python - 匹配数字按顺序列出python

Question

我的问题是试图让两个列表按特定顺序彼此相等

例如 [1,3,5,7,4,3,2,5,7,10] == [1,3,5,7,4,3,2,5,7 ,10]

我想继续计算 10 个数字的随机列表，直到它们按特定顺序彼此相等。

我遇到的问题是如何使它们按特定顺序彼此相等？我知道我不能做硬编码数字，因为它们可以是任何地方的任何值。

我知道语句 i[] == j[] 是不正确的，那是我想使两个列表相等但不确定要放什么的地方

import random
hats = []
people = []
count = 0

while True:
    for i in range(11):
        hats.add(random.choice([1,2,3,4,5,6,7,8,9,10]))
    for j in range(11):
        people.add(random.choice([1,2,3,4,5,6,7,8,9,10]))
        if(i[] == j[]):
             stop == True
             break
        else:
             count+=1
print(f'Count is {count}')

Answer 1

这样就可以了，但是你可能会坐很长时间:

import random
hats = []
people = []
count = 0
while len(hats) != 10:
  k=0
  while k != 10:
      n = random.choice([1,2,3,4,5,6,7,8,9,10])
      m = random.choice([1,2,3,4,5,6,7,8,9,10])
      if(n == m):
        hats.append(n)
        people.append(m)
        k += 1
      else:
        hats   = []
        people = []
        count += 1
        k=10
print(count, hats, people)

如果它们不同，我不确定您是否必须重置列表，另一种更短的方法是:

import random
hats = []
people = []
count = 0
while len(hats) != 10:
  k=0
  while k != 10:
      n = random.choice([1,2,3,4,5,6,7,8,9,10])
      m = random.choice([1,2,3,4,5,6,7,8,9,10])
      if(n == m):
        hats.append(n)
        people.append(m)
        k += 1
      else:
        count += 1
        k=10

print(f"times: {count}", f"hats: {hats}", f"people: {people}")

输出示例:

times: 130 hats: [7, 7, 9, 5, 1, 5, 1, 7, 4, 8] people: [7, 7, 9, 5, 1, 5, 1, 7, 4, 8]