如何在python中编写一个字母bigram(aa,ab,bc,cd ... zz)频率分析计数器?

时间:2019-01-12 17:45:56

标签: python nltk counter

这是我当前的代码,它打印出输入文件中每个字符的频率。

from collections import defaultdict

counters = defaultdict(int)
with open("input.txt") as content_file:
   content = content_file.read()
   for char in content:
       counters[char] += 1

for letter in counters.keys():
    print letter, (round(counters[letter]*100.00/1234,3)) 

我希望它仅打印字母(aa,ab,ac ..zy,zz)的双字母组的频率,而不要打印标点符号。这该怎么做?

3 个答案:

答案 0 :(得分:0)

您可以围绕当前代码进行构建以处理对。通过添加另一个变量来跟踪2个字符,而不是仅跟踪1个字符,并进行检查以消除非字母。

from collections import defaultdict

counters = defaultdict(int)
paired_counters = defaultdict(int)
with open("input.txt") as content_file:
   content = content_file.read()
   prev = '' #keeps track of last seen character
   for char in content:
       counters[char] += 1
       if prev and (prev+char).isalpha(): #checks for alphabets.
           paired_counters[prev+char] += 1
       prev = char #assign current char to prev variable for next iteration

for letter in counters.keys(): #you can iterate through both keys and value pairs from a dictionary instead using .items in python 3 or .iteritems in python 2.
    print letter, (round(counters[letter]*100.00/1234,3)) 

for pairs,values in paired_counters.iteritems(): #Use .items in python 3. Im guessing this is python2.
    print pairs, values

(免责声明:我的系统上没有python2。如果代码中存在问题,请告知我。)

答案 1 :(得分:0)

有一种更有效的方法来计算二部图:Counter。首先阅读文本(假设文本不太大):

from collections import Counter
with open("input.txt") as content_file:
   content = content_file.read()

过滤掉非字母

letters = list(filter(str.isalpha, content))

您可能也应该将所有字母都转换为小写,但这取决于您:

letters = letters.lower()    

用剩余的字母自己构建一个zip,将其移动一个位置,然后计算二部图:

cntr = Counter(zip(letters, letters[1:]))

规范字典:

total = len(cntr)
{''.join(k): v / total for k,v in cntr.most_common()}
#{'ow': 0.1111111111111111, 'He': 0.05555555555555555...}

通过更改计数器,可以轻松地将解决方案推广到三边形等。

cntr = Counter(zip(letters, letters[1:], letters[2:]))

答案 2 :(得分:0)

如果您使用的是nltk

from nltk import ngrams
list(ngrams('hello', n=2))

[输出]:

[('h', 'e'), ('e', 'l'), ('l', 'l'), ('l', 'o')]

进行计数:

from collections import Counter
Counter(list(ngrams('hello', n=2)))

如果要使用python本机解决方案,请查看: