请求实现无input的Python猜数字程序（附尝试代码）

我有一段依赖input()的猜数字Python程序，现在想改成不依赖input的版本，打算用列表或变量来实现，但自己写的代码没成功，想请人帮忙完成。

原依赖input的代码 #

import random
number = random.randint(1, 10)

player_name = "doo"
number_of_guesses = 0
print('I\'m glad to meet you! {} \nLet\'s play a game with you, I will think a number between 1 and 10 then you will guess, alright? \nDon\'t forget! You have only 3 chances so guess:'.format(player_name))

while number_of_guesses < 3:
    guess = int(input())
    number_of_guesses += 1
    if guess < number:
        print('Your estimate is too low, go up a little!')
    if guess > number:
        print('Your estimate is too high, go down a bit!')
    if guess == number:
        break
if guess == number:
    print( 'Congratulations {}, you guessed the number in {} tries!'.format(player_name, number_of_guesses))
else:
    print('Close but no cigar, you couldn\'t guess the number. \nWell, the number was {}.'.format(number))

需求 #

• 实现无input()的版本
• 用列表或变量替代用户输入的部分

我尝试的代码（未成功） #

import random
list=[1, 2, 3, 4, 5, 8, 9, 10]
print('I am Guessing a number between 1 and 10:\n')
for number in lis:
    number_of_guesses = 0
    while number_of_guesses<3:
        guess_number=random.randint(1,10)
        if number<guess_number:
            number_of_guesses+=1
            print('Your guess number is high '+str(guess_number))
        elif number>guess_number:
            number_of_guesses+=1
            print('Your guess number is low '+str(guess_number))
        else:
            print("You guess Right The number is: "+str(guess_number)+"\nNumber of guess taken "+str(number_of_guesses+1))
            break
    if number_of_guesses==3:
        print("Sorry your chances of guessing is over! You can not guess the number correct")

希望能帮忙修正这段代码，实现无input()的猜数字功能。

内容的提问来源于stack exchange，提问作者dodo87