fatesses
2019-07-14 19:05
采纳率: 98%
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如何修改这段python代码?

def test(got, expected):
    if got == expected:
        prefix = '正确!'
    else:
        prefix = '错误!'
    print('%s 你的结果: %s 应该返回的结果: %s' % (prefix, repr(got), repr(expected)))

def match_ends(words):
    for item in match_ends:
        if len(item)>2 and item[0]==item[-1]:
            number = number+1
        else :
            continue
    return

test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)

其中错误提示是:

TypeError            Traceback (most recent call last)
<ipython-input-59-28f9cc682b29> in <module>
     14     return
     15 
---> 16 test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
     17 test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
     18 test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)

<ipython-input-59-28f9cc682b29> in match_ends(words)
      7 
      8 def match_ends(words):
----> 9     for item in match_ends:
     10         if len(item)>2 and item[0]==item[-1]:
     11             number = number+1

TypeError: 'function' object is not iterable

该如何将这个函数更改为可迭代的呢?

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1条回答 默认 最新

  • fatesses 2019-07-14 19:28
    已采纳

    def match_ends(words):
    number = 0
    for item in words:
    if len(item)>=2 and item[0]==item[-1]:
    number = number+1
    else :
    continue
    return number

    
    
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