按嵌套数组中的第一个元素对数组进行排序

Question

This is an incredibly basic question that I could do in Python in a matter of seconds- but I'm new to Javascript and maybe I just don't know the nomenclature for the language, but my research hasn't quite answered it.

I'm making an API call; and in response I've got:

let unordered_ranges = [[
    [1461913200000, 57, 69],
    [1380006000000, 75, 79],
    [1321344000000, 78, 79],
    [1276585200000, 69, 75],
    [1252998000000, 68, 76],
    [1234512000000, 79, 81],
    [1423814400000, 77, 78],
    [1489820400000, 69, 79]
]];

The first element in the nested arrays are timestamps in milliseconds. How do I sort the parent array chronologically using the nested timestamps?

So far I've got:

let ranges= unordered_ranges.sort(function (a, b) {
    return a > b
});

I understand .sort() is lexicographic; so I need to pass my own function to sort it; however this function doesn't quite do it.

Answer 1

You'll want to access the first array entry (since you only have one) and then Array.prototype.sort() with a standard numeric comparator using the first array item of each entry.

 let unordered_ranges = [[ [1461913200000, 57, 69], [1380006000000, 75, 79], [1321344000000, 78, 79], [1276585200000, 69, 75], [1252998000000, 68, 76], [1234512000000, 79, 81], [1423814400000, 77, 78], [1489820400000, 69, 79] ]]; let ranges = unordered_ranges[0].sort((a, b) => a[0] - b[0]) console.info(ranges) 

---

To explain the comparator, it's best to read the documentation...

If compareFunction is supplied, the array elements are sorted according to the return value of the compare function. If a and b are two elements being compared, then:

• If compareFunction(a, b) is less than 0, sort a to an index lower than b , ie a comes first.
• If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements. Note: the ECMAscript standard does not guarantee this behaviour, and thus not all browsers (eg Mozilla versions dating back to at least 2003) respect this.
• If compareFunction(a, b) is greater than 0, sort b to an index lower than a , ie b comes first.
• compareFunction(a, b) must always return the same value when given a specific pair of elements a and b as its two arguments. If inconsistent results are returned then the sort order is undefined.