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javascriptjsoncomparelodash
Murshid Hassen
Murshid Hassen

Reputation: 149

JSON comparison neglecting data/values & considering only properties/keys

Comparing two json objects neglecting the data (values) and considering only the keys/properties (structure). I used isEqual from lodash. which is good. It will neglect the keys order. What I would need is a way to neglect the values and comparing only the structure of the two jsons.

That is

  1. Removing the data
  2. Comparing the structures

This comparison should ideally return true.

var _ = require('lodash');

const object1 = {
  name: 'Tom',
  address: 'USA'
};
    
const object2 = {
  name: 'Kumar',
  address: 'India',
};

const comparison = _.isEqual(object1 ,object2)
console.log('Is comparison ?', comparison)

Upvotes: 0

Views: 105

Answers (1)

Ori Drori
Ori Drori

Reputation: 191976

Lodash doesn't have an out of the box function that compare keys equality, but you can create one by comparing the number of keys in each objects, and if they are the same check that they are equal to the combined keys number (using a Set):

const isEqualKeys = (a, b) => {
  const keysA = Object.keys(a);
  const keysB = Object.keys(b);
  
  return keysA.length === keysB.length // has the same length of keys
    && new Set([...keysA, ...keysB]).size === keysA.length // has the same length to the combined keys
};

const o1 = { name: 'Tom', address: 'USA' };
const o2 = { name: 'Kumar', address: 'India' };
const o3 = { name: 'Tom', country: 'USA' };
const o4 = { name: 'Tom', address: 'USA', gender: "f" };
const o5 = { name: 'Kumar' };

console.log(isEqualKeys(o1, o2)); // true
console.log(isEqualKeys(o1, o3)); // false
console.log(isEqualKeys(o1, o4)); // false
console.log(isEqualKeys(o1, o5)); // false

Recursive solution to compare nested objects. The check is only performed if both values are objects, if one of them is an object and the other not, we ignore the values:

const isEqualKeys = (a, b) => {
  const keysA = Object.keys(a);
  const keysB = Object.keys(b);
  
  const combinedKeys = new Set([...keysA, ...keysB]);
  
  return keysA.length === combinedKeys.size 
    && keysB.length === combinedKeys.size
    && [...combinedKeys].every(key => {
      const typeA = typeof(a[key]);
      const typeB = typeof(b[key]);
      
      if(typeA === 'object' && typeB === 'object') return isEqualKeys(a[key], b[key]);
      
      return true;
    })
};

const o1 = { name: 'Tom', address: 'USA' };
const o2 = { name: 'Kumar', address: 'India' };
const o3 = { name: 'Tom', country: 'USA' };
const o4 = { name: 'Tom', address: 'USA', gender: "f" };
const o5 = { name: 'Kumar' };

const o6 = { name: 'Kumar', b: { c: 5 } };
const o7 = { name: 'Kumar', b: { c: 7 } };
const o8 = { name: 'Kumar', a: 1 };
const o9 = { name: 'Kumar', b: 1 };
const o10 = { name: 'Kumar', b: { d: 7 } };

console.log(isEqualKeys(o1, o2)); // true
console.log(isEqualKeys(o1, o3)); // false
console.log(isEqualKeys(o1, o4)); // false
console.log(isEqualKeys(o1, o5)); // false
console.log(isEqualKeys(o6, o7)); // true
console.log(isEqualKeys(o6, o8)); // false
console.log(isEqualKeys(o6, o9)); // true
console.log(isEqualKeys(o6, o10)); // false

Upvotes: 1

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