我对编程很陌生，对Python更陌生，我很难弄清楚这一点。我正在尝试制作一个简单的文本hangman游戏，并尝试验证player1的输入。这是我迄今为止所拥有的。
a = 0
int = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
d = input()
d = list(d)
print(d)
if any(int) == any(d):
print("Invalid input. Please choose a non-number with no spaces or special characters.")
我不明白为什么，无论我的响应中是否包含一个数字，它总是满足我的any(int(==any(d(条件。我知道我一定只是误解了any((的工作原理。
您是对的：问题在于如何使用any()函数。any()返回True或False，这取决于其可迭代参数中的任何元素是否为True，换句话说，如果它们存在并且不是False或空列表或空字符串或0。对于int，这将始终计算为True，因为int确实包含一个True元素，而对于d，也会发生同样的情况，除非在input()函数提示时您没有提供任何输入(您在不键入任何内容的情况下点击return(。你的条件基本上是问以下问题：
if True==True
要解决此问题，只需将代码更改为以下内容：
a = 0
int = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
d = input()
print(d)
for letter in d:
if letter in int:
print("Invalid input. Please choose a non-number with no spaces or special characters.")
break
然而，最简单的解决方案根本不涉及int列表。事实上，python中有一种特殊的方法可以衡量字符串中是否有数字(或特殊字符(，而这种方法.isalpha()可以用来进一步简化这个过程。以下是如何实现此解决方案：
while True:
d = input("Please choose any word by typing it here: n")
print(d)
if not d.isalpha():
print("Invalid input. Please choose a non-number with no spaces or special characters.")
试试这个：
a = 0
ints = ["1","2","3","4","5","6","7","8","9","0"]
while a == 0:
print("Please choose any word by typing it here:")
word = input()
word_list = list(word)
if any(letter in ints for letter in word_list):
print("Invalid input. Please choose a non-number with no spaces or special characters.")