Question

而不是显示有多少元音出现在最后计算它们。我该如何解决这个问题？

def main():  
    sentence=input("Please type a sentence")  
    countA=0     
    for i in sentence:    
        if i =="a":  
            countA+=1    
            print("The letter a appears:", countA,"times")    
    countE=0           
    for i in sentence:  
        if i == "e":  
            countE+=1  
            print("The letter e appears:", countE,"times")  
    countI=0  
    for i in sentence:  
        if i == "i":  
            countI+=1  
            print("The letter i appears:", countI,"times")  
    countO=0  
    for i in sentence:    
        if i == "o":  
            countO+=1  
            print("The letter o appears:", countO,"times")  
    countU=0  
    for i in sentence:  
        if i == "u":  
            countU+=1  
            print("The letter u appears:", countU,"times")  
    countY=0  
    for i in sentence:  
        if i == "y":  
            countY+=1  
            print("The letter y appears:", countY,"times")  


main()

Answer 1

Sentence.count()会更容易，但我为你解决了这个问题。从for循环中取出打印命令。

sentence=input("Please type a sentence")
countE=0           
for i in sentence:  
    if i == "e" or i == "E":  
        countE+=1  
print("The letter e appears:", countE,"times")  
countI=0  
for i in sentence:  
    if i == "i" or i == "I":  
        countI+=1  
print("The letter i appears:", countI,"times")  
countO=0  
for i in sentence:    
    if i == "o" or i == "O":  
        countO+=1  
print("The letter o appears:", countO,"times")  
countU=0  
for i in sentence:  
    if i == "u" or i == "U":  
        countU+=1  
print("The letter u appears:", countU,"times")  
countY=0  
#Or use sentence.count(letter)
#for i in sentence:  
 #   if i == "y" or i == "Y":  
  #      countY+=1  
sentence.lower()
print("The letter y appears:", sentence.count("y"),"times")

计算输入中的元音

1 个答案: