Is there an Array equality match function that ignores element position in jest.js?

Question

I get that .toEqual() checks equality of all fields for plain objects:

expect(
    {"key1":"pink  wool","key2":"diorite"}
).toEqual(
    {"key2":"diorite","key1":"pink wool"}
);

So this passes.

But the same is not true for arrays:

expect(["pink wool", "diorite"]).toEqual(["diorite", "pink wool"]);

There does not seem to be a matcher function that does this in the jest docs, i.e. that tests for the equality of two arrays irrespective of their elements positions. Do I have to test each element in one array against all the elements in the other and vice versa? Or is there another way?

Answer 1

There is no built-in method to compare arrays without comparing the order, but you can simply sort the arrays using .sort() before making a comparison:

expect(["ping wool", "diorite"].sort()).toEqual(["diorite", "pink wool"].sort());

You can check the example in this fiddle.

Answer 2

Put the elements into a set. Jest knows how to match these.

expect(new Set(["pink wool", "diorite"])).toEqual(new Set(["diorite", "pink wool"]));

Answer 3

As already mentioned expect.arrayContaining checks if the actual array contains the expected array as a subset. To check for equivalence one may

• either assert that the length of both arrays is the same (but that wouldn't result in a helpful failure message)
• or assert the reverse: That the expected array contains the actual array:

// This is TypeScript, but remove the types and you get JavaScript
const expectArrayEquivalence = <T>(actual: T[], expected: T[]) => {
  expect(actual).toEqual(expect.arrayContaining(expected));
  expect(expected).toEqual(expect.arrayContaining(actual));
};

This still has the problem that when the test fails in the first assertion one is only made aware of the elements missing from actual and not of the extra ones that are not in expected.

Answer 4

this does not answer the question exactly, but still may help people that end up here by google search:

if you only care that a subset of the array has certain elements, use expect.arrayContaining() https://jestjs.io/docs/en/expect#expectarraycontainingarray

e.g.,

expect(["ping wool", "diorite"])
  .toEqual(expect.arrayContaining(["diorite", "pink wool"]));

Answer 5

Another way is to use the custom matcher .toIncludeSameMembers() from jest-community/jest-extended.

Example given from the README

test('passes when arrays match in a different order', () => {
    expect([1, 2, 3]).toIncludeSameMembers([3, 1, 2]);
    expect([{ foo: 'bar' }, { baz: 'qux' }]).toIncludeSameMembers([{ baz: 'qux' }, { foo: 'bar' }]);
});

It might not make sense to import a library just for one matcher but they have a lot of other useful matchers I've find useful.

Answer 6

What about checking the content and the length?

  expect(resultArray).toEqual(expect.arrayContaining(expectedArray));
  expect(resultArray.length).toEqual(expectedArray.length);

Answer 7

If you want to compare two arrays in JEST use the bellow model.

Official link: https://jestjs.io/docs/en/expect#expectarraycontainingarray

const array1 = ['a', 'b', 'c'];
const array2 = ['a', 'b', 'c'];
const array3 = ['a', 'b'];


it("test two arrays, this will be true", () => { 
    expect(array1).toEqual(expect.arrayContaining(array2));
});

it("test two arrays, this will be false", () => { 
    expect(array3).toEqual(expect.arrayContaining(array1));
});

Answer 8

You can combine using sets as stated in this answer with checking length of actual result and expectation. This will ignore element position and protect you from duplicated elements in the same time.

const materials = ['pink wool', 'diorite'];
const expectedMaterials = ['diorite', 'pink wool'];

expect(new Set(materials)).toEqual(new Set(expectedMaterials));
expect(materials.length).toBe(expectedMaterials.length);

EDIT: As there is suggested in comment below, this will only work for arrays with unique values.

Answer 9

If you don't have array of objects, then you can simply use sort() function for sorting before comparison.(mentioned in accepted answer):

expect(["ping wool", "diorite"].sort()).toEqual(["diorite", "pink wool"].sort());

However, problem arises if you have array of objects in which case sort function won't work. In this case, you need to provide custom sorting function. Example:

const x = [
{key: 'forecast', visible: true},
{key: 'pForecast', visible: false},
{key: 'effForecast', visible: true},
{key: 'effRegForecast', visible: true}
]

// In my use case, i wanted to sort by key
const sortByKey = (a, b) => { 
  if(a.key < b.key) return -1; 
  else if(a.key > b.key) return 1; 
  else return 0; 
  }

x.sort(sortByKey)
console.log(x)

Hope it helps someone someday.

Answer 10

Still a work in progress, but this should work albeit, the error messages may not be clear:

expect.extend({
  arrayContainingExactly(receivedOriginal, expected) {
    const received = [...receivedOriginal];

    if (received.length !== expected.length) return {
      message: () => `Expected array of length ${expected.length} but got an array of length ${received.length}`,
      pass: false,
    };

    const pass = expected.every((expectedItem, index) => {
      const receivedIndex = findIndex(received, receivedItem => {
          if (expectedItem.asymmetricMatch) return expectedItem.asymmetricMatch(receivedItem);
          return isEqual(expectedItem, receivedItem);
      });
      if (receivedIndex === -1) return false;
      received.splice(receivedIndex, 1);
      return true;
    });

    return {
      message: () => 'Success',
      pass,
    }
  }
});

Then use it like this:

expect(['foo', 'bar']).arrayContainingExactly(['foo']) // This should fail

or

expect({foo: ['foo', 'bar']}).toEqual({
  foo: expect.arrayContainingExactly(['bar', 'foo'])
}) // This should pass

We are looping through each value and removing it from the received array so that we can take advantage of the asymmetric matching provided by Jest. If we just wanted to do direct equivalency this could be simplified to just compare the 2 sorted arrays.

Note: This solution uses findIndex and isEqual from lodash.