如何通过一次迭代检查数组是否具有多个特定元素？

Question

How to check if array has more than one specific value by one iteration For example, we have two arrays:

   const arr = [{prop: 'test'}, {prop: "test1"}, {prop: "test3"}];
   const arr2 = [{prop: 'test3'}, {prop: "test"}, {prop: "test2"}];

I need to get true if array object elements have prop with value test and test1 when I put it in if statement for example.I need to get true only if both values exists, If we have not at least one of them I need to get false :

   const arr = [{prop: 'test'}, {prop: "test1"}, {prop: "test3"}]; //true
   const arr2 = [{prop: 'test3'}, {prop: "test"}, {prop: "test2"}]; //false

Answer 1

To do it in a single pass, build a structure that keeps track of the prop values that must be present (must not be missing). Run through the array and tick them off as they are found. This runs in O(n) where n is the length of the big array...

 // array is a long array of objects // prop is the prop to check in the long array // mustHaves is a short array of values that must be present in prop function check(array, prop, mustHaves) { let missing = {} for (const key of mustHaves) missing[key] = true array.forEach(e => { if (missing[e[prop]]) { missing[e[prop]] = false } }) return Object.values(missing).every(v => !v) } const arr = [{prop: 'test'}, {prop: "test1"}, {prop: "test3"}]; const arr2 = [{prop: 'test3'}, {prop: "test"}, {prop: "test2"}]; const propToCheck = 'prop' const mustHaves = ['test', 'test1'] console.log(check(arr, propToCheck, mustHaves)) console.log(check(arr2, propToCheck, mustHaves))

Answer 2

You can use Array.prototype.every() to check a pre-existing whitelist (ie test and test1 ) against your array. You can conver your array of objects objects into plain array that contains just the prop values for easy comparison, using Array.prototype.map() :

const mappedArray = arr.map(v => v.prop);

And then, you can simply check if all elements in your whitelist is found in this mapped array:

['test', 'test1'].every(x => mappedArray.includes(x));

See proof-of-concept below:

 const arr = [{prop: 'test'}, {prop: "test1"}, {prop: "test3"}]; const arr2 = [{prop: 'test3'}, {prop: "test"}, {prop: "test2"}]; function mustContainValues(arr, whitelist) { const mappedArr = arr.map(v => v.prop); return whitelist.every(x => mappedArr.includes(x)); } console.log(mustContainValues(arr, ['test', 'test1'])); // true console.log(mustContainValues(arr2, ['test', 'test1'])); // false

Answer 3

You can try to build an object using reduce:

   const func = (arr, value1, value2) =>
    {
      let obj = arr.reduce((acc, rec) => {
    if(rec.prop === value1) acc['hasValue1'] = true
    if(rec.prop === value2) acc['hasValue2'] = true
    return acc
    }, {hasValue1: false, hasValue2: false})
    return obj['hasValue1'] && obj['hasValue2']
    }

It will build an object like {hasValue1: false, hasValue2: false} and then return boolean AND for found flags. Reduce function will iterate through an array only once Invoke this function using:

console.log(func(arr, 'test', 'test1'))
console.log(func(arr2, 'test', 'test1'))