返回字典中单词的频率

Question

我认为我对如何解决此功能有正确的想法，但我不确定 为什么我没有得到文档字符串中显示的所需结果。谁能帮我解决这个问题？

def list_to_dict(word_list):
'''(list of str) -> dict
Given a list of str (a list of English words) return a dictionary that keeps 
track of word length and frequency of length. Each key is a word length and 
the corresponding value is the number of words in the given list of that 
length.
>>> d = list_to_dict(['This', 'is', 'some', 'text'])
>>> d == {2:1, 4:3}
True
>>> d = list_to_dict(['A', 'little', 'sentence', 'to', 'create', 'a', 
'dictionary'])
>>> d == {1:2, 6:2, 8:1, 2:1, 10:1}
True
'''
d = {}
count = 0
for i in range(len(word_list)):
length = len(i)
if length not in d:
    count = count + length
    d[length] = {count}
    count += 1
return d

Answer 1

使用Counter 绝对是最好的选择：

In [ ]: from collections import Counter
   ...: d = Counter(map(len, s))
   ...: d == {1:2, 6:2, 8:1, 2:1, 10:1}
Out[ ]: True

不使用“花哨的东西”，我们使用生成器表达式，我认为它同样花哨：

Counter(len(i) for i in s)

如果“通常”是指使用 for 循环，我们可以这样做：

d = {}
for i in s:
    if len(i) not in d:
        d[len(i)] = 1
    else:
        d[len(i)] += 1

Answer 2

只需对列表中的任何单词进行循环。在每次迭代中，如果长度不在字典中作为键，则创建值为 1 的新键，否则增加键的先前值：

def list_to_dict(word_list):
   d = dict()
   for any_word in word_list:
      length = len(any_word)  
      if length not in d:
          d[length] = 1
      else:
          d[length] += 1
   return d

Answer 3

您可以使用字典推导来迭代 s，它现在包含其原始元素的长度：

s = ['A', 'little', 'sentence', 'to', 'create', 'a', 'dictionary']
final_s = {i:len([b for b in s if len(b) == i]) for i in map(len, s)}

输出：

{1: 2, 6: 2, 8: 1, 2: 1, 10: 1}

简单地说：

new_d = {} 
for i in s:
    new_d[len(i)] = 0
for i in s:
    new_d[len(i)] += 1

输出：

{8: 1, 1: 2, 2: 1, 10: 1, 6: 2}