我使用Claypl进行了两组地址之间的模糊匹配。 文档说默认值是：
如果未给出成本，则所有默认为10％，另一个默认为10％
转换号码范围默认为全部。组件名称可以
缩写。
但是，与阅读此Q＆amp; a 这个示例，这似乎不匹配。这是一个示例：
agrepl("cold", "cool")
#> [1] FALSE
agrepl("cool", "cold")
#> [1] TRUE
从描述中，我想计算10％的10个字母单词中会有1个更改，但这是4中的1个。这是如何计算的？
I used Agrepl for fuzzy matching between two sets of addresses. The documentation says that the default is:
If cost is not given, all defaults to 10%, and the other
transformation number bounds default to all. The component names can
be abbreviated.
However, reading this q&a with this example, that doesn't seem to match up. Here is that example:
agrepl("cold", "cool")
#> [1] FALSE
agrepl("cool", "cold")
#> [1] TRUE
From the description, I'd imagine that calculating the 10% would be having 1 change in a 10 letter word, but this is 1 in 4. How exactly is this calculated?
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诚然，这是非常令人困惑的（至少对我来说！），但这是我的尝试。 链接答案说：
我们如何从0.1（成本）×4（图案长度）到1？好吧，
？cyspl指出max.dist表示为（添加了强调）；我将括号的子句表示意味着最大转换数为
上限（0.1*4）= 1。我们需要一个带有长度≥11的模式，以便forforfortote> tatter_length ）要从1增加到2 ...如果您想找出实际实现的位置，则必须深入研究C源代码，即功能，我们看到的地方
This is admittedly very confusing (at least to me!), but here's my attempt to explain it. The linked answer says:
How do we get from 0.1 (cost) × 4 (pattern length) to 1? Well,
?agreplnotes that themax.distis expressed as(emphasis added); I take the parenthetical clause to mean that the maximum number of transformations is
ceiling(0.1*4)= 1. We would need a pattern with length ≥ 11 in order forceiling(0.1*pattern_length)to increase from 1 to 2 ...If you want to find out where this is actually implemented, you have to dig fairly deep into the C source code, i.e. lines 59-60 of agrep.c, in the
amatch_regparamsfunction, where we see