當(dāng)我在下面的代碼塊中的 print 函數(shù)中使用 if-else 語(yǔ)句的常規(guī)語(yǔ)法時(shí)，出現(xiàn)如下所述的錯(cuò)誤，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我所說(shuō)的常規(guī) if-else 語(yǔ)句的意思是，if total_candies>1: return "candies" else: return "candy")使用三元運(yùn)算符的相同語(yǔ)句，print("Splitting", total_candies, "candy" if total_candies == 1 else "candies")
1 回答
侃侃爾雅
TA貢獻(xiàn)1801條經(jīng)驗(yàn) 獲得超16個(gè)贊
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，請(qǐng)注意，在傳遞函數(shù)方面，Python 不如 Javascript 通用——lambda它有其局限性，特別是它只是一個(gè)單行函數(shù)。相反，我建議只在 print 語(yǔ)句之外一起定義您的函數(shù)。
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)
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