javascript - 将超集数组叠加到顺序/顺序很重要的子集数组上(在 Javascript 中)

Question

我有一组形式的数组

A= [1,2,3,4,5,6]
B= [7,8,9]
C= [5,6]
D= [9]

我想在子集(子序列)上“覆盖”右侧(后缀)超集(严格来说，超序列)，以便结果集看起来像这样:

A= [1,2,3,4,5,6] (unchanged, because not a subset of anything)
B= [7,8,9] (unchanged, because not a subset of anything)
C= [1,2,3,4,5,6] (C overlayed with A, because C is a subset of A)
D= [7,8,9] (D overlayed with B, because D is a subset of B)

我在 node.js 中这样做。我认为部分原因是我未能掌握的逻辑问题。我

现实世界的用例是合并路径名称，以便规范分类层次结构，该分类层次结构包含许多项目，其中混合了完整路径和 chop 路径，例如/Science/Biology 和/Biology 标准化为/Science/Biology

非常感谢您提供有关如何执行此操作的任何指示。

Answer 1

首先用 Haskell 编写这个只是为了让算法下降。

import Data.List (maximumBy, tails)
import Data.Map (Map, findWithDefault)
import qualified Data.Map.Strict as Map
import Data.Ord (comparing)

main :: IO()
main = putStrLn $ show $ normalize [[1..6], [7..9], [5..6], [9]]

normalize :: Ord a => [[a]] -> [[a]]
normalize xxs = map (\xs -> findWithDefault xs xs index) xxs
  where index = suffixIndex xxs

suffixIndex :: Ord a => [[a]] -> Map [a] [a]
suffixIndex xxs = Map.fromListWith (maxBy length) entries
  where entries = [ (suf, xs) | xs <- xxs, suf <- suffixes xs ]
        suffixes xs = drop 1 $ filter (not . null) $ tails xs

maxBy :: Ord b => (a -> b) -> a -> a -> a
maxBy f x y = maximumBy (comparing f) [x, y]

suffixIndex 将每个后缀映射到具有该后缀的最长列表。因此，例如 [[1,2,3], [2,3]] 导致索引看起来像 [2,3] -> [1,2,3] , [3] -> [1,2,3]。

建立索引后，只需应用映射(如果存在映射)，每个列表就会“规范化”(用你的话)。

现在使用 Javascript。

console.log(JSON.stringify(normalize([[1,2,3,4,5,6], [7,8,9], [5,6], [9]])));

function normalize(xxs) {
    var index = suffixIndex(xxs);
    return xxs.map(function (xs) {
        str = JSON.stringify(xs);
        return index.hasOwnProperty(str) ? index[str] : xs;
    });
}

function suffixIndex(xxs) {
    var index = {};
    xxs.forEach(function (xs) {
        suffixes(xs).forEach(function (suffix) {
            var str = JSON.stringify(suffix);
            index[str] = index.hasOwnProperty(str)
                ? maxBy(lengthOf, index[str], xs)
                : xs;
        });
    });
    return index;
}

function suffixes(xs) {
    var i, result = [];
    for (i = 1; i < xs.length; i++) result.push(xs.slice(i));
    return result;
}

function lengthOf(arr) { return arr.length; }

function maxBy(f, x, y) { return f(x) > f(y) ? x : y; }