计算同现的更有效方法

Question

I want to visualise a network of relationships between people.

My data looks like this :

let networkData = [
    ["John","Dylan","Brian"],
    ["Dylan","Brian"],
];

And I want an output like this :

let networkMatrix.links = [
    {source: "John",  target: "Dylan", weight: 1},
    {source: "John",  target: "Brian", weight: 1},
    {source: "Brian", target: "Dylan", weight: 2}
];

John and Brian share one group so their weight is 1. The same is true for John and Dylan. As Dylan and Brian share in two groups, their relationship gets a weight of 2.

Now here is where I think I need help. My way of doing that was to go through each line of networkData , then through each element of the array. And for each element, go through all the elements that come after, and increment their score in networkMatrix . Here is my code.

var i = 0, j = 0;
networkData.map(function(d) {
    d.forEach(function(val, ind, tab) {
        for (let k = ind + 1; k < tab.length; k++) {
            while ((i = networkMatrix.person1.indexOf(val, i + 1)) != -1) {
                while ((j = networkMatrix.person2.indexOf(tab[k], j + 1)) != -1) {
                    if (i === j) {
                        networkMatrix.score[i]++;
                    }
                }
            }
        }
    })
});

Here is a jsfiddle : https://jsfiddle.net/0t81jg3b/2/

As you can see it doesn't even work on jsfiddle. But it works more or less on my computer, I don't know why :

If feel like this is getting waaaaay too complicated for such a simple task, could someone give me some indications on how to get out of this mess?

Answer 1

You could do it as follows:

• Only consider name pairs where the source is alphabetically less than the target
• Build a map keyed by source, and for each a nested map keyed by target, and assign a counter to those nested keys. You could use an ES6 Map for that, or else plain object

 const groupes = [ ["John", "Dylan", "Brian"], ["Dylan", "Brian"], ["John", "Kate"] ], // Get unique list of names names = [...new Set([].concat(...groupes))], // Create a 2D-map with a counter set to zero map = new Map(names.map(name => [name, new Map(names.map(name2 => [name2, 0]))])), result = []; // Count the pairs for (const groupe of groupes) { for (const source of groupe) { const targetMap = map.get(source); for (const target of groupe) { if (source < target) targetMap.set(target, targetMap.get(target)+1); } } } // Convert nested maps to result structure for (const [source, targetMap] of map.entries()) { for (const [target, weight] of targetMap.entries()) { if (weight) result.push({source, target, weight}); } } console.log(result); 

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