Javascript / Typescript - Remove from object

Question

How can I remove all objects which contain filter1?

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
}

what is your expected output? is it { 2: { filter2: { ... } } } — Mμ.
– Mμ., Commented Jul 20, 2017 at 3:58

dork · Accepted Answer · 2017-07-20 04:57:01Z

3

You can use reduce to get a new object:

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

console.log(newObject);

Or as @zerkms said, using entries:

const entries = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

console.log(newObject);

edited Jul 20, 2017 at 4:57

answered Jul 20, 2017 at 3:58

dork

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3 Comments

zerkms Over a year ago

If you're using non-standard syntax anyway, why not use Object.entries?

dork Over a year ago

@zerkms wouldn't it be longer though?

Object.entries(obj).reduce((newObj, entry) => {   if (entry[1].filter1) {     return newObj;   }      return {     ...newObj,     [entry[0]]: entry[1],   } }, {});

zerkms Over a year ago

But then you would have code without free variables. PS: (newObj, [k, v]) perhaps: Object.entries(obj).reduce((acc, [k, v]) => k.filter1 ? acc : { ...acc, [k]: v }, {});

Samuel Toh · Accepted Answer · 2017-07-20 04:17:47Z

3

Easy.

To remove an object you can use the delete API.
To check whether an object has a specific key you can use the hasOwnProperty API.

What you could do here is. First loop all of the keys in obj and then check whether its value (an object) has the filter1 property. If they do then delete it away.

Example:

const obj = {
    0: {
        'filter1': {
            'key1': {
                'email1': 'email1'
            }
        }
    },
    1: {
        'filter1': {
            'key12': {
                'email12': 'email12'
            }
        }
    },
    2: {
        'filter2': {
            'key2': {
                'email2': 'email2'
            }
        }
    }
};

for (var key in obj) {
    var objectInKey = obj[key];
    if (objectInKey.hasOwnProperty("filter1")) {
        delete obj[key]
    }
}

console.log(JSON.stringify(obj, null, 4));

console.log(obj[0]); // <-- Should be undefined.
console.log(obj[2]); // <-- Should not be undefined as obj wasn't deleted.

edited Jul 20, 2017 at 4:17

answered Jul 20, 2017 at 4:02

Samuel Toh

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6 Comments

brk Over a year ago

I have a doubt.What will happen if you don't stringify obj. In that case it will still show the old object and its keys

Samuel Toh Over a year ago

@brk I have updated the test code to try accessing the deleted key to prove that they are indeed removed. Let me know if it clears your doubt.

Paleo Over a year ago

If you use a loop for ... in then you have to check if (!obj.hasOwnProperty(key)) continue.

Samuel Toh Over a year ago

@Paleo maybe I am missing something. My output is consistent with what OP was asking. I'm not sure why !obj.hasOwnProperty(key) continue is required.

Paleo Over a year ago

@SamuelToh If you don't, iterations can occur in the loop for inherited members. In your code it is without consequence because delete obj[key] will do nothing with these wrong keys. But a future modification in the body of the loop could easily introduce a bug. It is a good practice to always protect for ... in with hasOwnProperty. Here is an example of code that adds an inherited key for all objects: Object.prototype['abc'] = {'filter1': 123}.

|

david25272 · Accepted Answer · 2017-07-20 04:30:20Z

0

To modify the object in place (rather than copying it, or creating arrays), I'd just use a simple loop and a delete:

let k = Object.keys(obj);
for (let i = 0; i < k.length ; i++ ) {
    if (obj[k[i]].hasOwnProperty('filter1')) delete obj[k[i]];
}

answered Jul 20, 2017 at 4:30

david25272

9546 silver badges12 bronze badges

Comments

score 0 · Accepted Answer · 2017-07-20 04:56:29Z

0

You could use Object.entries to get an array of key-value pairs, then use Array#reduce to filter the data and translate it back into an object format.

const obj = {
  0: { 'filter1': { 'key1':  { 'email1':  'email1'  } } },
  1: { 'filter1': { 'key12': { 'email12': 'email12' } } },
  2: { 'filter2': { 'key2':  { 'email2':  'email2'  } } }
};

console.log(Object.entries(obj).reduce((m, [k, v]) => (v.filter1 || (m[k] = v), m), {}));

// { "2": { "filter2": { "key2": { "email2": "email2" } } } }

Further reading:

edited Jul 20, 2017 at 4:56

answered Jul 20, 2017 at 4:24

user4639281

Comments

samAlvin · Accepted Answer · 2017-07-20 04:00:07Z

-1

You can use filter,

obj = obj.filter(o => o.filter1 == undefined);

See filter reference here.

p.s. the obj in this example is a var not a const, see snippet below

var obj = [
  {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
];

console.log(obj); // initial obj
obj = obj.filter(o => o.filter1 == undefined);
console.log(obj); // element with `filter1` property is now deleted

edited Jul 20, 2017 at 4:00

answered Jul 20, 2017 at 3:53

samAlvin

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2 Comments

zerkms Over a year ago

OP does not have an array. It's also not obvious what you mean by "the obj in this example is a var not a const"

david25272 Over a year ago

Setting the properties value to undefined doesn't actually remove the property. For instance Object.hasOwnProperty() will still return true. To completely remove it, use delete.

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