Question

我有这个问题，它需要一个字典，并试图找到一个字母的每个实例，如果它存在并打印出来

 get_letter_frequency('all animals are equal but some animals are more equal than others')

会打印：

a appears 10 times
b appears 1 time
e appears 7 times
h appears 2 times
i appears 2 times
l appears 6 times
m appears 4 times
n appears 3 times
o appears 3 times
q appears 2 times
r appears 4 times
s appears 4 times
t appears 3 times
u appears 3 times

到目前为止，我的get_letter_frequency函数

def get_letter_frequency(a_string):
    dictionary = {}
    words = a_string.split()
    for letters in words:
      if letters != " ":
          if letters in dictionary:
              dictionary[letters] += 1
          else:
              dictionary[letters] = 1
      dictionary_keys = dictionary.keys()
      new_list = list(dictionary_keys)
      new_list.sort()
      for alphabet in new_list:
          if dictionary[alphabet] == 1:
              print(alphabet, "appears", dictionary[alphabet], "time")
          else:
              print(alphabet, "appears", dictionary[alphabet], "times")

但这反过来给了我字典中的所有字母并告诉我它出现了多少次。

all appears 1 time
animals appears 2 times
are appears 2 times
but appears 1 time
equal appears 2 times
more appears 1 time
others appears 1 time
some appears 1 time
than appears 1 time

你能帮忙吗？谢谢。

Answer 1

代码中的问题是：

for letters in words:

由于words是单词列表，变量letters将依次包含每个单词，而不是这些单词中的字母。

集合模块中的Counter类可以满足您的需求：

In [1]: from collections import Counter

In [2]: ctr = Counter('all animals are equal but some animals are more equal than others')

In [3]: list(ctr)
Out[3]: ['a', ' ', 'b', 'e', 'i', 'h', 'm', 'l', 'o', 'n', 'q', 's', 'r', 'u', 't']

In [4]: dict(ctr)
Out[4]: 
{' ': 11,
 'a': 10,
 'b': 1,
 'e': 7,
 'h': 2,
 'i': 2,
 'l': 6,
 'm': 4,
 'n': 3,
 'o': 3,
 'q': 2,
 'r': 4,
 's': 4,
 't': 3,
 'u': 3}

不使用集合模块

如果必须使用for循环并且没有模块执行此操作：

def get_letter_frequency(a_string):
    dictionary = {}
    for letter in a_string.replace(' ', ''):
        dictionary[letter] = dictionary.get(letter, 0) + 1
    for letter in sorted(dictionary.keys()):
        print(letter, "appears", dictionary[letter], ["time", "times"][dictionary[letter]>1])

例如：

>>> get_letter_frequency('all animals are equal but some animals are more equal than others')
a appears 10 times
b appears 1 time
e appears 7 times
h appears 2 times
i appears 2 times
l appears 6 times
m appears 4 times
n appears 3 times
o appears 3 times
q appears 2 times
r appears 4 times
s appears 4 times
t appears 3 times
u appears 3 times

Answer 2

错误在以下几行：

words = a_string.split()
for letters in words:

您期望它迭代words中的所有字母，但它实际上遍历单个单词。在Python中，您可以使用字符串，就好像它们是list个字符一样，因此将`for line更改为以下内容将起作用：

for letters in a_string:

Answer 3

我认为使用collections.Counter为您提供最优雅的解决方案，但如果您更喜欢使用自己的功能，请尝试

letters = list(a_string)

而不是str.split()。

Answer 4

假设你有这样的字符串（从你的例子中取出）

statement = 'all animals are equal but some animals are more equal than others'
characters = list(statement)

then

from collections import Counter

Counter(characters)

Output:

Counter({'a': 10,
         'b': 1,
         'e': 7,
         'h': 2,
         'i': 2,
         'l': 6,
         'm': 4,
         'n': 3,
         'o': 3,
         'q': 2,
         'r': 4,
         's': 4,
         't': 3,
         'u': 3})

似乎所有的答案或多或少都在同一条线上：D

Answer 5

标准库提供了这样的计数器：

>>> from collections import Counter
>>> words = "apple banana apple strawberry banana lemon cupcake battery staple"
>>> Counter(words)
Counter({' ': 8,
         'a': 12,
         'b': 4,
         'c': 2,
         'e': 7,
         'k': 1,
         'l': 4,
         'm': 1,
         'n': 5,
         'o': 1,
         'p': 6,
         'r': 4,
         's': 2,
         't': 4,
         'u': 1,
         'w': 1,
         'y': 2})

Answer 6

words = a_string.split()将字符串拆分为单词列表。然后循环遍历所有这些单词而不是字母。由于您似乎不关心空格" "，您可以通过

替换它

no_whitespace_str = a_string.replace(" ", "")
for letters in no_whitespace_str:
    ...

较短的版本是

def get_letter_frequency(str):
    occ = {v: str.count(v) for v in str.replace(" ","")}
    for letter, num in sorted(occ.iteritems()):
        print "{} appears {} times".format(letter, num)

Answer 7

sent = "all animals are equal but some animals are more equal than others"
dic = {}

for ele in sent:
    dic.setdefault(ele,0) += 1

这使得dic成为使用（print dic）

{'a': 10, ' ': 11, 'b': 1, 'e': 7, 'i': 2, 'h': 2, 'm': 4, 'l': 6, 'o': 3, 'n': 3, 'q': 2, 's': 4, 'r': 4, 'u': 3, 't': 3}

在python string中是list，因此您可以迭代字符串中的单个字符

字典中出现了多少个字母实例

7 个答案:

不使用集合模块