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?
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: