python - 如何在pygame中的两个列表之间画线？

Question

import pygame as pg
import sys
pg.init()
screen = pg.display.set_mode((800, 600))
clock = pg.time.Clock()
screenGray = pg.Color('gray80')
RotorFont = pg.font.SysFont("malgun gothic",17)
textColour = pg.Color('navy')
background = pg.Surface(screen.get_size())
background.fill(screenGray)

ACTIONPRINT = False

while True:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            pg.quit()
            sys.exit()
        if event.type == pg.MOUSEBUTTONDOWN:
            ACTIONPRINT = True

    screen.blit(background,(0,0))
    rotorAA = ['G','N','Z','M','V','B','F','L','Q','R','Y','P','I','C','E','A','D','K','J','W','X','S','H','U','O','T']#2
    rotorAB = ['L','Q','R','Y','D','K','J','W','X','S','H','U','O','P','I','C','F','A','G','N','Z','M','V','B','E','T']#3
    rotorBA = ['Q','W','E','R','T','Y','U','I','O','P','A','S','D','F','G','H','J','K','L','Z','X','C','V','B','N','M']#4
    rotorBB = ['Y','H','Q','V','L','T','C','W','K','P','S','N','X','E','O','M','B','U','G','F','A','J','D','R','Z','I']#5
    rotorCA = ['D','F','P','A','N','E','Y','C','S','G','K','J','M','X','O','V','L','W','Q','H','T','U','B','R','Z','I']#6
    rotorCB = ['Z','I','A','C','T','F','U','Q','N','V','P','B','D','O','L','R','S','X','M','G','H','J','W','E','K','Y']#7

    rotorList = [rotorAA,rotorAB,rotorBA,rotorBB,rotorCA,rotorCB]

    count = 0
    k = 0
    if ACTIONPRINT == True:
    ACTIONPRINT = False
    for i in range(0,len(rotorList)):
        if count % 2 == 0 and count != 0:
                k += 25
        for j in range(0,26):
            a = RotorFont.render(rotorList[i][j],1,textColour)
            background.blit(a,(25 + (i * 25) + k,90+(j * 16)))

            if rotorList[i][j] == letter and i + 2 < 6:
                correspondingLetter = rotorList[i+1][(rotorList[i].index(letter))]
                pg.draw.line(background,black,(25 + (i * 25) + k,90+(j * 16)),(25 + (i + 1 * 25) + 25,90+(rotorList[i+1].index(correspondingLetter) * 16)))
                letter = rotorList[i+2][rotorList[i+1].index(correspondingLetter)]
        count += 1

    pg.display.flip()
    clock.tick(60)

上面的代码将转子成对地打印到屏幕上。

我想要发生的是在输入中输入一个字母，并在相应的字母之间绘制一条线，从最右边的列表到左边的列表。

但是，我不知道如何开始。

提前致谢!

Answer 1

这是解决方案。在主循环开始之前，我将字母传输到背景表面，然后每帧传输背景。为了存储字母的坐标，我将字母添加到字典中，并以列表作为值，并将坐标附加到该列表中。

当用户按下某个键时，我调用 dict.get 来查找字典中的字母，并将相应的坐标列表分配给我然后使用的 coords 变量传递给 pygame.draw.lines 。如果 key 不存在，get 方法将返回 None，从而防止出现 KeyError。

import sys
import pygame as pg


pg.init()
screen = pg.display.set_mode((800, 600))
clock = pg.time.Clock()
screenGray = pg.Color('gray80')
RotorFont = pg.font.SysFont('malgun gothic', 17)
textColour = pg.Color('navy')
background = pg.Surface(screen.get_size())
background.fill(screenGray)

# You can create the lists and blit the letters before the main loop starts.
rotorAA = ['G','N','Z','M','V','B','F','L','Q','R','Y','P','I','C','E','A','D','K','J','W','X','S','H','U','O','T']#2
rotorAB = ['L','Q','R','Y','D','K','J','W','X','S','H','U','O','P','I','C','F','A','G','N','Z','M','V','B','E','T']#3
rotorBA = ['Q','W','E','R','T','Y','U','I','O','P','A','S','D','F','G','H','J','K','L','Z','X','C','V','B','N','M']#4
rotorBB = ['Y','H','Q','V','L','T','C','W','K','P','S','N','X','E','O','M','B','U','G','F','A','J','D','R','Z','I']#5
rotorCA = ['D','F','P','A','N','E','Y','C','S','G','K','J','M','X','O','V','L','W','Q','H','T','U','B','R','Z','I']#6
rotorCB = ['Z','I','A','C','T','F','U','Q','N','V','P','B','D','O','L','R','S','X','M','G','H','J','W','E','K','Y']#7
rotorList = [rotorAA,rotorAB,rotorBA,rotorBB,rotorCA,rotorCB]
# This dict will have the letters as the keys and the
# corresponding coordinates as the values.
rotor_dict = {}

k = 0

# You can use `enumerate` if you need the index and the item.
for i, sublist in enumerate(rotorList):
    if i % 2 == 0 and i != 0:
        k += 25
    for j, letter in enumerate(sublist):
        surface = RotorFont.render(letter, 1, textColour)
        x = 25 + (i*25) + k
        y = 90 + (j*16)
        background.blit(surface, (x, y))
        # If the letter isn't in the dict, add a new list with the
        # first coordinates.
        if letter not in rotor_dict:
            rotor_dict[letter] = [(x+4, y+11)]
        # Otherwise append the next coordinates to the list.
        else:
            rotor_dict[letter].append((x+4, y+11))

coords = None  # The list of currently selected coordinates.

while True:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            pg.quit()
            sys.exit()
        elif event.type == pg.KEYDOWN:  # If the user pressed a key.
            # Use the `get` method which returns `None` by
            # default if the key doesn't exist.
            # `event.unicode` is the letter (string).
            # So if the letter is in the dict, this assigns the coords
            # list in the dict to the `coords` variable.
            coords = rotor_dict.get(event.unicode.upper())

    screen.blit(background, (0, 0))
    # If a key was pressed and a list was assigned to `coords`.
    if coords is not None:
        # Pass the coords list to draw lines.
        pg.draw.lines(screen, (200, 100, 0), False, coords, 2)

    pg.display.flip()
    clock.tick(60)

我还会使用collections.defaultdict而不是普通的字典，但我不知道你是否已经熟悉它们。

<小时/>

如果您想按照评论中描述的方式连接字母，您需要找出以下列表中字母的索引，并使用索引依次获取下一个字母。

然后，您可以压缩字母及其坐标，将它们传输到背景表面上，并为每个字母再次将坐标列表添加到 rotor_dict 中。

import sys
import pygame as pg


pg.init()
screen = pg.display.set_mode((800, 600))
clock = pg.time.Clock()
screenGray = pg.Color('gray80')
RotorFont = pg.font.SysFont('malgun gothic', 17)
textColour = pg.Color('navy')
background = pg.Surface(screen.get_size())
background.fill(screenGray)

rotorAA = ['G','N','Z','M','V','B','F','L','Q','R','Y','P','I','C','E','A','D','K','J','W','X','S','H','U','O','T']
rotorAB = ['L','Q','R','Y','D','K','J','W','X','S','H','U','O','P','I','C','F','A','G','N','Z','M','V','B','E','T']
rotorBA = ['Q','W','E','R','T','Y','U','I','O','P','A','S','D','F','G','H','J','K','L','Z','X','C','V','B','N','M']
rotorBB = ['Y','H','Q','V','L','T','C','W','K','P','S','N','X','E','O','M','B','U','G','F','A','J','D','R','Z','I']
rotorCA = ['D','F','P','A','N','E','Y','C','S','G','K','J','M','X','O','V','L','W','Q','H','T','U','B','R','Z','I']
rotorCB = ['Z','I','A','C','T','F','U','Q','N','V','P','B','D','O','L','R','S','X','M','G','H','J','W','E','K','Y']
rotor_dict = {}

# Iterate over the letters in the first list.
for i, letter1 in enumerate(rotorAA):
    # Find the indices of the following letters.
    letter2 = rotorAB[i]  # The letter at index i in the AB list.
    j = rotorBA.index(letter2)  # Letter2's index in the BA list.
    letter3 = rotorBB[j]  # Letter at index j in the AB list.
    k = rotorCA.index(letter3)  # Letter3's index in the CA list.
    letter4 = rotorCB[k]  # Letter at index k in the CB list.
    # A list of the connected letters.
    letters = [letter1, letter2, letter2, letter3, letter3, letter4]
    # The coords of the letters in the `letters` list above.
    coords = [(25, 90 + i*16), (50, 90 + i*16),
              (100, 90 + j*16), (125, 90 + j*16),
              (175, 90 + k*16), (200, 90 + k*16),
              ]
    rotor_dict[letter1] = coords
    # Draw the letters. Check out the `zip` function.
    for letter, coord in zip(letters, coords):
        background.blit(RotorFont.render(letter, True, textColour), coord)

coords = None
offset = pg.math.Vector2(5, 11)  # Add this to the coords to center the lines.

while True:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            pg.quit()
            sys.exit()
        elif event.type == pg.KEYDOWN:
            # Add an offset, so that the lines start at the center points.
            # This is called a list comprehension if you haven't seen this before.
            coords = [coord+offset for coord in rotor_dict.get(event.unicode.upper())]

    screen.blit(background, (0, 0))
    if coords is not None:
        pg.draw.lines(screen, (200, 100, 0), False, coords, 2)

    pg.display.flip()
    clock.tick(60)