Question

我如何在Python中列出元组中元音的出现，然后根据元音数将其附加到字典？

这是我目前的尝试：

def vowelCount(input1):
    d = {'half vowels': None,
         'mostly consonant': None, 'mostly vowels': None}
    vowels = 'aeiou'
    con = 0
    for word in input1:
        print(word, len(word))
        for vowel in vowels:
            if vowel in word:
                print(word.count(vowel))
                # FIXME: I can't figure out how to
                # sum up the occurrences
                #if (word.count(vowel)) < len(word):
                #    print(word)

以下是对我的期望的描述：

its 3

1

a 1

1

death 5

1

1

trap 4

1

its 3

1

a 1

1

suicide 7

1

2

1

rap 3

1

we 2

1

gotta 5

1

1

get 3

1

out 3

1

1

while 5

1

1

were 4

2

young 5

1

1

Answer 1

我想我理解你的问题，但我并非百分百肯定，如果这不是你正在寻找的，那么道歉。元音数量有限，因此您可以通过定义元音并保持活动计数来确定有多少元音。

def vowel_count(s):
    """
    @param s is a string representing the contents of the list
    """

    vowels = ['a', 'e', 'i', 'o', 'u']
    counts = {}
    for v in vowels:
       counts[v] = 0
    for c in s:
       if c in vowels:
           counts[c] += 1
    return counts

这些方面的东西？

Answer 2

根据你的代码，我就是你想要的。对于每个单词，将vowelcount重置为0并继续查找单词中的元音数量。根据此计数，构建您的字典D.

def vowelCount(input1):

    d={'half vowels': [], 'mostly constant':[], 'mostlyVowels':[]}
    vowels='aeiou'
    for word in input1:
        print(word, len(word))
        vowelcount = 0
        for vowel in vowels:
            if vowel in word:
                vowelcount += word.count(vowel)
        if vowelcount == len(word)/2:
           d["half vowels"].append(word)
        elif vowelcount < len(word)/2:
            d["mostly constant"].append (word)
        else:
             d["mostlyVowels"].append(word)  
    return d

然而，这段代码很乱。尝试将这些功能分离为独立的可测试功能：

def vowelcount(word):
    # ...
    return number of vowels in word

def constructSummary(listOfWords):
    d = { }
    for word in words:
        count = vowelcount(word)
        # .. If else ladder based on count 
        # construct d based on count
    return d

Answer 3

以您的代码为基础，您只需修改一些内容，如下所述。结果是你的isLoggedIn字典，所有单词都正确排序。这个版本将一个句子作为输入，在空格处分成一个列表：

<ul class="nav navbar-nav navbar-right" ng-controller="loginController">
    <li class="dropdown">
        <a href="#" class="dropdown-toggle" 
                    data-toggle="dropdown" 
                    ng-show="isLoggedIn" 
                    ng-click="logout()">
          <b>Logout</b></a>
        <a href="#" class="dropdown-toggle"
                    data-toggle="dropdown" 
                    ng-show="!isLoggedIn">
          <b>Login</b> <span class="caret"></span>
        </a>
    </li>
</ul>

此代码将提供以下结果：

计算元音并附加到字典

3 个答案: