How to differentiate arrays of object in javascript?

Question

How can I differentiate if an array is a simple array?

f.e

let simpleArr = ["hi","1",3,"5","this is a string",29999]

or an array of objects

f.e

let objectArr = [
 {el1: "hi", el2: "hi"},
 {el1: "hi2", el2: "hi2"},
 {el1: "hi3", el2: "hi3"},
 {el1: "hi4", el2: "hi4"},
]

Is the below enough?

const isObject = (obj) => typeof obj === 'object' && obj !== null

isArray(objectArr) && objectArr.every(isObject)

If you want to check if all members are objects - yes. But this will return false when the array contains objects and other data, e.g. [{}, true, {}]. Is that expected? — VLAZ
– VLAZ, Commented May 14, 2021 at 10:23
const isSimpleArray = arr => !arr.some(x => typeof x === 'object') — boxdox
– boxdox, Commented May 14, 2021 at 10:24
@georg yeah, that as well. If you want any objects, then you need to include functions as they are also objects. — VLAZ
– VLAZ, Commented May 14, 2021 at 10:24

Nguyễn Văn Phong · Accepted Answer · 2021-05-14 10:47:26Z

1

You can do this.

const isObject = item => typeof item === "object" 
                          && !Array.isArray(item) 
                          && item !== null;

const simpleArr = ["hi","1",3,"5","this is a string",29999];
const array_of_objects = [{el1:"hi",el2:"hi"},{el1:"hi2",el2:"hi2"},{el1:"hi3",el2:"hi3"},{el1:"hi4",el2:"hi4"}];

console.log(simpleArr.every(isObject)); // false
console.log(array_of_objects.every(isObject)); // true

answered May 14, 2021 at 10:47

Nguyễn Văn Phong

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Comments

Krzysztof Krzeszewski · Accepted Answer · 2021-05-14 10:27:35Z

0

Maybe it's not a straight forward approach, but you can try doing a deep copy of the Array. You see in javascript comparison between non primitive values like objects will return false even if they both have the exact same properties.

{foo: "bar"} === {foo: "bar"} // false

const simpleArr = ["hi","1",3,"5","this is a string",29999]
const objectArr = [
 {el1: "hi", el2: "hi"},
 {el1: "hi2", el2: "hi2"},
 {el1: "hi3", el2: "hi3"},
 {el1: "hi4", el2: "hi4"},
]

function deepCopy(array) {
  return JSON.parse(JSON.stringify(array));
}

function isSimple(array) {
  const copy = deepCopy(array);
  return array.every((el, index) => el === copy[index]); 
}

console.log("simpleArr:", isSimple(simpleArr))
console.log("objectArr:", isSimple(objectArr))

answered May 14, 2021 at 10:27

Krzysztof Krzeszewski

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1 Comment

VLAZ Over a year ago

Don't do that. Too much work and it's prone to failure as you can lose data which can skew your results. jsbin.com/zatucuk/edit?js,console

Behemoth · Accepted Answer · 2021-05-14 11:02:16Z

0

Yes, you can do it like so. Just make sure to do a check for Function as well.

let simpleArr = [
  "hi",
  "1",
  3,
  "5",
  "this is a string",
  29999,
  Function,
];

let objectArr = [{
    el1: "hi",
    el2: "hi",
  },
  {
    el1: "hi2",
    el2: "hi2",
  },
  {
    el1: "hi3",
    el2: "hi3",
  },
  {
    el1: "hi4",
    el2: "hi4",
  },
];

const isSimpleArray = arr =>
  !arr.some(
    x => x != null && (typeof x === "object" || typeof x === "function")
  );

console.log(isSimpleArray(simpleArr));
console.log(isSimpleArray(objectArr));

answered May 14, 2021 at 11:02

Behemoth

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