Initialize an array of object with unknown keys in a better way using es6

Question

I have an object as follows:

{
    "counts": {
        "created_by_ref": {
            "00f5c303-32c0-4f5f-ac61-bf6577a8f4ed": 1,
            "0dde2de1-7951-45b1-8bf3-013f8dbfc4ae": 1
        },
        "file.hashes.MD5": {
            "qqq": 200
        },
        "ipv4_addr.value": {
            "127.0.0.1": 200,
            "192.168.1.10": 200
        },
        "network_traffic.dst_port": {
            "xxx": 200
        },
        "network_traffic.dst_ref.resolves_to_refs.value": {
            "yyy": 200
        },
        "network_traffic.dst_ref.value": {
            "rrr": 200
        }
    }
}

I need to initialize all the values to 0 as follows:

{
    "counts": {
        "created_by_ref": {
            "00f5c303-32c0-4f5f-ac61-bf6577a8f4ed": 0,
            "0dde2de1-7951-45b1-8bf3-013f8dbfc4ae": 0
        },
        "file.hashes.MD5": {
            "qqq": 0
        },
        "ipv4_addr.value": {
            "127.0.0.1": 0,
            "192.168.1.10": 0
        },
        "network_traffic.dst_port": {
            "xxx": 0
        },
        "network_traffic.dst_ref.resolves_to_refs.value": {
            "yyy": 0
        },
        "network_traffic.dst_ref.value": {
            "rrr": 0
        }
    }
}

and all the keys except count is unknown and dynamically changing so cannot be hard code. What I did is:

  for (const key in getSearchDataAllCounts) {
    if (getSearchDataAllCounts.hasOwnProperty(key)) {
      Object.keys(getSearchDataAllCounts[key]).forEach((innerkey) => {
        getSearchDataAllCounts[key][innerkey] = 0;
      });
    }
  }

But I really think there is a better way doing it in ES6. Is there any improvement I can make to make it more es6 compatible and also do it with one loop preferebly?

I'm curious why for-in for one loop but Object.keys().forEach for the other...? — T.J. Crowder
– T.J. Crowder, Commented Mar 26, 2020 at 17:05
Right, which avoids the need for the hasOwnProperty check. It's just, it looks like an intentional choice, using for-in for the outer loop and Object.keys().forEach for the inner one. If it's not intentional, no worries, I just thought it might be. :-) — T.J. Crowder
– T.J. Crowder, Commented Mar 26, 2020 at 17:07
You can do it with what looks like one loop, but is really nested loops like the above. I'll add that to the answer. :-) — T.J. Crowder
– T.J. Crowder, Commented Mar 26, 2020 at 17:08
@T.J.Crowder Thanks a million will wait for your update in the answer. I am learning a lot :D — Learner
– Learner, Commented Mar 26, 2020 at 17:11

T.J. Crowder · Accepted Answer · 2020-03-26 18:04:08Z

2

ES2015+ doesn't really give you much to improve that code, other than for-of and Object.values:

for (const entry of Object.values(getSearchDataAllCounts.counts)) {
  for (const key of Object.keys(entry)) {
    entry[key] = 0;
  }
}

In a comment you asked:

So we cannot do it with one loop right?

You can do it with what looks like one loop, but is really nested loops like the above (more loops, in fact):

for (const [key, entry] of Object.entries(getSearchDataAllCounts.counts)) {
    getSearchDataAllCounts.counts[key] = Object.fromEntries(
      Object.keys(entry).map(key => [key, 0])
    );
}

That uses Object.entries to get an array of [key, value] pairs for the outer object, then replaces that object with a new object we create by

Getting its keys
Mapping them to [key, 0] entries
Using Object.fromEntries to create an object from those entries

There's an important difference there, though: It creates new objects to store on the properties of the outermost object. The first one just updates the properties of the existing objects.

edited Mar 26, 2020 at 18:04

answered Mar 26, 2020 at 17:06

T.J. Crowder

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5 Comments

Learner Over a year ago

Also can explain if there is a specific reason your outer loop is over value rather than key? what if we have two unique key with the same value?

T.J. Crowder Over a year ago

@Learner - That doesn't matter, we'll see the value twice. But if you might really have two properties that refer to the same object, the second solution will replace them with two distinct objects.

T.J. Crowder Over a year ago

@Learner - Apologies, I'm getting called away. If you have further follow-up questions, I'll answer them when I'm back.

Learner Over a year ago

Thanks should I not change the first loop to: for (const entry of Object.values(getSearchDataAllCounts.counts)) {. ???

T.J. Crowder Over a year ago

@Learner - Doh! Yes. :-) Edited.

xdeepakv · Accepted Answer · 2020-03-26 17:21:23Z

You dont need ES6. Using JSON replacer, u can do it on the tree traversing.

Util

const zero = obj =>
   JSON.parse(JSON.stringify(obj, (_, v) => (typeof v === "object" ? v : 0)));

console.log(zero(d));

How to use

console.log(zero(d));

var d = {
  counts: {
    created_by_ref: {
      "00f5c303-32c0-4f5f-ac61-bf6577a8f4ed": 1,
      "0dde2de1-7951-45b1-8bf3-013f8dbfc4ae": 1
    },
    "file.hashes.MD5": {
      qqq: 200
    },
    "ipv4_addr.value": {
      "127.0.0.1": 200,
      "192.168.1.10": 200
    },
    "network_traffic.dst_port": {
      xxx: 200
    },
    "network_traffic.dst_ref.resolves_to_refs.value": {
      yyy: 200
    },
    "network_traffic.dst_ref.value": {
      rrr: 200
    }
  }
};
const zero = obj => JSON.parse(JSON.stringify(obj, (_, v) => (typeof v === "object" ? v : 0)));
console.log(zero(d));

.as-console-row {color: blue!important}

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