Python 刽子手游戏错误类型错误：'；非类型'；对象是不可编辑的

我是一个初学者程序员有两周的时间，这是我学习了一些基础知识后的第一个程序。这是一个刽子手游戏，但它是一个非常粗糙的版本，没有视觉效果（这将在稍后发布）。当我运行它时，它允许我选择一个字母，但在这之后它会给我错误“noneType对象不可编辑”

有人能解释为什么会这样吗？ 很抱歉代码的质量非常糟糕，但是当我浏览类似错误的其他问题时，代码太复杂了，我无法应用于我的情况

import time
import random

"""This is a hang man game
"""

name = input("what is your name?")


print("Hello" + " " + name)
time.sleep(1)
print("Ok, lets paly hangman, let me choose a word......")
time.sleep(1)
print("Thinking, thinking, thinking....")
time.sleep(3)
print("AHA! got it, ok " + name + ", take your first guess")
time.sleep(1)
def hangman_game():

    word_list = ["herd", "heaven", "manly", "startle"]
    number = random.randint(0, len(word_list) - 1)
    secret_word = word_list[number] #chooses a word from the word list


    blank_char = "-"
    word_length = len(secret_word) * blank_char
    word_length_split = list(word_length)
    print("Try and guess the word: ", word_length) #displays the length of the word in '_'


    previous_guesses = []
    guesses_taken = 0
    turns = 10
    secret_word_list = list(secret_word)

    while turns > 0:
        letter_guessed = input("Pick a letter: ")
        if letter_guessed in previous_guesses:
                print("You have already guessed this letter")
        else:
            previous_guesses = previous_guesses.append(letter_guessed)
            if letter_guessed in secret_word_list:
                secret_word_list = secret_word_list.remove(letter_guessed)
                guesses_taken = guesses_taken + 1
                print("That is correct," + letter_guessed + " is in the word")
                time.sleep(1)
            else:
                turns = turns - 1
                guesses_taken = guesses_taken + 1
                print("Try again")

    print("Sorry mate, no turns left, play again?")

hangman_game()

---

更改

previous\u猜测=previous\u猜测。追加（字母\u猜测）

到前面的猜测。附加（字母猜测）

另外

secret\u word\u list=secret\u word\u list.删除（字母猜测）

到

secret\u word\u列表。删除（猜测字母）

如果您尝试以下操作：

In [5]: l = [1,2,3,4]

In [6]: print (type(l.remove(1)))
<class 'NoneType'>

---

`更改

previous\u猜测=previous\u猜测。追加（字母\u猜测）

到前面的猜测。附加（字母猜测）

另外

secret\u word\u list=secret\u word\u list.删除（字母猜测）

到

secret\u word\u列表。删除（猜测字母）

如果您尝试以下操作：

In [5]: l = [1,2,3,4]

In [6]: print (type(l.remove(1)))
<class 'NoneType'>

---

`

前一个空的\u猜到问题了吗？前一个空的\u猜到问题了吗？