我可以在 print() 函数或常规 if-else 语句中定义函数并且不在 Python

Question

当我在下面的代码块中的 print 函数中使用 if-else 语句的常规语法时，出现如下所述的错误，def to_smash(total_candies):"""Return the number of leftover candies that must be smashed after distributingthe given number of candies evenly between 3 friends.>>> to_smash(91)1"""print("Splitting", total_candies, (def plural_or_singular(total_candies):    if total_candies>1:        return "candies"    else:        return "candy"),plural_or_singular(total_candies))return total_candies % 3to_smash(1)to_smash(15)#################################################################################### Output:File "<ipython-input-76-b0584729b150>", line 10    (def plural_or_singular(total_candies):       ^SyntaxError: invalid syntax我所说的常规 if-else 语句的意思是，if total_candies>1:        return "candies"    else:        return "candy")使用三元运算符的相同语句，print("Splitting", total_candies, "candy" if total_candies == 1 else "candies")

Answer 1

def to_smash(total_candies):

    print("Splitting", total_candies, (lambda total_candies: "candies" if total_candies > 1 else "candy")(total_candies))        

    return total_candies % 3

to_smash(1)

to_smash(15)

但是，请注意，在传递函数方面，Python 不如 Javascript 通用——lambda它有其局限性，特别是它只是一个单行函数。相反，我建议只在 print 语句之外一起定义您的函数。

def to_smash(total_candies):

    def plural_or_singular(total_candies):

        if total_candies>1:

            return "candies"

        else:

            return "candy"

    print("Splitting", total_candies, plural_or_singular(total_candies))

    return total_candies % 3

to_smash(1)

to_smash(15)