.append('x')

    display.blit(text_render, (10, 10))

    pygame.display.update()
    
    if len(input_sequence) == len(sequence):
        if input_sequence == sequence:
            print('correct sequence')
            pygame.quit()
            exit()
        else:
            print('incorrect sequence, try again')
            input_sequence = []

最重要的是，现在你可以在序列列表中输入你想要的任何序列，它将匹配用户输入，不管列表有多长。除了一些调整的情况下，一个长长的名单可以与字体大小（自动化的过程），也许一些时间可以添加

下面是一个更短的选项：

import pygame

pygame.init()
pygame.font.init()
display = pygame.display.set_mode((500, 300))
font = pygame.font.SysFont('arial', 20)

keys = {pygame.K_UP: 'up', pygame.K_DOWN: 'down', pygame.K_RIGHT: 'right', pygame.K_LEFT: 'left',
        pygame.K_z: 'z', pygame.K_x: 'x'}
sequence = ['up', 'up', 'down', 'right', 'left', 'z', 'x', 'z', 'up', 'down']
text_render = font.render(' '.join(sequence), False, (255, 255, 255))
input_sequence = []

clock = pygame.time.Clock()

while True:
    clock.tick(60)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.quit()

        if event.type == pygame.KEYDOWN:
            for key, name in keys.items():
                if event.key == key:
                    input_sequence.append(name)
                    break

    display.blit(text_render, (10, 10))

    pygame.display.update()
    
    if len(input_sequence) == len(sequence):
        if input_sequence == sequence:
            print('correct sequence')
            pygame.quit()
            exit()
        else:
            print('incorrect sequence, try again')
            input_sequence = []