无法使用pygame在python 2.7中的受控庄园中有效产生精灵

Question

我对此问题的目标是能够获得一个答案，该答案解释了如何做到这一点，以便当我的角色四处走动以拾取“坏蛋”时，坏蛋将自动重新出现在屏幕上。 随时将屏幕上的精灵总数保持在屏幕上，以供玩家在40点时拿起。

我在控制'pikachu.jpg'精灵生成的方式时遇到问题。 我可以通过手动在程序中输入一个整数来仅以组的方式生成精灵，或者我尝试写错测试，并且程序支持绘制坏家伙精灵的无限循环。 我建立了一个系统，我认为该系统应该可以使40个精灵的坏人保持稳定，尽管它无法正常工作。 我对python的了解还不是很好，我非常感谢您提供有关此问题的帮助。

import necessary pygame modules
import sys
import random
import time
import pygame
from pygame.locals import *
# initialise pygame
pygame.init()
# define window width and height variables for ease of access
WindowWidth = 600
WindowHeight = 500
global PLAY
PLAY = False
x = 600 / 3
y = 460 / 2 - 25

# load and define image sizes

enemyImage = pygame.image.load('pikachu.jpg')
enemyStretchedImage = pygame.transform.scale(enemyImage, (40,40))
screen = pygame.display.set_mode((WindowWidth, WindowHeight))


def play(PLAY):
    playerImage = pygame.image.load('player.jpg')
    playerStretchedImage = pygame.transform.scale(playerImage, (40, 40))
    movex,movey = 0,0
    charx, chary=300, 200
    enemyy, enemyx = 10, 10
    moveright = False
    moveleft = False
    spawn = True
    direction = 'down'
    baddie = []


    for i in range(20):
        baddie.append(pygame.Rect(random.randint(0, WindowWidth - 40), random.randint(0, 0), 40, 40))
    while True:
        for event in pygame.event.get():
            if event.type == KEYDOWN:
                if event.key == ord('m'):
                    pygame.mixer.music.stop()
                    music = False
                if event.key == ord('n'):
                    pygame.mixer.music.play()
                    music = True
                if event.key == ord('a'):
                    moveleft = True

                if event.key == ord('d'):
                    moveright = True

                if event.key == ord('w'):
                    movey = -0.5
                if event.key == ord('s'):
                    movey = 0.5
                if event.key == ord('p'):
                    time.sleep(5)

            if event.type ==KEYUP:
                if event.key == ord('a'):
                    moveleft = False
                    if moveleft == False:
                        movex = 0
                if event.key == ord('d'):
                    moveright = False
                    if moveright == False:
                        movex = 0
                if event.key == ord('w'):
                    movey = 0
                if event.key == ord('s'):
                    movey = 0
            elif event.type == QUIT:
                pygame.quit()
                sys.exit()
            elif event.type == MOUSEBUTTONDOWN:
                pygame.quit()
                sys.exit()


        screen.fill(pygame.Color("red"))
        player = pygame.Rect(charx, chary, 40, 40)
        if direction == 'down':
            enemyy += 0.1
        if moveright ==True:
            movex = 0.5
        if moveleft ==True:
            movex = -0.5

        if player.bottom > WindowHeight:
            chary = WindowHeight - 40
            movey = 0
        if player.top < 0:
            chary = 1
            movey = 0
        if player.left < 0:
            charx = 1
            movex = 0
        if player.right > WindowWidth:
            charx = WindowWidth  - 40
            movex = 0

        for bad in baddie[:]:    #Here is where my attempt of testing was.
            if bad < 40:
                spawn = True
            if bad >= 40:
                spawn = False
            if spawn == True:
                baddie.append(pygame.Rect(random.randint(0, WindowWidth - 40), random.randint(0,0), 40, 40))



        screen.blit(playerStretchedImage, player)
        charx+=movex
        chary+=movey

        for bad in baddie[:]:
            if player.colliderect(bad):
                baddie.remove(bad)

        for bad in baddie:
            screen.blit(enemyStretchedImage, bad)
        pygame.display.update()

def presskey():
    myfont = pygame.font.SysFont("monospace", 30)
    label = myfont.render("press space to play!", 1, (255,125,60))
    while True:
        for event in pygame.event.get():
            if event.type == QUIT:
                pygame.quit()
                sys.exit()
            elif event.type == KEYDOWN:
                if event.key == K_SPACE:
                    PLAY = True
                if PLAY == True:
                    play(PLAY)
                    return
        screen.fill(pygame.Color("cyan"))  
        screen.blit(label, (x,y))
        pygame.display.update()



presskey()

Answer 1

在这里，您尝试填充最多40个坏人，但bad是一个矩形，并且将它用作许多伙伴：

for bad in baddie[:]:    #Here is where my attempt of testing was.
            if bad < 40:
                spawn = True
            if bad >= 40:
                spawn = False
            if spawn == True:
                baddie.append(pygame.Rect(random.randint(0, WindowWidth - 40), random.randint(0,0), 40, 40))

我建议您将这些行替换为：

if len(baddie)<40:
    baddie.append(pygame.Rect(random.randint(0, WindowWidth - 40), 

您可以使用len(baddie)获得[]列表中的项目数。 您应该在每个游戏循环周期获得一个新的Baddie。