波特词干算法实现问题？ [英] porter stemming algorithm implementation question?

问题描述

我正在尝试实现搬运程序阻止算法，但此时此刻我被困住了：

I am trying to implement porter stemming algorithm but i am stuck at this point:

步骤1b

Step 1b

(m>0) EED -> EE                    feed      ->  feed
                                   agreed    ->  agree
(*v*) ED  ->                       plastered ->  plaster
                                   bled      ->  bled
(*v*) ING ->                       motoring  ->  motor
                                   sing      ->  sing

feed的m等于1吗？ feed >> [c] vvc [] >> [c] vc []。

Isn't the m of feed equal 1? feed >> [c]vvc[] >>[c]vc[].

如果是这样，他为什么不将feed转换为费用
i知道这是错误的，任何人都可以解决吗？

If it was so why didn't he convert feed to fee i know it is wrong ,can any one clear that up?

您可以在此处 http://tartarus.org/~martin/PorterStemmer/def.txt

谢谢

推荐答案

m的提要确实为1。但是，您需要仔细阅读文档。条件中的m是指茎的尺寸，即您需要在替换后的 中进行计算。就您而言，检查 feed->费用有效，您计算出m（fee）= 0，因此无需进行替换。

m of 'feed' is indeed 1. Yet, you need to re-read the document carefully. The m in the condition refers to the measure of the stem, that is you need to calculate in after the replacement. In your case to check if feed -> fee is valid, you calculate m(fee) = 0, hence you don't do the replacement.

还要感谢算法！真有趣！

Also thanks for the algorithm! It was interesting!

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