Question

我正在处理任何.txt文件的霍夫曼编码，所以首先我需要分析这个文本文件。我需要阅读它，然后分析。我需要像表一样“退出”：

信|频率（后者重复多少次）|霍夫曼代码（稍后会出现）

我开始时：

 f = open('test.txt', 'r')    #open test.tx
 for lines in f:
     print lines          #to ensure if all work...

如何按字母顺序从文件中订购字符：

with open("test.txt") as f_in:
    for line in f_in:
        for char in line:
            frequencies[char] += 1

非常感谢

Well I tried like this:
frequencies = collections.defaultdict(int)
with open("test.txt") as f_in:
    for line in f_in:
        for char in line:
            frequencies[char] += 1


 frequencies = [(count, char) for char, count in frequencies.iteritems()]
 frequencies.sort(key=operator.itemgetter(1))

但是编译器给我一个“错误” 在这里输入代码

我需要这个字母顺序 in for loop ，而不是频道结束...

Answer 1

要获得您的频率表，我会使用defaultdict。这只会迭代数据一次。

import collections
import operator

frequencies = collections.defaultdict(int)
with open(filename) as f_in:
    for line in f_in:
        for char in line:
            frequencies[char] += 1


frequencies = [(count, char) for char, count in frequencies.iteritems()]
frequencies.sort(key=operator.itemgetter(1))

Answer 2

with open('test.txt') as f: data = f.read()
table = dict((c, data.count(c)) for c in set(data))

Answer 3

我使用collections.Counter()：

制作了此解决方案

import re
import collections


if __name__ == '__main__':
    is_letter = re.compile('[A-Za-z]')

    frequencies = collections.Counter()
    with open(r'text.txt') as f_in:
        for line in f_in:
            for char in line:
                if is_letter.match(char):
                    frequencies[char.lower()] += 1

    # Sort characters 
    characters = [x[0] for x in frequencies.most_common()]
    characters.sort()
    for c in characters:
        print c, '|', str(frequencies[c])

正则表达式is_letter用于仅过滤我们感兴趣的字符。它提供的输出看起来像这样。

a | 177
b | 29
c | 7
d | 167
e | 374
f | 58
g | 100
h | 44
i | 135
j | 21
k | 64
l | 125
m | 85
n | 191
o | 105
p | 34
r | 185
s | 130
t | 146
u | 34
v | 68
x | 1
y | 14

阅读.txt文件并进行分析

3 个答案: