如何在巨大的 json 文件之间执行最优模式搜索？

Question

Looking for solution to perform pattern search between JSON files which could traverse the huge JSON file without much impact on performance. Following are the few test cases.

Search criteria

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1. 'cabin_1' matches with 'cabin_1'
2. 'cabin_3' matches with 'cabin 3' or '3 cabin'
3. 'first cabin' matches with '1st cabin'

Test case files

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you can find the test json files here

My Idea

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for each json1Property in json1
     for each json2Property in json2
        isMatch = regex('somepattern', json1property , json2property)
        if (isMatch) 
           return true 
        else 
           return false

Answer 1

This is rather basic and I'm no algorithm expert, but basically, the goal is to build a simple index for each array. You simplify and map values to something more easy/fast to compare later. I think one way or another, you have to iterate on arrays.

Here, you iterate once on each array to build the indices, while in your first attempt, you have a double loop.

The double loop is a bit existing in the second phase, comparing indices with filter / includes but I think it would be lighter because the arrays' length has decreased and the data is simpler to check.

 const data = { "Building": { "floor": [ { "space": [ "cabin_1", "cabin_2", "cabin_3", "mycabin" ] }, { "space": [ "first cabin", "xyz's cabin", "Zone c", "Zone d" ] } ] } }; const spaces = data.Building.floor; const indices = spaces.reduce((acc, item) => { acc.push(item.space.map(it => { return it.replace(/ ?cabin[_ ]?/g, '') //Remove cabin, trailing spaces and underscores. .replace(/1st|first/g, '1') //Map things that are not numbers to numbers. .replace(/2nd|second/g, '2') .replace(/3rd|third/g, '3'); }).filter(it => !isNaN(it))); //Removes every thing that is not processed by the index engine. return acc; }, []); console.log(indices); let shorterArray, longerArray; if(indices[0].length > indices[1].length) { shorterArray = indices[1]; longerArray = indices[0]; } else { shorterArray = indices[0]; longerArray = indices[1]; } const sharedItems = shorterArray.filter(it => longerArray.includes(it)); console.log('Shared items found', !!sharedItems.length, sharedItems);