Object containing arrays of objects and arrays of strings

Question

I have an object that looks like this:

var obj = {
  "array1": [
    {"label": "something1", "ref": "option2a"},
    {"label": "something2", "ref": "option2b"},
    {"label": "something3", "ref": "option2a"},
    {"label": "something4", "ref": "option2a"},
    {"label": "something5 is the longest", "ref": "option2a"}
  ],
  array2: [
    "arrayItem1",
    "array Item 2"
  ]
}

This object contains arrays of objects and arrays of strings. I would like to traverse through each object that has a label and return the longest label. In the previous example, the expected output id:

"something5 is the longest".

I have tried the following:

function getLongest(object, key) {
  return Object.values(object).reduce((l, v) => {
    if (object.hasOwnProperty(key)) {
      if (key in v)
        return Math.max(l, v[key].length);
      if (v && typeof v === 'object')
        return Math.max(l, getLongest(v, key));
      return l;
    }
  }, 0);
}

However this gives me an error because it cannot find the property label for the items in array2.

@Mandalina, are 'array1' and 'array2' to be compared against each other? Separately? Or only arrays with "label" keys (such as 'array1')? — Sunny Patel
– Sunny Patel, Commented Jan 11, 2019 at 19:01
Hey @SunnyPatel yes I only need to get arrays with "label" keys; like array1. — Mandalina
– Mandalina, Commented Jan 11, 2019 at 19:07

Shidersz · Accepted Answer · 2019-01-11 20:18:49Z

1

Here you have one approach that used anidated reduce() methods over the Object.Values():

var obj = {
    "array1": [
        {"label": "something1", "ref": "option2a"},
        {"label": "something2", "ref": "option2b"},
        {"label": "something3", "ref": "option2ahgf"},
        {"label": "something4", "ref": "option2a"},
        {"label": "I'm not the longest", "ref": "option2a"}
    ],
    array2: [
        "arrayItem1",
        {"label": "something5 is the longest", "ref": "option2a"},
        "array Item 2"
    ],
    array3: [
        {"label": "somethingX", "ref": "option2X"},
        {"label": "somethingY", "ref": "option2Y"}
    ]
}

const getLongest = (obj, key) =>
{
    let r = Object.values(obj).reduce((res, curr) =>
    {
        if (!Array.isArray(curr))
            return res;

        let newMax = curr.reduce(
            (r, c) => c[[key]] && (c[[key]].length > r.length) ? c[[key]] : r,
            ""
        );

        res = newMax.length > res.length ? newMax : res;
        return res;

    }, "");

    return r;
}

console.log(getLongest(obj, "label"));
console.log(getLongest(obj, "ref"));

edited Jan 11, 2019 at 20:18

answered Jan 11, 2019 at 19:41

Shidersz

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1 Comment

Shidersz Over a year ago

I just updated it to wrap the logic inside a function and generalize it a little more.

abc123 · Accepted Answer · 2019-01-11 18:53:45Z

0

Example:

// object declaration
var obj = {
  "array1": [{
      "label": "something1",
      "ref": "option2a"
    },
    {
      "label": "something2",
      "ref": "option2b"
    },
    {
      "label": "something3",
      "ref": "option2a"
    },
    {
      "label": "something4",
      "ref": "option2a"
    },
    {
      "label": "something5 is the longest",
      "ref": "option2a"
    }
  ],
  array2: [
    "arrayItem1",
    "array Item 2"
  ]
};

// iterate over all items in obj
for (var propertyName in obj) {
  // if the item is an object and is an array
  if (typeof obj[propertyName] === 'object' && Array.isArray(obj[propertyName])) {
    // if the array has at least one item and a property of label
    if (obj[propertyName].length > 0 && typeof obj[propertyName][0].label !== 'undefined') {
      // sort the array by length of label
      obj[propertyName].sort((a,b) => (a.label.length < b.label.length) ? 1 : ((b.label.length < a.label.length) ? -1 : 0));
      // console.log print the longest label
      console.log('longest label name:', obj[propertyName][0].label);
    }
  }
}

// output the whole obj
console.log(obj);

edited Jan 11, 2019 at 18:53

answered Jan 11, 2019 at 18:50

abc123

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3 Comments

Sunny Patel Over a year ago

I don't think OP literally wants to sort them. Just get back the label with longest value.

Code Maniac Over a year ago

` I would like to sort through each object that has a label and return the label that is the longest. In this case it would return "something5 is the longest" ` @SunnyPatel

Sunny Patel Over a year ago

"sort through each" can be considered as "iterate through each". As in "sorting through papers", you would go "through" them.

Sunny Patel · Accepted Answer · 2019-01-11 19:27:22Z

Since you mentioned jQuery, I've included a solution that uses jQuery.map to iterate through your outer object and array.reduce to yield only one result from each array value inside obj.

var obj = {
  "array1": [{
      "label": "something1",
      "ref": "option2a"
    },
    {
      "label": "something2",
      "ref": "option2b"
    },
    {
      "label": "something3",
      "ref": "option2a"
    },
    {
      "label": "something4",
      "ref": "option2a"
    },
    {
      "label": "something5 is the longest",
      "ref": "option2a"
    }
  ],
  array2: [
    "arrayItem1",
    "array Item 2"
  ]
}
function getLongest(curr, next) {
  return !Object.keys(next).includes("label") || curr.label.length > next.label.length ? curr : next;
}

var reduced = $.map(obj, item => item.reduce(getLongest, {"label": ""}))
console.log(reduced);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

This basically iterates through each key's value inside obj and reduces the array down to a single object that has a label with the longest label, if it even has a label key. If it does not, it returns a default object without a label and works with arrays with mixed objects/strings.

So it returns an object with the matching keys of obj.

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