Exploring Javascript object properties in an array and matching them against a specific value

Question

Exploring Javascript object properties in an array and matching them against a specific value

Can someone provide a clear explanation of my question using an example? I have an array of objects structured like this:

[{a:"a",b:"",c:"c"}, {a:"a",b:"b",c:""}, {a:"",b:"b",c:"c"}, {d:""} ]

and [a,b]. Initially, I need to verify that the objects in the first array contain the properties from the second array, and then check if those properties are empty strings or undefined. If they are, I want to set them to "something". The final result should appear as follows:

[{a:"a",b:"something",c:"c"}, {a:"a",b:"b",c:""}, {a:"something", b:"b",c:"c"}, {a:"something", b:"something", d:""} ]

Please note that while I currently have a functional code, it is not very aesthetically pleasing and I am looking for a more elegant approach.

javascript typescript

Answer 1

Answer №1

If you need a unique collection of keys and want to map values for new entries, here's a solution:

const 
    array = [{ a: "a", b: "", c: "c" }, { a: "a", b: "b", c: "" }, { a: "", b: "b", c: "c" }, { d: "" }],
    keys = ['a', 'b'],
    result = array.map(obj => Object.fromEntries(
        [...new Set([...Object.keys(obj), ...keys])].map(key => [key, keys.includes(key) ? obj[key] || 'something' : obj[key]])
    ));

console.log(result);

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

Alternatively, here's another way to achieve this without using Set:

const 
    array = [{ a: "a", b: "", c: "c" }, { a: "a", b: "b", c: "" }, { a: "", b: "b", c: "c" }, { d: "" }],
    keys = ['a', 'b'],
    result = array.map(obj => Object.assign(
        {},
        obj,
        ...keys.map(k => ({ [k]: obj[k] || 'something' }))
    ));

console.log(result);

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

Answer 2

If you need a unique collection of keys and want to map values for new entries, here's a solution:

const 
    array = [{ a: "a", b: "", c: "c" }, { a: "a", b: "b", c: "" }, { a: "", b: "b", c: "c" }, { d: "" }],
    keys = ['a', 'b'],
    result = array.map(obj => Object.fromEntries(
        [...new Set([...Object.keys(obj), ...keys])].map(key => [key, keys.includes(key) ? obj[key] || 'something' : obj[key]])
    ));

console.log(result);

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

Alternatively, here's another way to achieve this without using Set:

const 
    array = [{ a: "a", b: "", c: "c" }, { a: "a", b: "b", c: "" }, { a: "", b: "b", c: "c" }, { d: "" }],
    keys = ['a', 'b'],
    result = array.map(obj => Object.assign(
        {},
        obj,
        ...keys.map(k => ({ [k]: obj[k] || 'something' }))
    ));

console.log(result);

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

Answer 3

Answer №2

let arr1 = [{a:"a",b:"",c:"c"}, {a:"a",b:"b",c:""}, {a:"",b:"b",c:"c"}, {d:""} ];
let arr2 = ['a','b'];

for (let item of arr1) {
    for (let keyItem of arr2) {
        if (arr2.includes(keyItem) && !item[keyItem]) {
            item[keyItem] = 'Something';
        }
    }
}
console.log('arr1 = ' + JSON.stringify(arr1));

Answer 4

let arr1 = [{a:"a",b:"",c:"c"}, {a:"a",b:"b",c:""}, {a:"",b:"b",c:"c"}, {d:""} ];
let arr2 = ['a','b'];

for (let item of arr1) {
    for (let keyItem of arr2) {
        if (arr2.includes(keyItem) && !item[keyItem]) {
            item[keyItem] = 'Something';
        }
    }
}
console.log('arr1 = ' + JSON.stringify(arr1));

Answer 5

Answer №3

Verify that the elements in the initial array contain all the attributes found in the second array

Iterate through each element in the "first" Array
Go through each key in the "second" Array
Ensure that every key is included at least once within the values of the element

Determine whether any of them are blank strings (or undefined) and replace them with a placeholder like "something"

Iterate through each element in the "first" Array
Examine each key/value pair within the element
If any of the values are empty Strings or undefined, update the value to be "something"

const arr = [{a:"a",b:"",c:"c"}, {a:"a",b:"b",c:""}, {a:"",b:"b",c:"c"}, {d:""}];
const required = ["a", "b"];
const replacement = "something";

const hasRequired = arr.every(e => required.every(req => Object.values(e).includes(req)));

const transformed = arr.map(e => {
  return Object.entries(e).reduce((res, [k, v]) => {
    if (v === "" || v === undefined) {
      e[k] = replacement;
    }

    return res;
  }, e);
});

console.log(hasRequired);
console.log(transformed);

Answer 6

Verify that the elements in the initial array contain all the attributes found in the second array

Iterate through each element in the "first" Array
Go through each key in the "second" Array
Ensure that every key is included at least once within the values of the element

Determine whether any of them are blank strings (or undefined) and replace them with a placeholder like "something"

Iterate through each element in the "first" Array
Examine each key/value pair within the element
If any of the values are empty Strings or undefined, update the value to be "something"

const arr = [{a:"a",b:"",c:"c"}, {a:"a",b:"b",c:""}, {a:"",b:"b",c:"c"}, {d:""}];
const required = ["a", "b"];
const replacement = "something";

const hasRequired = arr.every(e => required.every(req => Object.values(e).includes(req)));

const transformed = arr.map(e => {
  return Object.entries(e).reduce((res, [k, v]) => {
    if (v === "" || v === undefined) {
      e[k] = replacement;
    }

    return res;
  }, e);
});

console.log(hasRequired);
console.log(transformed);

Answer 7

Answer №4

One clever trick we can use is leveraging the falsy nature of empty strings and undefined.

While there are more concise methods available, here's one approach you could take. Loop through each object in the array, check the specified keys, and assign them a value of 'something' if they are falsy; otherwise, keep their current value.

const objects = [{a:'a',b:'',c:'c'}, {a:'a',b:'b',c:''}, {a:'',b:'b',c:'c'}, {d:''}];
const keys = ['a', 'b'];

objects.map(o => {
  for (const key of keys) {
    o[key] = o[key] || 'something';
  }
  return o;
});

Just be cautious if your values might include false, 0, NaN, etc., as those will require specific validation.

Answer 8

One clever trick we can use is leveraging the falsy nature of empty strings and undefined.

While there are more concise methods available, here's one approach you could take. Loop through each object in the array, check the specified keys, and assign them a value of 'something' if they are falsy; otherwise, keep their current value.

const objects = [{a:'a',b:'',c:'c'}, {a:'a',b:'b',c:''}, {a:'',b:'b',c:'c'}, {d:''}];
const keys = ['a', 'b'];

objects.map(o => {
  for (const key of keys) {
    o[key] = o[key] || 'something';
  }
  return o;
});

Just be cautious if your values might include false, 0, NaN, etc., as those will require specific validation.

Exploring Javascript object properties in an array and matching them against a specific value

Answer №1

Answer №2

Answer №3

Answer №4

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