如何判断语义搜索引擎的准确性？
我的表弟创建了一个语义搜索引擎，他声称他的搜索引擎是最准确的。
我见过很多语义搜索引擎，它们对我来说看起来都一样，因为它们的设计目的不是根据您输入的关键字为您提供结果。
那么如果你正在创建一个语义搜索引擎，如何确定其结果的准确性呢？
my cousin has created a semantic search engine and he claims that his search engine is the most accurate.
I've seen many semantic search engine and they all look the same to me, because they are not designed to give you results based on the keyword you type.
So if you are creating a semantic search engine, how to to determine the accuracy of its results?
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事实上，萨诺德的建议并不离谱。
您通常会做的是让一大群人尝试一堆标准查询。或者，如果您想让实验更加公平，您可以让每个用户选择自己的查询，以避免任何偏见指控（因为您可以选择您知道您的引擎擅长回答的标准查询）。
对于每个查询，用户将查看前 10 个左右的结果，并说明他们是否认为每个结果相关（您可能希望让用户按比例评分，而不仅仅是是/否）。
然后，对于每个查询，您可以计算准确度分数，具体取决于您设置实验的方式精确度和召回率< /a> 可能是最合适的措施，尽管这些措施依赖于已知的预期答案，而您可能不一定有。计算简单的百分比精度可能更简单、更合适。
要确定您的搜索引擎是否比竞争对手更好，您需要让相同的人在这些搜索引擎上执行相同的查询，并以相同的方式评分。完成此操作后，您可以计算搜索引擎的分数并将其与您自己的分数进行比较。
Actually sarnold's suggestion is not far off the mark.
What you would typically do is to take a whole bunch of people and have them try out a bunch of standard queries. Or if you wanted to make the experiment fairer you might let each user pick their own queries to avoid any accusation of bias (because you could pick standard queries you knew your engine was good at answering).
For each query the user would look through the first 10 or so results and say whether they thought each result was relevant or not (you may want to have users score on a scale rather than just yes/no).
Then for each of the queries you can calculate accuracy scores, depending on exactly how you set up the experiment Precision and Recall may be the most appropriate measures though these rely on having a known expected answer which you may not necessarily have. It may be simpler and more appropriate to calculate a simple percentage accuracy.
To determine whether your search engine was better than your competitors you'd have the same people perform the same queries on those search engines scoring in the same way. Having done this you can then calculate and compare the scores for the search engines against your own.