确定密文的字母频率 [英] Determining Letter Frequency Of Cipher Text

问题描述

我想要一个工具，找到某些类型的密文中的字母的频率。
假设它都是小写的a-z没有数字。编码的邮件在一个txt文件中

我试图构建一个脚本来帮助破解替换或可能的转置密码。

到目前为止的代码：

  cipher = open（'cipher.txt'，'U'）。 （）
 cipherfilter = cipher.lower（）
 cipherletters = list（cipherfilter）
 
 alpha = list（'abcdefghijklmnopqrstuvwxyz'）
 occurrences = {} 
在字母中的字母：
 occurrences [letter] = cipherfilter.count（letter）
 for letter in occurrences：
打印字母，出现[letter] 
  

到目前为止，它只显示一封信出现的次数。
如何打印此文件中找到的所有字母的频率。

解决方案

  import collections 
 
d = collections.defaultdict（int）
 for'test'：
d [c] + = 1 
 
 print d ＃defaultdict（< type'int'>，{'s'：1，'e'：1，'t'：2}）
   
 
从文件：
  myfile = open（'test.txt'）
在myfile中的行：
 line = line.rstrip（'\\\
'）
 for c in line：
d [c] + = 1 
  
对于 defaultdict 容器，我们必须给予感谢和赞美。否则，我们都会做这样的事情：
  s =andnowforsomethingcompletelydifferent
d = {} 
表示s中的字母：
如果字母不在d：
d [letter] = 1 
 else：
d [letter] + = 1 
  
 
I am trying to make a tool that finds the frequencies of letters in some type of cipher text. 
Lets suppose it is all lowercase a-z no numbers. The encoded message is in a txt file

I am trying to build a script to help in cracking of substitution or possibly transposition ciphers.

Code so far:
cipher = open('cipher.txt','U').read()
cipherfilter = cipher.lower()
cipherletters = list(cipherfilter)

alpha = list('abcdefghijklmnopqrstuvwxyz')
occurrences = {} 
for letter in alpha:
    occurrences[letter] = cipherfilter.count(letter)
for letter in occurrences:
    print letter, occurrences[letter]
All it does so far is show how many times  a letter appears.
How would I print the frequency of all letters found in this file.
解决方案
import collections

d = collections.defaultdict(int)
for c in 'test':
    d[c] += 1

print d # defaultdict(<type 'int'>, {'s': 1, 'e': 1, 't': 2})
From a file:
myfile = open('test.txt')
for line in myfile:
    line = line.rstrip('\n')
    for c in line:
        d[c] += 1
For the genius that is the defaultdict container, we must give thanks and praise. Otherwise we'd all be doing something silly like this:  
s = "andnowforsomethingcompletelydifferent"
d = {}
for letter in s:
    if letter not in d:
        d[letter] = 1
    else:
        d[letter] += 1


                        
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