python之匹配数字按顺序列出python

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}') 
 

请您参考如下方法:

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]