python - 在 LZW 中压缩时计数？

Question

注意:这不是 LZW 压缩的正确用法。我只是在玩弄它。

问题

在一次传递中，是否可以同时更新字典中元素的频率计数？

我的实现

import sys
from collections import defaultdict
import re

# The silliest string!
inputString = "this is the first sentence in this book the first sentence is really the most interesting the first sentence is always first"
inputString = inputString.lower().split()

StringTable = defaultdict(int)
FreqTable = defaultdict(int)

def DoPass():
    global inputString
    global StringTable
    global FreqTable

    print ""
    print "INPUT STRING:"
    print inputString

    CODE = 256

    STRING = inputString[0]

    output = []

    StringTable[STRING] = CODE
    CODE += 1

    total = len(inputString)

    for i in range(1, total):
        WORD = inputString[i]

        if STRING + " " + WORD in StringTable:
            STRING += " " + WORD
        else:
            if STRING in StringTable:
                output.append(str(StringTable[STRING]))
            else:
                output.append(STRING)
            StringTable[STRING + " " + WORD] = CODE
            CODE += 1
            STRING = WORD

    StringTable[STRING] = CODE
    CODE += 1
    output.append(str(StringTable[STRING]))

    print ""
    print "OUTPUT STRING:"
    print output

    print ""
    print "Dictionary Built..."
    for i in sorted(StringTable.keys(), key=lambda x: len(x)):
        print i, StringTable[i]

    print ""
    print "Frequencies:"
    for i in sorted(FreqTable.keys(), key=lambda x: len(x)):
        print i, FreqTable[i]

def main():
    DoPass()

if __name__ == "__main__":
    main()

输出

INPUT STRING:
['this', 'is', 'the', 'first', 'sentence', 'in', 'this', 'book', 'the', 'first', 'sentence', 'is', 'really', 'the', 'most', 'interesting', 'the', 'first', 'sent
ence', 'is', 'always', 'first']

OUTPUT STRING:
['256', 'is', 'the', 'first', 'sentence', 'in', '256', 'book', '259', 'sentence', 'is', 'really', 'the', 'most', 'interesting', '265', 'is', 'always', '275']

Dictionary Built...
this 256
first 275
is the 258
in this 262
this is 257
book the 264
the most 269
this book 263
is always 273
is really 267
the first 259
really the 268
sentence in 261
sentence is 266
always first 274
first sentence 260
interesting the 271
most interesting 270
the first sentence 265
the first sentence is 272

Frequencies:
#### I am trying to fill this

我想用它找到的任何模式的频率计数填充 FreqTable。出于明显的原因，我没有将我的方法放在这里 - 它不起作用并且它给了我错误的计数。关于这是否可能的任何建议都会很棒。

Answer 1

不确定是否理解您的问题。如果您只需要频率表，那么这应该很简单:每次找到一个模式时，您只需将 +1 添加到它的频率计数。所以真正的问题应该是找到规律。

如果你想保持模式排序，这也应该很容易，因为你一直保持表格排序，它最终是一个插入排序操作，这是非常快的。

现在，找到正确的模式是另一回事。你需要一棵树，或者一个哈希表，然后是树，或者列表，或者其他什么，来找到你最好的匹配序列。这就是使此类算法执行起来更加复杂的原因。

显然，对于非常小的数据集，“简单”搜索(一项一项地测试所有条目)可能会给出一些结果。但随着数据集的扩大，搜索成本将变得令人望而却步。