我怎样才能找出字母和字母转换频率?

How can I find out the alphabet and alphabetic transition frequency?

示例:'peter piper picked a peck of pickled peppers'

首先,计算出一个字母出现的频率。

abc = 'peter piper picked a peck of pickled peppers'
set(abc)
freq = {}
for i in abc:
    freq[i] = abc.count(i)
freq

但是,我找不到如上转换的字母数字,如何获取?

  1. p>e : 5
  2. e>t : 1
  3. t>e : 1
  4. e>r : 3
  1. 使用 zip 将相同的字符串移位 1 以获得当前字符和下一个字符对
  2. 使用 collections.Counter 获取计数。它还可以将您的代码简化为一行。
from collections import Counter
    
Counter(zip(abc, abc[1:]))

Returns:

Counter({('p', 'e'): 5,
         ('e', 't'): 1,
         ('t', 'e'): 1,
         ('e', 'r'): 3,
         ('r', ' '): 2,
         (' ', 'p'): 5,
         ('p', 'i'): 3,
         ('i', 'p'): 1,
         ('i', 'c'): 2,
         ('c', 'k'): 3,
         ('k', 'e'): 1,
         ('e', 'd'): 2,
         ('d', ' '): 2,
         (' ', 'a'): 1,
         ('a', ' '): 1,
         ('e', 'c'): 1,
         ('k', ' '): 1,
         (' ', 'o'): 1,
         ('o', 'f'): 1,
         ('f', ' '): 1,
         ('k', 'l'): 1,
         ('l', 'e'): 1,
         ('e', 'p'): 1,
         ('p', 'p'): 1,
         ('r', 's'): 1})

collections.Counter是你的朋友:

from collections import Counter

sentence = 'peter piper picked a peck of pickled peppers'
letter_pairs = [sentence[i:i+2] for i in range(len(sentence) - 1)]
pair_freq = Counter(letter_pairs)

给予

Counter({'pe': 5, ' p': 5, 'er': 3, 'pi': 3, 'ck': 3, 'r ': 2, 'ic': 2, 'ed': 2, 'd ': 2, 'et': 1, 'te': 1, 'ip': 1, 'ke': 1, ' a': 1, 'a ': 1, 'ec': 1, 'k ': 1, ' o': 1, 'of': 1, 'f ': 1, 'kl': 1, 'le': 1, 'ep': 1, 'pp': 1, 'rs': 1})

您可以像这样过滤掉带有空格的对:

internal_pair_freq = {k: v for k, v in pair_freq.items() if ' ' not in k}

这当然也适用于单个字母:

letter_freq = Counter(sentence)

给予

Counter({'p': 9, 'e': 8, ' ': 7, 'r': 3, 'i': 3, 'c': 3, 'k': 3, 'd': 2, 't': 1, 'a': 1, 'o': 1, 'f': 1, 'l': 1, 's': 1})