Question

编写一个名为symmetry的函数，它将字符串text作为参数并返回字典d。每个字母都是文本中某个单词的第一个和最后一个字母，应该是d中的一个键。例如，如果文本包含单词'shucks'，则's'应该是d中的键。 d中's'的值是以's'开头和结尾的单词数。参数文本仅包含大小写字符和空格。

以下是正确输出的示例：

t = '''The sun did not shine
it was too wet to play
so we sat in the house
all that cold cold wet day

I sat there with Sally
we sat there we two
and I said how I wish
we had something to do'''

print(symmetry(t))
{'d': 1, 'i': 3, 't': 1}

我已经多次尝试过这个问题的答案，但我似乎无法得到它。使用我的代码，每个字母都放入字典中，而不是仅用于开始和结束单词的字母。能帮助我更好地理解这个吗？

到目前为止，这是我的代码：

def symmetry(text):
   d = {}
   text.split()
   for word in text:
      if word[0] == word[-1]:
         d[word[0]] =+ 1
      else:
   return word
return d

text = ''' The sun did not shine
it was too wet to play
so we sat in the house
all that cold cold wet day

I sat there with Sally
we sat there we two
and I said how I wish
we had something to do'''

print(symmetry(text))

Answer 1

from collections import defaultdict
result = defaultdict(lambda: 0)
for letter in [word[0] for word in t.split() if word[0] == word[-1]]:
    result[letter] += 1

应该给你一个想法：

现在，您需要更改if条件，以便它适用于小写和大写字母。

了解列表推导，它将解释表达式如何：

[word[0] for word in t.split() if word[0] == word[-1]]

可以形成，它们是什么意思。

Answer 2

在split()上致电text后，您需要存储所获得的字词列表。现在，它并没有保存该列表，因此word实际上采用了文本中每个字母的值。

Answer 3

你快到了。首先定义一个默认字典，第一个字母为键，0为值。然后迭代拆分列表，只有当项目的第一个字母和最后一个字母相同时，才将项目附加到定义的字典中。最后使用dict理解来过滤值大于0的dict项。

text = ''' The sun did not shine
    it was too wet to play
    so we sat in the house
    all that cold cold wet day

    I sat there with Sally
    we sat there we two
    and I said how I wish
    we had something to do'''
def symmetry(t):
    j = t.split()
    d = {}
    for i in  j:
        if i[0] == i[-1]:
            d[i[0]] = d.get(i[0], 0) + 1
    return {i:j for i,j in d.items() if j > 0}

>>> print symmetry(text)
{'I': 3, 'd': 1, 't': 1}
>>>

Answer 4

              def symmetry(text):
               d = {}
               list = text.split(" ")
               for word in list:
                  word = word.strip()
                  if word and word[0] == word[len(word)-1]:
                     try:
                        d[word[0]] = d[word[0]] + 1
                     except:
                        d[word[0]] = 1
               return d

            text = ''' The sun did not shine
            it was too wet to play
            so we sat in the house
            all that cold cold wet day 

            I sat there with Sally
            we sat there we two
            and I said how I wish
            we had something to do'''

            print(symmetry(text))

Answer 5

最常用的解决方案是使用Counter模块中的collections。

from collections import Counter

def symmetry(text):
    return Counter(w[0] for w in text.lower().split() if w[0] == w[-1])

如果你真的需要一个字典而不是类字典对象（Counter实际上是从dict继承），你可以这样做：

def symmetry(text):
    return dict(Counter(w[0] for w in text.lower().split() if w[0] == w[-1]))

Related article I wrote comparing the different methods for counting things

字典文本以相同的字母开头和结尾

5 个答案: