python 确定密文的字母频率
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Determining Letter Frequency Of Cipher Text
提问by Veedrac
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
我正在尝试制作一种工具，可以在某种类型的密文中查找字母的频率。让我们假设它都是小写的 az 没有数字。编码后的消息位于 txt 文件中
I am trying to build a script to help in cracking of substitution or possibly transposition ciphers.
我正在尝试构建一个脚本来帮助破解替换或可能的换位密码。
Code so far:
到目前为止的代码：
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.
到目前为止，它所做的只是显示一个字母出现的次数。我将如何打印在此文件中找到的所有字母的频率。
回答by mechanical_meat
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:
从一个文件：
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 defaultdictcontainer, we must give thanks and praise. Otherwise we'd all be doing something silly like this:
对于defaultdict容器的天才，我们必须感谢和赞美。否则我们都会做这样的蠢事：
s = "andnowforsomethingcompletelydifferent"
d = {}
for letter in s:
if letter not in d:
d[letter] = 1
else:
d[letter] += 1
回答by Veedrac
The modern way:
现代方式：
from collections import Counter
string = "ihavesometextbutidontmindsharing"
Counter(string)
#>>> Counter({'i': 4, 't': 4, 'e': 3, 'n': 3, 's': 2, 'h': 2, 'm': 2, 'o': 2, 'a': 2, 'd': 2, 'x': 1, 'r': 1, 'u': 1, 'b': 1, 'v': 1, 'g': 1})
回答by jacob
If you want to know the relative frequencyof a letter c, you would have to divide number of occurrences of c by the length of the input.
如果您想知道字母 c的相对频率，您必须将 c 出现的次数除以输入的长度。
For instance, taking Adam's example:
例如，以亚当的为例：
s = "andnowforsomethingcompletelydifferent"
n = len(s) # n = 37
and storing the absolute frequence of each letter in
并将每个字母的绝对频率存储在
dict[letter]
we obtain the relative frequencies by:
我们通过以下方式获得相对频率：
from string import ascii_lowercase # this is "a...z"
for c in ascii_lowercase:
print c, dict[c]/float(n)
putting it all together, we get something like this:
把它们放在一起，我们得到这样的东西：
# get input
s = "andnowforsomethingcompletelydifferent"
n = len(s) # n = 37
# get absolute frequencies of letters
import collections
dict = collections.defaultdict(int)
for c in s:
dict[c] += 1
# print relative frequencies
from string import ascii_lowercase # this is "a...z"
for c in ascii_lowercase:
print c, dict[c]/float(n)
