Python 返回列表的第二个、第三个等变量的索引
我正在做一个简单的刽子手游戏来练习我的Python技能。 我正在尝试获取单词中第二个或更多字符的索引 例如,单词是“eggsalad”,我需要得到g和a的索引 短代码段:Python 返回列表的第二个、第三个等变量的索引,python,indexing,pygame,python-3.7,Python,Indexing,Pygame,Python 3.7,我正在做一个简单的刽子手游戏来练习我的Python技能。 我正在尝试获取单词中第二个或更多字符的索引 例如,单词是“eggsalad”,我需要得到g和a的索引 短代码段: while x < len(letterlist): if letterlist[x-1] == "g": pos_ = (AdjustMeUWU + (27 * LTGL.index(letterlist[x-1])), 226) e
while x < len(letterlist):
if letterlist[x-1] == "g":
pos_ = (AdjustMeUWU + (27 * LTGL.index(letterlist[x-1])), 226)
else:
pos_ = (AdjustMeUWU + (27 * LTGL.index(letterlist[x-1])), 230)
lettersguessed.append(letterlist[x-1])
print(x)
if LTGL == lettersguessed:
print("you won")
running = False
x += 1
而x
完整代码:
from random_word import RandomWords
r = RandomWords()
import pygame
import sys
def main():
pygame.init()
clock = pygame.time.Clock()
size = [256, 356]
white = [255, 255, 255]
black = [0, 0, 0]
#ListToGuess = list("{0}".format(r.get_random_word(hasDictionaryDef="true", minLength=4, maxLength=8)))
ListToGuess = list("abcdefgg")
LTGL = ListToGuess
print(ListToGuess)
pygame.display.set_caption('Hang man')
screen = pygame.display.set_mode(size)
font = pygame.font.Font('freesansbold.ttf', 28)
def DisplayLetter(LTGL, letterlist, lettersguessed):
if len(LTGL) == 4:
AdjustMeUWU = 78
if len(LTGL) == 5:
AdjustMeUWU = 63
if len(LTGL) == 6:
AdjustMeUWU = 53
if len(LTGL) == 7:
AdjustMeUWU = 37
if len(LTGL) == 8:
AdjustMeUWU = 25
x = 0
print(letterlist)
while x < len(letterlist):
letter = font.render(letterlist[x-1], True, black, white)
if letterlist[x-1] == "g":
pos_ = (AdjustMeUWU + (27 * LTGL.index(letterlist[x-1])), 226)
else:
pos_ = (AdjustMeUWU + (27 * LTGL.index(letterlist[x-1])), 230)
screen.blit(letter, pos_)
screen.blit(BottomDashes, (0, 256))
lettersguessed.append(letterlist[x-1])
print(x)
if LTGL == lettersguessed:
print("you won")
running = False
x += 1
YouLost = pygame.image.load(r'C:\Users\Wit\Desktop\Programming shit\img\Hangman\Youlost.png')
WhitePng = pygame.image.load(r'C:\Users\Wit\Desktop\Programming shit\img\Hangman\white100px.png')
BottomDashes = pygame.image.load(r'C:\Users\Wit\Desktop\Programming shit\img\Hangman\Dashes{0}.png'.format(len(ListToGuess)))
def HangmanDrawImg(Mistakes):
HangmanImg = pygame.image.load(r'C:\Users\Wit\Desktop\Programming shit\img\Hangman\Hangman{0}.png'.format(Mistakes))
screen.blit(HangmanImg, (30, -50))
screen.fill(white)
button = pygame.Rect(103, 296, 50, 50) # creates a rect object
pygame.draw.rect(screen, [255, 0, 0], button) # draw button
screen.blit(BottomDashes, (0, 256))
pygame.display.flip()
lettersguessed = []
running = True
Mistakes = 0
while running is True:
try:
letter = ""
if Mistakes == 9:
screen.blit(WhitePng, (0, 259))
screen.blit(YouLost, (20, 286))
pygame.display.flip()
break
for event in pygame.event.get():
if event.type == pygame.QUIT:
running = False
return running
if event.type == pygame.MOUSEBUTTONDOWN:
mouse_pos = event.pos
if button.collidepoint(mouse_pos):
Mistakes += 1
if event.type == pygame.KEYDOWN:
current_text = letter
letter += event.unicode
if current_text == letter:
pass
elif letter not in ListToGuess:
Mistakes += 1
elif letter in ListToGuess:
RunMe = True
letterlist = []
while RunMe:
lettercount = ListToGuess.count(letter)
if lettercount > 0:
letterlist.append(letter)
letterfind = ListToGuess.index(letter)
ListToGuess.pop(letterfind)
if lettercount == 0:
RunMe = False
LTGL.insert(letterfind, letter)
DisplayLetter(LTGL, letterlist, lettersguessed)
HangmanDrawImg(Mistakes)
pygame.display.flip()
clock.tick(50)
except UnboundLocalError:
pass
pygame.quit()
sys.exit
if __name__ == '__main__':
main()
从随机词导入随机词
r=随机词()
导入pygame
导入系统
def main():
pygame.init()
clock=pygame.time.clock()
大小=[256356]
白色=[255,255,255]
黑色=[0,0,0]
#ListToGuess=list(“{0}”。格式(r.get_random_word(hasdirectionarydef=“true”,minLength=4,maxLength=8)))
ListToGuess=list(“abcdefgg”)
LTGL=ListToGuess
打印(ListToGuess)
pygame.display.set_标题('hangman')
screen=pygame.display.set_模式(大小)
font=pygame.font.font('freesansbold.ttf',28)
def显示字母(LTGL、字母列表、字母标记):
如果len(LTGL)=4:
调整MEUWU=78
如果len(LTGL)=5:
调整MEUWU=63
如果len(LTGL)=6:
调整MEUWU=53
如果len(LTGL)=7:
调整MEUWU=37
如果len(LTGL)=8:
调整MEUWU=25
x=0
打印(信单)
当x0:
信函列表。附加(信函)
letterfind=ListToGuess.index(字母)
ListToGuess.pop(letterfind)
如果lettercount==0:
RunMe=False
LTGL.插入(字母查找,字母)
显示字母(LTGL、字母列表、字母组合)
HangmanDrawImg(错误)
pygame.display.flip()
时钟滴答(50)
除UnboundLocalError外:
通过
pygame.quit()
系统出口
如果uuuu name uuuuuu='\uuuuuuu main\uuuuuuu':
main()
PS:记住这还没有结束,所以有很多bug
PS2:如果您看到任何不正确的代码,请随时指出它首先您会看到出现多次的字母:
duplicates = [item for item, count in collections.Counter('eggsalad').items() if count > 1]
然后循环复制,并获得所有副本的索引:
for duplicate in duplicates:
print(f'{duplicate}: {[i for i, e in enumerate('eggsalad') if e == duplicate]}')
整体而言:
>>> import collections
>>>
>>>
>>> word = 'eggsalad'
>>> duplicates = [item for item, count in collections.Counter(word).items() if count > 1]
>>> for duplicate in duplicates:
... print(f'{duplicate}: {[i for i, e in enumerate(word) if e == duplicate]}')
...
g: [1, 2]
a: [4, 6]
使用for循环和enumerate内置函数构建元组列表。也许这对初学者有帮助:
res = []
word = "eggsalad"
for i, l in enumerate(word):
res.append((l, i))
print(res)
您的意思是想获得单词中多次出现的字符的索引吗?