特殊排序对象数组

Question

So theres tons of posts on sorting something like this:

var arr = [{a: 1}, {b: 2}] alphabetically by key, but what if you have something like var arr = [{a: 100}, {a: 50}] , what I want is to then say "oh you're the same? lets sort you then by value (which will always be a number).

I am unsure how to do either in lodash or any other similar javascript way.

The end result should be either:

[{b: 2}, {a: 1}] // Keys are different (alphabetical)
// or:
[{a: 50}, {a: 100}] // Keys are the same (lowest to highest)

Everything I have seen on stack overflow becomes a quick mess (code wise), and I was thinking, there are sort methods on lodash, but how exactly do I achieve what I want given the circumstances ??

Any ideas?

Some one asked a good question, are there more keys? Are they only a and b ?

There would only be two objects in this array at any given time and yes the keys would only ever be strings, in this case the strings could be anything, cat, dog, mouse, apple, banana ... What ever.

The values will only ever be numbers.

Clearing the air

If the keys match, only sort the array by value, if the keys do not match, only sort the array by key. There will only ever be two objects in this array. Apologies for the misunderstanding.

Answer 1

In case you always have one property in your objects you can first sort by key using localeCompare and then by value of that property.

 var arr = [{b: 2}, {b: 10}, {a: 1}, {c: 1}, {a: 20}] arr.sort(function(a, b) { var kA = Object.keys(a)[0] var kB = Object.keys(b)[0] return kA.localeCompare(kB) || a[kA] - b[kB] }) console.log(arr) 

Before sorting you can create array of unique keys that you can use to check if all object have the same key by checking if length is > 1 and use that in sort function.

 var arr = [{b: 10}, {b: 2}, {a: 1}, {c: 1}, {a: 20}, {b: 22}] var arr2 = [{a: 10}, {a: 2}, {a: 1}, {a: 22}] function customSort(data) { var keys = [...new Set([].concat(...data.map(e => Object.keys(e))))] data.sort(function(a, b) { var kA = Object.keys(a)[0] var kB = Object.keys(b)[0] return keys.length > 1 ? kA.localeCompare(kB) : a[kA] - b[kB] }) return data; } console.log(customSort(arr)) console.log(customSort(arr2)) 

Answer 2

You can use only one function to perform the two types of sorting (works for your case, in which you have only an array with two items, but it is completely generic regarding the array length):

 var arr1 = [{a: 30}, {a: 2}]; var arr2 = [{b: 30}, {a: 2}]; function sortArr(arr) { return arr.sort((a, b) => { var aKey = Object.keys(a)[0]; var bKey = Object.keys(b)[0]; return (aKey !== bKey) ? aKey.localeCompare(bKey) : a[aKey] - b[bKey]; }); } var sortedArr1 = sortArr(arr1); var sortedArr2 = sortArr(arr2); console.log(sortedArr1); console.log(sortedArr2); 

Answer 3

 var arr = [{b: 2}, {a: 1}, {b: 1}, {a: 100}, {a: 20}]; arr.sort(function(a, b) { var aKey = a.hasOwnProperty("a")? "a": "b", // get the first object key (if it has the property "a" then its key is "a", otherwise it's "b") bKey = b.hasOwnProperty("a")? "a": "b"; // same for the second object return aKey === bKey? // if the keys are equal a[aKey] - b[aKey]: // then sort the two objects by the value of that key (stored in either aKey or bKey) aKey.localeCompare(bKey); // otherwise sort by the strings aKey and bKey (the keys of the two objects) }); console.log(arr);