python - 单词之间只保留一个空格，用replace为什么会出错

Question

在python中，字符串是没有contains函数的，不过可以使用find来代替。现在有一串单词，有的相邻两个单词之间可能不止一个空格，现在我只想保留一个空格，使用下面的办法就是无效的，不知道为什么这样会出现死循环。我知道可以使用' '.join(a.split())来解决问题，但是下面的代码不知道为什么会出错，是我的思想错误，还是其他原因？在java等其他语言里面，使用这种思想可行么？
def StringUnderLine(str):

str.strip();
n=0;
while str.find('  '):
    str.replace('  ',' ')
    n=n+1
    print(n,end='')
print(str,end='')

a="1 2 3 4 5 "
b="1 "
c=" 2"
d="abc def"
StringUnderLine(a)
StringUnderLine(b)
StringUnderLine(c)
StringUnderLine(d)

Answer 1

str.find(' ') returns -1 if not found and the while loop is True to loop infinitely, so while -1 will loop infinitely
Also, to make str.replace(' ', ' ') effective,
should

a = '1    3   5'
while a.find('  ') != -1:
    a = a.replace('  ', ' ')
   
如果只是a.replace('  ', ' ')，那么循环执行的永远只是a = '1    3   5'， 也就是初始字符串，

Answer 2

str = str.replace(' ',' ');
In JAVA, C#, the string variable itself is immutable. In other words, replace does not change the original string.

Answer 3

Let me digress first, you brother, if you ask python, you only label python, why do you label so many languages ​​​​?

---

If you are not sure about the length of the space, you can use regular matching. I don’t know python, so I will give you an idea.

I checked it out and found that all non-zero numbers in python represent true. If str.find cannot find a match, it will return -1. , if found, it will return the first matching position, so the brother above is right, if you use it as a condition, it will be equivalent to true when it is not found, but it will be regarded as false when it finds the match at position 0, so I want to see If str.find finds a match, check it out! = -1 is correct

Answer 4

str has not been changed by assignment. replace returns a new string instead of modifying str directly