我正在为混合不同压缩算法的短字符串构建一个压缩器,而RLE就是其中之一,它正在解决这个问题。
我现在拥有的剧本如下,目前还不完整:
# -*- coding: utf-8 -*-
import re
dictionary = {'hello':'\§', 'world':'\°', 'the': '\@', 'for': '\]'}
a_test_string = 'hello******** to the world****!'
def compress(string, dictionary):
pattern = re.compile( '|'.join(dictionary.keys() ))
result = pattern.sub(lambda value: dictionary[value.group() ], string)
'''
Here I should also implement a snippet to check for characters beginning with "\" so that they won't get replaced and screw up the result.
'''
for character in string:
occurrence = string.count(character*2)
there_is_more_than_one_occurrence = occurrence > 1
if there_is_more_than_one_occurrence:
second_regex_pass_for_multiple_occurrences = re.sub('\*\*\*+', '/'+character+str(occurrence), result)
result = second_regex_pass_for_multiple_occurrences
print 'Original string:', string
print 'Compressed string:', result
print 'Original size:', len(string)
print 'Compressed size:', len(result)
compress(a_test_string, dictionary)
当我运行该功能时,我得到了这个:
Original string: hello******** to the world****!
Compressed string: \§/*6 to \@ \°/*6!
Original size: 31
Compressed size: 20
但我应该得到:
Original string: hello******** to the world****!
Compressed string: \§/*8 to \@ \°/*4!
Original size: 31
Compressed size: 20
我在这里做错了我将 6 作为重复字符的计数?
答案 0 :(得分:0)
我不会试图准确理解你在做什么,但一个好的调试方法是在你的for循环中添加一些“print”语句或者使用python调试器并查看实际发生的情况。尝试自己运行其中一些调用,看看返回的是什么。
我认为你的主要问题是“string.count”返回整个字符串的计数,因此当它第一次看到所有12个*
时(或者技术上所有6个{{1}模式) 1}})。当for循环检查下一组**
时,它仍在检查整个字符串。希望这会有所帮助。