Python比较列表中的连续项，如果它们相同，则移除其中一个
Python比较列表中的连续项，如果它们相同，则移除其中一个
提问于 2018-07-01 04:32:34
这是一个魔方加扰发生器的尝试。有时我会轮换同一张脸，例如(R，R')。我试图使用for和while循环来解决这个问题，但它不起作用。
import random
def getscramble():
moves = ["R","F","U","L","B","D"]
scramble = []
finascramble = []
for x in range(25):
scramble.append(random.choice(moves))
for x in range(0,len(scramble)-1):
if scramble[x] == scramble[x+1]:
scramble[x] = random.choice(moves)
for x in range(0,len(scramble)-1):
if scramble[x] == "R" and scramble[x+1] == "L":
scramble[x+1] = random.choice(moves)
if scramble[x] == "U" and scramble[x+1]== "D":
scramble[x+1] == random.choice(moves)
if scramble[x] == "F" and scramble[x+1] == "B":
scramble[x+1] == random.choice(moves)
modifiers = ["","2","'"]
for x in range(25):
randy = random.randint(0,2)
finascramble.append(scramble[x]+modifiers[randy])
return " ".join(finascramble)
t = True
while t == True:
again = input()
if again == "s":
print()
print(getscramble())回答 3
Stack Overflow用户
发布于 2018-07-01 04:50:02
我给你提供这个解决方案：
import random
MOVES = ['R', 'L', 'D', 'U', 'B', 'F']
MODFIERS = ['', '2', "'"]
def getScramble(length=25):
return ''.join([random.choice(MOVES) for _ in range(length)])
def isValid(scramble):
for seq in ['LR', 'DU', 'BF'] + [move * 2 for move in MOVES]:
if seq in scramble or seq[::-1] in scramble:
return False
return True
while True:
scramble = getScramble()
if isValid(scramble):
break
scramble = ' '.join([move + random.choice(MODFIERS) for move in scramble])
print(scramble)Stack Overflow用户
发布于 2018-07-01 05:03:52
我已经重写了你的加扰函数，使之更高效，完全符合你的期望。
import random
moves = ("R", "F", "U", "L", "B", "D")
modifiers = ("", "2", "'")
def getscramble(n=25):
scramble = random.choices(moves, k=n)
def is_valid(c, c2):
together = c + c2
return (c != c2 and together not in
("RL", "UD", "FB", "LR", "DU", "BF"))
for x in range(1, n):
before = scramble[x - 1]
current = scramble[x]
while not is_valid(before, current):
current = scramble[x] = random.choice(moves)
for i, x in enumerate(random.choices(modifiers, k=n)):
scramble[i] += x
return " ".join(scramble)Stack Overflow用户
发布于 2018-07-01 05:19:43
如果追加的最后一个值相同，则不能追加列表。
下面是一个更好的while循环，它可以执行您想要的操作：
i = len(scramble)
while i < 25:
current_face = random.choice(moves)
if i == 0:
scramble.append(current_face)
elif scramble[i - 1] == current_face:
continue
else:
scramble.append(current_face)
i += 1
print(' '.join(scramble))此循环确保列表具有固定长度。您可以将修改器添加到此列表中。
但是..。如果你不熟悉使用修饰符的概念，为了随机性和简单性，最好的做法是使用上面的循环和所有可能的排列。
import random
def getscramble():
moves = ["R", "F", "U", "L", "B", "D",
"R'", "F'", "U'", "L'", "B'", "D'",
"R2", "F2", "U2", "L2", "B2", "D2"]
scramble = []
i = len(scramble)
while i < 25:
curent_face = random.choice(moves)
if i == 0:
scramble.append(curent_face)
elif (scramble[i - 1])[0] == curent_face[0]:
continue
else:
scramble.append(curent_face)
i += 1
print(' '.join(scramble))
getscramble()以上代码的输出为：
L' R D2 U L' U' F R2 L' U2 D' R' F' B2 F2 L2 R' D' L2 F' U2 F U2 R D
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51118278
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