期望数组无效顺序

Question

使用Jasmine有一种方法可以测试2个数组是否包含相同的元素，但不一定是相同的顺序吗？即

array1 = [1,2,3];
array2 = [3,2,1];

expect(array1).toEqualIgnoreOrder(array2);//should be true

Answer 1

如果它只是整数或其他原始值，你可以在比较之前sort()。

expect(array1.sort()).toEqual(array2.sort());

如果是其对象，则将其与map()函数组合以提取将要比较的标识符

array1 = [{id:1}, {id:2}, {id:3}];
array2 = [{id:3}, {id:2}, {id:1}];

expect(array1.map(a => a.id).sort()).toEqual(array2.map(a => a.id).sort());

Answer 2

茉莉花版2.8及以后版本

jasmine.arrayWithExactContents()

这要求数组按任何顺序包含列出的元素。

array1 = [1,2,3];
array2 = [3,2,1];
expect(array1).toEqual(jasmine.arrayWithExactContents(array2))

见https://jasmine.github.io/api/3.4/jasmine.html

Answer 3

jest-extended包提供了很少的断言来简化我们的测试，它不那么冗长，而且对于测试失败，错误更明确。

对于这种情况，我们可以使用toIncludeSameMembers

expect([{foo: "bar"}, {baz: "qux"}]).toIncludeSameMembers([{baz: "qux"}, {foo: "bar"}]);

Answer 4

对于它的价值，在Jest中你可以做到：

expect(new Set([1,2,3])).toEqual(new Set([3,2,1]))

Answer 5

// check if every element of array2 is element of array1
// to ensure [1, 1] !== [1, 2]
array2.forEach(x => expect(array1).toContain(x))

// check if every element of array1 is element of array2
// to ensure [1, 2] !== [1, 1]
array1.forEach(x => expect(array2).toContain(x))

// check if they have equal length to ensure [1] !== [1, 1]
expect(array1.length).toBe(array2.length)

Answer 6

简单...

array1 = [1,2,3];
array2 = [3,2,1];

expect(array1).toEqual(jasmine.arrayContaining(array2));

Answer 7

//Compare arrays without order
//Example
//a1 = [1, 2, 3, 4, 5]
//a2 = [3, 2, 1, 5, 4]
//isEqual(a1, a2) -> true
//a1 = [1, 2, 3, 4, 5];
//a2 = [3, 2, 1, 5, 4, 6];
//isEqual(a1, a2) -> false


function isInArray(a, e) {
  for ( var i = a.length; i--; ) {
    if ( a[i] === e ) return true;
  }
  return false;
}

function isEqArrays(a1, a2) {
  if ( a1.length !== a2.length ) {
    return false;
  }
  for ( var i = a1.length; i--; ) {
    if ( !isInArray( a2, a1[i] ) ) {
      return false;
    }
  }
  return true;
}

Answer 8

function equal(arr1, arr2){
    return arr1.length === arr2.length
    &&
    arr1.every((item)=>{
        return arr2.indexOf(item) >-1
    }) 
    &&
    arr2.every((item)=>{
        return arr1.indexOf(item) >-1
    })
}

这里的想法是首先确定两个数组的长度是否相同，然后检查所有元素是否都在另一个数组中。

Answer 9

这是一个适用于任何数字或数组的解决方案

https://gist.github.com/tvler/cc5b2a3f01543e1658b25ca567c078e4

const areUnsortedArraysEqual = (...arrs) =>
  arrs.every((arr, i, [first]) => !i || arr.length === first.length) &&
  arrs
    .map(arr =>
      arr.reduce(
        (map, item) => map.set(item, (map.get(item) || 0) + 1),
        new Map(),
      ),
    )
    .every(
      (map, i, [first]) =>
        !i ||
        [...first, ...map].every(([item]) => first.get(item) === map.get(item)),
    );

一些测试（这个问题的一些答案不考虑具有相同值的多个项的数组，因此[1,2,2]和[1,2]将错误地返回true）

[1, 2] true
[1, 2], [1, 2] true
[1, 2], [1, 2], [1, 2] true
[1, 2], [2, 1] true
[1, 1, 2], [1, 2, 1] true
[1, 2], [1, 2, 3] false
[1, 2, 3, 4], [1, 2, 3], [1, 2] false
[1, 2, 2], [1, 2] false
[1, 1, 2], [1, 2, 2] false
[1, 2, 3], [1, 2], [1, 2, 3] false

Answer 10

此算法非常适用于每个项目都是唯一的数组。如果没有，你可以添加一些东西来检查重复...

tests = [
  [ [1,0,1] , [0,1,1] ],
  [ [1,0,1] , [0,0,1] ], //breaks on this one...
  [ [2,3,3] , [2,2,3] ], //breaks on this one also...
  [ [1,2,3] , [2,1,3] ],
  [ [2,3,1] , [1,2,2] ],
  [ [2,2,1] , [1,3,2] ]
]

tests.forEach(function(test) {
  console.log('eqArraySets( '+test[0]+' , '+test[1]+' ) = '+eqArraySets( test[0] , test[1] ));
});


function eqArraySets(a, b) {
	if ( a.length !== b.length ) { return false; }
	for ( var i = a.length; i--; ) {
		if ( !(b.indexOf(a[i])>-1) ) { return false; }
		if ( !(a.indexOf(b[i])>-1) ) { return false; }
	}
	return true;
}

Answer 11

这种方法具有更差的理论最坏情况运行时性能，但是，因为它不对阵列执行任何写操作，所以在许多情况下它可能更快（尚未测试性能）：

警告：正如Torben在评论中指出的那样，这种方法只有在两个数组都具有唯一（非重复）元素时才有效（就像这里的其他几个答案一样）。

/**
 * Determine whether two arrays contain exactly the same elements, independent of order.
 * @see https://stackoverflow.com/questions/32103252/expect-arrays-to-be-equal-ignoring-order/48973444#48973444
 */
function cmpIgnoreOrder(a, b) {
  const { every, includes } = _;
  return a.length === b.length && every(a, v => includes(b, v));
}

// the following should be all true!
const results = [
  !!cmpIgnoreOrder([1,2,3], [3,1,2]),
  !!cmpIgnoreOrder([4,1,2,3], [3,4,1,2]),
  !!cmpIgnoreOrder([], []),
  !cmpIgnoreOrder([1,2,3], [3,4,1,2]),
  !cmpIgnoreOrder([1], []),
  !cmpIgnoreOrder([1, 3, 4], [3,4,5])
];

console.log('Results: ', results)
console.assert(_.reduce(results, (a, b) => a && b, true), 'Test did not pass!');

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>

Answer 12

目前有一个针对此用例的匹配器：

https://github.com/jest-community/jest-extended/pull/122/files

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