根据数值条带填充数组
我需要根据一个人的“年龄”创build一组数据。 这条带不是线性的，而是分组的
1,2,3,4,5,6,7,8,9,10,11-15,16-20,21-30,31-50,51 +
所以对于任何一组数据，我都需要通过这个把这个人放到合适的年龄“斗”
在循环结束时，我想看看每个乐队中有多less人。
现在，我已经通过使用非常粗糙的技术（又名如果这个..然后..如果..）;），但想知道如果我可以学习新技术，如果有任何图书馆或function，可能需要一个范围并相应地填充桶
这就是我会做的：
let buckets = '1,2,3,4,5,6,7,8,9,10,11-15,16-20,21-30,31-50,51+'.split(','), bucketByAge = {}; // our bucket-age connector where we search later on // prepare the index buckets.forEach(function(bucket, index) { // get the range (+ will be replaced with 1000; for the age it is infinity) // and transform them to numbers let range = bucket.replace('+', '-1000').split('-').map(Number); // if it was not a range, simulate one range[1] = range[1] || range[0]; // go through the range and fill our connector while (range[0] <= range[1]--) { bucketByAge[range[1]] = index; } }); // search function getBucket(age) { return { age: age, bucketRange: buckets[bucketByAge[age]], bucketIndex: bucketByAge[age] }; } console.log(getBucket(1)); console.log(getBucket(12)); console.log(getBucket(61)); 这里有什么好处：你build立索引一次，然后取得结果。 非常高效。
一个非常粗糙的o（n ^ 2）实现
const buckets = [ { lowerLim: 1, upperLim: 1, persons: 0 }, { lowerLim: 2, upperLim: 2, persons: 0 }, { lowerLim: 3, upperLim: 5, persons: 0 }, { lowerLim: 6, upperLim: 10, persons: 0 }, { lowerLim: 11, persons: 0 } ]; const persons = [{name: 'john', age: 1},{name: 'john', age: 2}, {name: 'john', age: 6}, {name: 'john', age: 20}, {name: 'john', age: 40}, {name: 'john', age: 7}, {name: 'john', age: 1}]; persons.forEach(person => { buckets.forEach(bucket => { const age = person.age; const bucketLowerLimit = bucket.lowerLim; const bucketUpperLimit = bucket.upperLim || Number.MAX_SAFE_INTEGER; if(age >= bucketLowerLimit && age <= bucketUpperLimit) { bucket.persons++; } }) }); buckets.forEach(bucket => { console.log(bucket.persons); });
您可以使用中间值和单独的最后一个值中的所有值的引用。
var slots = '1,2,3,4,5,6,7,8,9,10,11-15,16-20,21-30,31-50,51+'.split(','), last = slots.pop(), reference = Object.create(null), histogram = Object.create(null), i, value; slots.forEach(function (a) { var temp = a.split('-'), left = temp[0], right = temp[1] || temp[0]; histogram[a] = 0; reference[right] = a; while (left < right) { reference[left] = a; left++; } }); histogram[last] = 0; for (i = 0; i < 1000000; i++) { value = Math.floor(Math.random() * 100); histogram[reference[value] || last]++; } console.log(histogram); console.log(reference);
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