Python - 你能"refresh"一个变量来用新的子变量重新初始化吗？

Question

假设我想声明一个字符串供整个类使用，但该字符串的一部分稍后可能会更改。是否可能只需要声明一次字符串；刷新"；晚些时候？

示例：

substring = ""
my_long_string = f"This is a long string using another {substring} in it."

def printMyString(newString):
substring = newString
print(my_long_string)
printMyString(newString="newer, better substring")

我是否可以调整此代码，以便最终打印出This is a long string using another newer, better substring in it.，，而不必稍后再次声明长字符串？

我唯一的猜测是这样的：

substring = ""
def regenerate_string(substring):
return f"This is a long string using another {substring} in it."

def printMyString(newString):
print(regenerate_string(newString))
printMyString(newString="newer, better substring")

但我想问一下，Python中是否有一个内置函数可以做到这一点？

Answer 1

不要使用f-string，用占位符{}声明模板字符串，必要时使用format()

my_long_string = "This is a long string using another {} in it."
print(my_long_string.format("newer, better substring")) # This is a long string using another newer, better substring in it.
print(my_long_string.format("some other substring")) # This is a long string using another some other substring in it.

你可以通过这种方式插入任意多个子字符串

string = "With substrings, this {} and this {}."
print(string.format("first", "second")) # With substrings, this first and this second.

您也可以使用变量

string = "With substrings, this {foo} and this {bar}."
print(string.format(foo="first", bar="second")) # With substrings, this first and this second.
print(string.format(bar="first", foo="second")) # With substrings, this second and this first.

Answer 2

这不起作用的原因是，您将my_long_string声明为函数f"This is a long string using another {substring} in it."的返回值，而不是函数本身。

如果要像在示例中那样使用子字符串，则需要将my_long_string声明为lambda : f"This is a long string using another {substring} in it."，这将在substring更改时更改返回值。

但正如盖伊所说，my_long_string.format(substring)是完全合理的。