Question

例如：“两个鳟鱼吃吐司”将返回数字2（鳟鱼，吐司）

到目前为止，这是我的代码。做这个程序的最好方法是什么？

string = input("Enter a string ")

words = string.split()

number = 0

if (word[0].lower() == word[len(word)-1].lower()):
number += 1

print(number)

Answer 1

您非常接近所需的内容。您需要遍历words来测试每个单词：

string = input("Enter a string: ")
words = string.split()

number = 0
# iterate over `words` to test each word.
for word in words:
    # word[len(word)-1] can be replaced with just word[-1].
    if (word[0].lower() == word[-1].lower()):
        number += 1
print(number)

使用给定的示例运行上述程序会产生结果：

Enter a string:  Two trout eat toast
2

执行上述操作的一种更干净的方法可能是在迭代输入字符串并使用sum之前将输入字符串小写：

words = input("Enter a string: ").lower().split()
number = sum(word[0] == word[-1] for word in words) 
print(number) # 2

Answer 2

如果您想要单线：

print(sum([1 for word in input("Enter a string").lower().split() if word[0] == word[-1]]))

Answer 3

b = a.split()                                                                                              
c = 0                                                                                                          
for item in b:                                                                                                    
    if item[0] == item[-1]:                                                                                       
        c+=1 
print(c)

Answer 4

如果您想要一个简单的细分...

s = 'i love you Dad'
    l =[]
    l = s.split(' ')
    count = 0
    for i in l:
        if len(i)==1:
            print(i)
        else:
            if i[0].upper()==i[len(i)-1].upper():
                count = count+1
                print(i)
print(count)
Output: i
Dad
1

Answer 5

您在这里

example = "Two trout eat toast"
example = (example.lower()).split()

count = 0
for i in example:
    i = list(i)
    if i[0] == i[-1]:
        count += 1
print(count)

只需取example字符串，将其制成lower.()进行比较，然后将其拆分，即可得到words。

将每个word变成一个list并比较该list的开头和结尾之后

给定一个字符串，编写一个程序来计算以相同字母开头和结尾的单词数

5 个答案: