添加例外莱文斯坦 - 距离样算法 [英] Adding exceptions to Levenshtein-Distance-like algorithm

问题描述

我想计算出有类似的多达6个变量序列。目前我使用的是集合计数器返回不同的变量作为我的编辑距离的频​​率。

I'm trying to compute how similar a sequence of up to 6 variables are. Currently I'm using a Collections Counter to return the frequency of different variables as my edit-distance.

默认情况下，在编辑一个变量的距离（添加/分/变更）为1或0。我想改变的距离取决于变量，我​​设置了该变量。什么价值

所以我可以说某些变量类似于其他变量，并为他们多么相似的值。 我还想说的某些变量值得比平时少跌多的距离。

So I can say certain variables are similar to other variables, and provide a value for how similar they are. I also want to say certain variables are worth less or more distance than usual.

下面是我的previous岗位作为背景：<一href="http://stackoverflow.com/questions/32456943/modify-levenshtein-distance-to-ignore-order">Modify莱文斯坦 - 距离忽视订单

Here is my previous post as context: Modify Levenshtein-Distance to ignore order

例如：

# 'c' and 'k' are quite similar, so their distance from eachother is 0.5 instead of 1
>>> groups = {['c','k'] : 0.5}

# the letter 'e' is less significant, and 'x' is very significant
>>> exceptions = {'e': 0.3, 'x': 1.5}

>>> distance('woke', 'woc')
0.8

说明：

woke
k -> c = 1
woce
-e = 1
woc
Distance = 2

# With exceptions:
woke
k -> c = 0.5
woce
-e = 0.3
woc
Distance = 0.8

我怎么能做到这一点？这将是可以实现这个反算法？

目前的code（谢谢你，大卫Eisenstat）

Current code (thank you David Eisenstat)

def distance(s1, s2):
    cnt = collections.Counter()
    for c in s1:
        cnt[c] += 1
    for c in s2:
        cnt[c] -= 1
    return sum(abs(diff) for diff in cnt.values()) // 2 + \
        (abs(sum(cnt.values())) + 1) // 2

推荐答案

我结束了将过程分成几个阶段，然后通过每个阶段的串进行迭代。我不知道它是有效的，因为它可以，但它的工作原理。

I ended up dividing the process into a few stages then iterating through the strings for each stage. I'm not sure if its as efficient as it could be but it works.

总结我试图实现（与编辑距离算法）

Summing up what I was trying to achieve (in relation to Edit-distance algorithms)

• 从一个字母到另一个距离为1。变化的J - ＆GT; K = 1
• 0是没有任何区别的。例如变化的J - ＆GT; J = 0
• 类似的信件可以值得小于1（由我指定）如 C 和 K 音同，因此 C，K = 0.5 ，变动c - ＆GT; K = 0.5
• 在某些字母可能价值更多或更少（由我指定）如 X 是罕见的，所以我希望它有更多的重量， X = 1.4 ， x更改 - ＆GT; K = 1.4

• Distance from one letter to another is 1. change j -> k = 1
• 0 being no difference at all. e.g. change j -> j = 0
• Similar letters can be worth less than 1 (specified by me) e.g. c and k sound the same, therefore c, k = 0.5, change c -> k = 0.5
• Certain letters could be worth more or less (specified by me) e.g. x is uncommon so I want it to have more weight, x = 1.4, change x -> k = 1.4

创建2字典，1 类似于书信，1 的异常

Created 2 dictionaries, 1 for similar letters, 1 for exceptions

1. 填充计数器 - 遍历两个字符串
2. 匹配相似的项 - ，迭代字符串1，如果在类似快译通，迭代字符串2，如果在同类词典
3. 更新计数 - 删除类似的项目，
4. 查找距离 - 加起来绝对频率，占区别在字符串的长度
5. 包含例外距离 - 异常值的基础上的字母频率帐户

1. Populate Counter - Iterate through both strings
2. Match similar items - Iterate string1, if in similar dict, iterate string2, if in similar dict
3. Update Counter - remove similar items,
4. Find Distance - add up absolute frequencies, account for difference in string length
5. Include exceptions distance - Account for exception values based on frequency of letters

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