How to compare the elements from one array in another array

Question

If any of the choice object from choices array is not found in any of the results array hide choice element in this case {"choice": "f"} and {"choice": "g"} is not available in any of those 3 results so I need to hide it. How can I do it?

{"choices":[
    {"choice": "a"},
    {"choice": "b"},
    {"choice": "c"},
    {"choice": "d"},
    {"choice": "e"},
    {"choice": "f"},
    {"choice": "g"}
],
"results":[
    {"result":["a", "b", "c"]},             
    {"result":["a", "b", "d"]},
    {"result":["a", "b", "e"]},

the question is unclear, the posted code is just a cut chunk of some code with no beginning and ending(voted to close) — RomanPerekhrest
– RomanPerekhrest, Commented Mar 7, 2017 at 15:17

Nina Scholz · Accepted Answer · 2017-03-07 15:58:04Z

2

You could collect all values from result and filter choices.

var object = { choices: [{ choice: "a" }, { choice: "b" }, { choice: "c" }, { choice: "d" }, { choice: "e" }, { choice: "f" }, { choice: "g" }], results: [{ result: ["a", "b", "c"] }, { result: ["a", "b", "d"] }, { result: ["a", "b", "e"] }] },
    hash = Object.create(null),
    notIn = [];

object.results.forEach(function (a) {
    a.result.forEach(function (b) {
        hash[b] = true;
    });
});

notIn = object.choices.filter(function (a) {
    return !hash[a.choice];
});

console.log(notIn);

.as-console-wrapper { max-height: 100% !important; top: 0; }

ES6 with Set

var object = { choices: [{ choice: "a" }, { choice: "b" }, { choice: "c" }, { choice: "d" }, { choice: "e" }, { choice: "f" }, { choice: "g" }], results: [{ result: ["a", "b", "c"] }, { result: ["a", "b", "d"] }, { result: ["a", "b", "e"] }] },
    results = object.results.reduce((s, a) => new Set([...s, ...a.result]), new Set),
    notIn = object.choices.filter(a => !results.has(a.choice));

console.log(notIn);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Mar 7, 2017 at 15:58

answered Mar 7, 2017 at 15:18

Nina Scholz

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3 Comments

Anis Over a year ago

+1 for the Set solution, maybe opting for let instead of var would be better for a full ES6 solution?

Nina Scholz Over a year ago

@Anis, actually the scope is not the problem.

Anis Over a year ago

I completely agree, and since it doesn't matter I just thought that let would be a better default in this case.

Anis · Accepted Answer · 2017-03-07 15:15:54Z

0

I'm not sure whether you are asking for generic implementation ideas, or pointers as to how to implement it in javascript. I'll go for the general idea. You can simply use a set to store all the possible results, then iterate through the choices and check for each choice whether it is inside the set. If not, you can take the action you want like hiding it, whatever that means.

answered Mar 7, 2017 at 15:15

Anis

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Comments

Nenad Vracar · Accepted Answer · 2017-03-07 15:17:16Z

0

You can use filter() on data.choices and then some() and includes() to check if choice exists in some of the objects in results array.

var data = {"choices":[{"choice":"a"},{"choice":"b"},{"choice":"c"},{"choice":"d"},{"choice":"e"},{"choice":"f"},{"choice":"g"}],"results":[{"result":["a","b","c"]},{"result":["a","b","d"]},{"result":["a","b","e"]}]}

data.choices = data.choices.filter(function(o, i) {
  var check = data.results.some(e => e.result.includes(o.choice))
  return check;
})

console.log(data)

answered Mar 7, 2017 at 15:17

Nenad Vracar

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Comments

kind user · Accepted Answer · 2017-03-07 15:30:16Z

0

var choices = {choices:[{choice:"a"},{choice:"b"},{choice:"c"},{choice:"d"},{choice:"e"},{choice:"f"},{choice:"g"}],results:[{result:["a","b","c"]},{result:["a","b","d"]},{result:["a","b","e"]}]},

    elems1 = choices.results.map(v => v.result).reduce((a,b) => a.concat(b)),
    filtered = [...new Set(elems1)];
    choices.choices = choices.choices.filter(v => filtered.some(c => c == v.choice) );
    console.log(choices);

edited Mar 7, 2017 at 15:30

answered Mar 7, 2017 at 15:21

kind user

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5 Comments

Heretic Monkey Over a year ago

Please add some explanation of why/how this code helps the OP. This will help provide an answer future viewers can learn from. See this Meta question and its answers for more information. This seems to be becoming a theme here, @Kinduser...

kind user Over a year ago

@MikeMcCaughan Actually I don't want to write anything because this is over complicated... I will wait for Nina response and I will decide to leave or to delete my answer. If I will decide to leave it, I will describe each step.

Nina Scholz Over a year ago

@Kinduser, you could reduce with Set, instead of mapping the values.

kind user Over a year ago

@NinaScholz Can you show me an example? Just a short one.

Nina Scholz Over a year ago

@Kinduser, please see my es6 example.

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