Javascript Arrays - Checking two arrays of objects for same contents, ignoring order

為{幸葍}努か 提交于 2019-12-17 19:19:50

问题


I have two JavaScript arrays (A and B) that contain objects that I created. I want to check that all the objects in array A are contained in array B, but not necessarily in the same order.

What is the best way to do this?

Edit:

They are all actual objects, not primitives, so I will need to compare their contents and structure as well (maybe using something like JSON.stringify).

I want to do this because I'm learning Test-Driven Development, and I want to test functions that return lists of objects. I need to test whether the returned lists have the expected objects in them or not (order doesn't matter in this case).


回答1:


Usage: isEqArrays(arr1, arr2)

//
// Array comparsion
//

function inArray(array, el) {
  for ( var i = array.length; i--; ) {
    if ( array[i] === el ) return true;
  }
  return false;
}

function isEqArrays(arr1, arr2) {
  if ( arr1.length !== arr2.length ) {
    return false;
  }
  for ( var i = arr1.length; i--; ) {
    if ( !inArray( arr2, arr1[i] ) ) {
      return false;
    }
  }
  return true;
}



回答2:


If the duplicates do not matter, you can make an one-liner by using lodash. Look if the difference between two arrays is empty.

if (_(array).difference(otherArray).isEmpty()) {
    // is equal ignoring order and duplicates
}

console.log(_([1,2,3]).difference([2,3,1]).isEmpty()) // -> true
console.log(_([1,2,3,3]).difference([2,3,1,1]).isEmpty()) // -> also true



回答3:


This is probably the simplest method if not the slowest.

var o = { PropA: 1, PropB: 2 };
var a = [1, 2, 3, 4, o];
var b = [2, 3, 4, 1];

var c = a.filter(function(value, index, obj) {
    return b.indexOf(value) > -1;
});

if (c.length !== a.length) {
    throw new Error("Array b is missing some elements!");
}

indexOf will only check that they refer to the same object. If you want to check value equivalence you will have to do a deep compare of the properties or use JSON.stringify as you mention in your question.




回答4:


With ES6 you could use every and some (and length).

let A = [1, 2, 3];
let B = [2, 3, 1];

// all objects in A are contained in B (A ⊆ B)
// you can compare a <-> b however you'd like (here just `a === b`)
let AsubB = A.every(a => B.some(b => a === b));

// A and B are the same length
let sameLength = A.length === B.length;

// their contents are as equal as previously tested:
let equal = AsubB && sameLength;


来源:https://stackoverflow.com/questions/3243275/javascript-arrays-checking-two-arrays-of-objects-for-same-contents-ignoring-o

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