游戏玩偶游戏

Question

我成功地完成了这场比赛，我相信你以前也遇到过。如果你运行它，就会有关于该做什么的说明。有什么代码建议/改进吗？

import pygame,random,math
from time import sleep

pygame.init()

#window resolution
screen_width = 800
screen_height = 800

screen = pygame.display.set_mode( ( screen_width,screen_height ) )
pygame.display.set_caption( 'The Flash Game - Marek Borik' )

n_squares_in_row = 7
line_thickness = 20 #pixels

wait_duration = 1.5 #seconds
flash_duration = 0.3 #seconds

fontsize = screen_width // 25
rounds = 10 #rounds of flashes

#defining colors
RED = ( 255,0,0 )
BLUE = ( 0,0,255 )
WHITE = ( 255,255,255 )
BLACK = ( 0,0,0 )


def TextObjects( string,font,color ):
    textSurface = font.render( string,True,color )
    return textSurface,textSurface.get_rect()


def DrawText( string,fontSizeX,color,x,y ):
    text = pygame.font.SysFont( "timesnewroman",int( math.ceil( fontSizeX ) ) )
    TextSurf, TextRect = TextObjects( string,text,color )
    TextRect.center = ( x,y )
    screen.blit( TextSurf,TextRect )


def GenerateColorInfo( n_squares_in_row ):
    n_squares_total = n_squares_in_row ** 2

    not5050 = True #Prevents from having the same amount of red and blue squares if applicable ( e.g. field with dimentions 7 x 7 squares squares will never have this issue )

    while( not5050 ):
        color_info = []

        for i in range( n_squares_total + 1 ):
            rand = random.randint( 0,1 )
            color_info.append( rand )

        counter_red = 0
        counter_blue = 0

        for i in color_info:
            if i == 0:
                counter_red += 1
            if i == 1:
                counter_blue += 1

        if counter_red == n_squares_total // 2 or counter_blue == n_squares_total // 2 or counter_red == counter_blue:
            not5050 = True
        else:
            not5050 = False


    return color_info,counter_red,counter_blue


def DrawSquares( n_squares_in_row,color_info,square_width,square_height ):
    for i in range( n_squares_in_row ):
        for j in range( n_squares_in_row ):
            if color_info[ i * n_squares_in_row + j ] == 0:
                pygame.draw.rect( screen,RED,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )
            else:
                pygame.draw.rect( screen,BLUE,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )


def DrawGrid( n_squares_in_row,line_thickness,square_width,square_height ):
    for i in range( 1,n_squares_in_row ):
        pygame.draw.line( screen,BLACK,( i * square_width,0 ),( i * square_width,screen_height ),line_thickness )

    for i in range( 1,n_squares_in_row ):
        pygame.draw.line( screen,BLACK,( 0,i * square_height ),( screen_height,i * square_width ),line_thickness )

    pygame.draw.line( screen,BLACK,( 0,0 ),( 0,screen_height ),line_thickness )
    pygame.draw.line( screen,BLACK,( 0,0 ),( screen_width,0 ),line_thickness )
    pygame.draw.line( screen,BLACK,( screen_width,0 ),( screen_width,screen_height ),line_thickness )
    pygame.draw.line( screen,BLACK,( 0,screen_height ),( screen_width,screen_height ),line_thickness )


def DrawStartScreen():
    end = False

    while not end:
        for event in pygame.event.get():
            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_SPACE:
                    end = True
            if event.type == pygame.QUIT:
                pygame.quit()
                quit()

        screen.fill( BLACK )

        DrawText( "The Flash Game - Created by Marek Borik",fontsize * 1.4,WHITE,screen_width * 0.5,screen_height * 0.05 )

        DrawText( "This game will test your subconscious perception.",fontsize,WHITE,screen_width * 0.5,screen_height * 0.2 )
        DrawText( ( "You will be shown a grid of red and blue squares " + str( rounds ) + " times." ),fontsize,WHITE,screen_width * 0.5, screen_height * 0.25 )
        DrawText( "Your task is to determine if you saw more red or blue circles.",fontsize,WHITE,screen_width * 0.5, screen_height * 0.3 )

        DrawText( "If you see more:",fontsize,WHITE,screen_width * 0.3,screen_height * 0.45 )
        DrawText( "If you see more:",fontsize,WHITE,screen_width * 0.7,screen_height * 0.45 )

        pygame.draw.rect( screen,RED,( screen_width * 0.2, screen_height * 0.5,screen_width * 0.2,screen_height * 0.2 ),0 )
        pygame.draw.rect( screen,BLUE,( screen_width * 0.6, screen_height * 0.5,screen_width * 0.2,screen_height * 0.2 ),0 )

        DrawText( "Press Left Arrow",fontsize,WHITE,screen_width * 0.3,screen_height * 0.75 )
        DrawText( "Press Right Arrow",fontsize,WHITE,screen_width * 0.7,screen_height * 0.75 )

        DrawText( "Press spacebar to start",fontsize,WHITE,screen_width * 0.5,screen_height * 0.93 )


        pygame.display.update()


def DoRound():
    left_arrow = False
    right_arrow = False
    turn = False

    #if the windows isn't square, then squares are not squares and we need to treat them like rectangles
    square_width = screen_width / n_squares_in_row
    square_height = screen_height / n_squares_in_row

    color_info,n_red,n_blue = GenerateColorInfo( n_squares_in_row )

    screen.fill( BLACK )
    pygame.display.update()

    sleep( wait_duration )

    #flicker squares for the flash duration

    DrawSquares( n_squares_in_row,color_info,square_width,square_height )
    DrawGrid( n_squares_in_row,line_thickness,square_width,square_height )

    pygame.display.update()

    sleep( flash_duration )
    screen.fill( BLACK )
    pygame.display.update()

    while not turn: # after flash wait for the turn to be completed by pressing either arrow to indicate the answer
       for event in pygame.event.get():
            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_LEFT:
                    left_arrow = True
                    turn = True
                if event.key == pygame.K_RIGHT:
                    right_arrow = True
                    turn = True

            if event.type == pygame.QUIT:
                pygame.quit()
                quit()

    if n_red >= n_blue and left_arrow == True:  #if the arrow coresponds to the correct answer, return 1, else 0
        return 1

    elif n_blue >= n_red and right_arrow == True:
        return 1

    else:
        return 0


def Game():
   guesses = []

    for i in range( rounds ):
        guesses.append( DoRound() )

    return guesses.count( 1 ) #count ones, meaning correct answers


def EndScreen( correct ):
    screen.fill( BLACK )

    DrawText( "You got " + str( correct ) + " / " + str(  rounds ) + " correct!",fontsize * 2,WHITE,screen_width * 0.5,screen_height * 0.5 )

    pygame.display.update()

    end = False

    while not end:
        for event in pygame.event.get():
            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_SPACE:
                    end = True
            if event.type == pygame.QUIT:
                pygame.quit()
                quit()


DrawStartScreen()
correct = Game()
EndScreen( correct )
pygame.quit()
quit()

Answer 1

在关注点分离方面做得好

我很高兴看到一个只处理逻辑的函数，还有一些处理绘图和交互的函数。它使我很容易测试，分析和简化它。

generate_color_info

这一职能过于复杂和冗长，难以简化其任务。

• not5050变量不是必需的(您可以直接使用该条件)。
• 使用详细的显式循环代替列表理解。
• 内置的list.count被忽略了。

我首先编写了一个测试(用doctest运行)，然后更简单地重新实现它：

def GenerateColorInfo( n_squares_in_row ):
    """
    >>> random.seed(0)
    >>> GenerateColorInfo( 5 )
    ([1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0], 10, 16)
    """
    while True:
        randoms = [random.randint(0, 1) for _ in range(n_squares_in_row ** 2 + 1)]
        if randoms.count(0) != randoms.count(1):
            break
    return randoms, randoms.count(0), randoms.count(1)

列表理解在Game

中的应用

Game不必要地长：

def Game():
    return [DoRound() for _ in range( rounds )].count(True)

布尔化简

if n_red >= n_blue and left_arrow == True:  #if the arrow coresponds to the correct answer, return 1, else 0
    return 1

elif n_blue >= n_red and right_arrow == True:
    return 1

else:
    return 0

是相同的

return (n_red >= n_blue and left_arrow) or \
       (n_blue >= n_red and right_arrow)

但是第二种方法显然更简单(而且它更好，因为它返回True / False而不是1和0，从而使它更加清楚，它是一个布尔函数)。

用户

的任意内存任务

右箭头和左箭头是如何分配给红色和蓝色的？让用户输入R ( Red )和B ( Blue )更合理。这样，连接是显而易见的，用户不需要使用内存。

代码处理矩形吗？

你的一些代码似乎是用来处理矩形的：

#if the windows isn't square, then squares are not squares and we need to treat them like rectangles
square_width = screen_width / n_squares_in_row
square_height = screen_height / n_squares_in_row

其他代码意味着正方形网格：

def DrawSquares( n_squares_in_row,color_info,square_width,square_height ):
    for i in range( n_squares_in_row ):
        for j in range( n_squares_in_row ):
            if color_info[ i * n_squares_in_row + j ] == 0:
                pygame.draw.rect( screen,RED,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )
            else:
                pygame.draw.rect( screen,BLUE,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )

请要么编写所有的代码来容纳矩形，要么一点也不写，只是其中的一些代码会造成复杂性和混乱，而不会带来任何好处。

重复

您的代码多次声明相同的概念：

            if color_info[ i * n_squares_in_row + j ] == 0:
                pygame.draw.rect( screen,RED,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )
            else:
                pygame.draw.rect( screen,BLUE,( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )

if和else除BLUE或RED作为颜色外是相等的。干原理是软件开发中最重要的原理之一。

下面是如何修复复制：

        color = RED if color_info[ i * n_squares_in_row + j ] == 0 else BLUE
        pygame.draw.rect( screen, color, ( j * square_width,i * square_height,j * square_width + square_width,i * square_height + screen_height ),0 )

现在只调用一次绘图代码，很明显，只有颜色发生了变化。

pygame.draw.line( screen,BLACK,( 0,0 ),( 0,screen_height ),line_thickness )
pygame.draw.line( screen,BLACK,( 0,0 ),( screen_width,0 ),line_thickness )
pygame.draw.line( screen,BLACK,( screen_width,0 ),( screen_width,screen_height ),line_thickness )
pygame.draw.line( screen,BLACK,( 0,screen_height ),( screen_width,screen_height ),line_thickness )

在这个代码块中，段pygame.draw.line( screen,BLACK, ... line_thickness)被重复4次。只有起点和终点发生变化，所以我们可以使用一个循环：

for (start, end) in [ ( ( 0,0 ),( 0,screen_height) ),
                      ( ( 0,0 ),( screen_width,0 ) ),
                      (( screen_width,0 ),( screen_width,screen_height )),
                      (( 0,screen_height ),( screen_width,screen_height )) ]:
    pygame.draw.line(screen, BLACK, start, end, line_thickness)

当DrawText fontsize是fontsize，color是WHITE时，可以将其定义为关键字默认参数，以减少重复，从而在需要其他颜色时显式化。

这些代码块类似：

while not turn: # after flash wait for the turn to be completed by pressing either arrow to indicate the answer
   for event in pygame.event.get():
        if event.type == pygame.KEYDOWN:
            if event.key == pygame.K_LEFT:
                left_arrow = True
                turn = True
            if event.key == pygame.K_RIGHT:
                right_arrow = True
                turn = True

        if event.type == pygame.QUIT:
            pygame.quit()
            quit()

和

while not end:
    for event in pygame.event.get():
        if event.type == pygame.KEYDOWN:
            if event.key == pygame.K_SPACE:
                end = True
        if event.type == pygame.QUIT:
            pygame.quit()
            quit()

应该可以编写一个函数一次，并使用它两次。

命名/间距

您可以称它为次要的，但是Python (PEP8)有一个非常广泛采用的样式指南，说明常量应该是ALL_UPPERCASE和所有其他名称lowercase_with_underscores。它有助于提高程序之间的一致性。我建议在您的代码上运行autopep8，以修正间距，使之与广泛接受的样式保持一致。