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)
上面的代码成对地将转子打印到屏幕上。
我想要的是在输入中输入一个字母，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，我不知道如何开始这个。
提前致谢！
答案 0 :(得分：1)
这是解决方案。在主循环开始之前，我将字母blit到背景表面上，然后每帧都将背景blit。为了存储字母的坐标，我将字母添加到字典中，列表作为值，并将坐标附加到此列表中。
当用户按下某个键时，我调用dict.get查找字典中的字母，并将相应的coords列表分配给coords变量，然后将其传递给pygame.draw.lines。如果密钥不存在，get方法将返回None，以防止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)
我也使用collections.defaultdict代替普通字典，但我不知道你是否已经熟悉它们。
如果您想以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。
然后你可以将字母和它们的坐标压缩在背景表面上，然后在每个字母上再次将坐标列表添加到rotor_dict。
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)