Given a javascript object array eg.
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}]
is there a faster way of getting all the b values from each object which meet a specific criteria such as a = 1 to return something like
b_consolidated = [2,3]
instead of looping through every object in the array?
You can use Array#filter function to get the items of your criteria, then use Array#map to get only b property.
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}];
let values = objArray.filter(item => item.a === 1).map(item => item.b);
console.log(values);
Or you can do this in one loop
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}];
let values = [];
objArray.forEach(item => {
if(item.a === 1) {
values.push(item.b);
}
});
console.log(values);
You could use Array#reduce in a single loop.
let array = [{ a: 1, b: 2, c: 3}, { a: 1, b: 3, c: 2 }, { a: 2, b: 5, c: 1 }],
result = array.reduce((r, o) => o.a === 1 ? r.concat(o.b) : r, []);
console.log(result);
Fastest version with for loop.
let array = [{ a: 1, b: 2, c: 3}, { a: 1, b: 3, c: 2 }, { a: 2, b: 5, c: 1 }],
i, l,
result = [];
for (i = 0, l = array.length; i < l; i++) {
if (array[i].a === 1) {
result.push(array[i].b);
}
}
console.log(result);
You only need to iterate over the array once, if you use reduce:
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}]
let result = objArray.reduce((arr, val) => {
if(val.a === 1)
arr.push(val.b);
return arr;
}, []);
console.log(result);
This is as fast as it'll get, short of a manual for loop:
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}]
let result = [];
for(var i = 0 ; i < objArray.length; i++){
if(objArray[i].a === 1)
result.push(objArray[i].b);
}
console.log(result);
Here's a JSPerf to illustrate the difference.
A manual for loop is by far the fastest.
More faster would be using .reduce
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}];
objArray.reduce(function(res,obj){
if(obj.a===1)
res.push(obj.b);
return res;
},[]);
// [2,3]
In Ramda
let objArray = [{a: 1, b: 2 , c:3},{a: 1, b:3, c:2},{a: 2, b:5, c:1}]
R.pipe(
R.filter(R.propEq('a', 1)),
R.pluck('b')
)(objArray)
// [2, 3]
Filter returns the array values matched by the condition.
Pluck returns a new list by plucking the same named property off all objects in the list supplied.
Edit 1:
Example of using the mentioned reduce pattern in Ramda:
R.reduce((acc, x) => R.ifElse(
R.propEq('a', 1),
(item) => R.pipe(R.prop('b'), R.append(R.__, acc))(item),
R.always(acc)
)(x), [])(objArray)
// [2, 3]
Related
compare array of object with array of keys, filter array of object with array keys.
Input:
let a = ['aa'];
let b = [{ aa: 1, bb: 2, c: 30 },{ aa: 2, bb: 3, c: 40}];
output:
b = [{bb: 2, c: 30 },{bb: 3, c: 40}];
original array should be mutate.
Much similiar to #SachilaRanawaka 's answer, but works without modifying the original b array:
let a = ['aa'];
let b = [{ aa: 1, bb: 2, c: 30 },{ aa: 2, bb: 3, c: 40}];
function removeKey(obj, key) {
let clone = Object.assign({}, obj); // <-- shallow clone
if (key in clone) {
delete clone[key];
}
return clone;
}
function removeKeys(keys, objs) {
return objs.map(o => keys.reduce(removeKey, o));
}
console.log(removeKeys(a, b));
You could take a destructuring with getting the rest approach.
This approach does not mutate the original data.
const
unwanted = ['aa'],
data = [{ aa: 1, bb: 2, c: 30 }, { aa: 2, bb: 3, c: 40 }],
result = data.map(o => unwanted.reduce((q, k) => {
const { [k]: _, ...r } = q;
return r;
}, o));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
You can simply achieve this requirement with the help of Array.forEach() method.
Live Demo :
let a = ['aa'];
let b = [{ aa: 1, bb: 2, c: 30 },{ aa: 2, bb: 3, c: 40}];
b.forEach(obj => {
Object.keys(obj).forEach(key => {
a.forEach(item => delete obj[item])
});
});
console.log(b);
It can probably be solved with less lines of code, but this was the first i could think of.
let keysToRemove = ['aa'];
let array = [{ aa: 1, bb: 2, c: 30 },{ aa: 2, bb: 3, c: 40}];
let result = array.map((item) => {
let filtered = Object.keys(item)
.filter((key) => !keysToRemove.includes(key))
.reduce((obj, key) => {
obj[key] = item[key];
return obj;
}, {});
return filtered;
});
console.log(result);
use the map operator and use delete to delete properties from the object
let a = ['aa'];
let b = [{ aa: 1, bb: 2, c: 30 },{ aa: 2, bb: 3, c: 40}];
const result = b.map(item => {
Object.keys(item).forEach(key => {
if(a.includes(key)){
delete item[key]
}
})
return item
})
console.log(result)
Say I have the following array:
let arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]
I would like to compute the cumulative sum of each key, but I would also like the output to be an array of the same length with the cumulative values at each step. The final result should be:
[{a: 1, b: 2}, {a: 3, b: 6}, {a: 11, b: 5}]
My issue is that I am not able to obtain the array as desired. I only get the final object with this:
let result = arr.reduce((accumulator, element) => {
if(accumulator.length === 0) {
accumulator = element
} else {
for(let i in element){
accumulator[i] = accumulator[i] + element[i]
}
}
return accumulator
}, [])
console.log(result); // {a: 11, b: 5}
What you're after sounds like the scan() higher-order function (borrowing the idea from ramda.js), which allows you to return an accumulated result for each element within your array. The scan method is similar to how the .reduce() method behaves, except that it returns the accumulator for each element. You can build the scan() function yourself like so:
let arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}];
const scan = ([x, ...xs], fn) => xs.reduce((acc, elem) => {
return [...acc, fn(acc.at(-1), elem)];
}, xs.length ? [x] : []);
const res = scan(arr, (x, y) => ({a: x.a+y.a, b: x.b+y.b}));
console.log(res);
You might consider further improvements such as providing an initial value to the scan method (similar to how reduce accepts one). Also, if you need better browser support the .at() method currently has limited browser support, so you may instead consider creating your own at() function:
const at = (arr, idx) => idx >= 0 ? arr[idx] : arr[arr.length + idx];
You can easily achieve the result using reduce as
let arr = [
{ a: 1, b: 2 },
{ a: 2, b: 4 },
{ a: 8, b: -1 },
];
const result = arr.reduce((acc, curr, i) => {
if (i === 0) acc.push(curr);
else {
const last = acc[i - 1];
const newObj = {};
Object.keys(curr).forEach((k) => (newObj[k] = curr[k] + last[k]));
acc.push(newObj);
}
return acc;
}, []);
console.log(result);
Something like this:
const arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]
const result = arr.reduce((accumulator, element, index) => {
if(accumulator.length === 0) {
accumulator.push(element)
} else {
const sum = {};
for(let i in element) {
sum[i] = element[i] + (accumulator[index - 1][i] || 0)
}
accumulator.push(sum)
}
return accumulator
}, [])
console.log(result);
Another option is keep sum result using a Map, it helps if keys in elements of the array are not always same.
const arr = [{a: 1, b: 2}, {a: 2}, {a: 8, b: -1}];
const map = new Map();
const result = arr.map((element) => {
const sum = {};
for (let i in element) {
sum[i]= element[i] + (map.get(i) || 0);
map.set(i, sum[i]);
}
return sum;
});
console.log(result);
Here is a bit more concise reduce, probably not as readable as a consequence...
array.reduce((y,x,i) => ( i===0 ? y : [...y, {a: x.a + y[i-1].a, b: x.b + y[i-1].b}]),[array[0]])
let array = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]
let culm = array.reduce((y,x,i) => ( i===0 ? y : [...y, {a: x.a + y[i-1].a, b: x.b + y[i-1].b}]),[array[0]])
console.log(culm)
Given:
const xs =
[ {a: 1, b: 2}
, {a: 2, b: 4}
, {a: 8, b: -1}];
Define a function sum such as:
const sum = ([head, ...tail]) =>
tail.reduce((x, y) =>
({a: (x.a+y.a), b: (x.b+y.b)}), head);
sum(xs);
//=> {a: 11, b: 5}
Then apply that function in a map on larger slices of xs:
xs.map((_, i, arr) => sum(arr.slice(0, i+1)));
//=> [ {a: 1, b: 2}
//=> , {a: 3, b: 6}
//=> , {a: 11, b: 5}]
I have an array with objects. I filtered the array and returned the values that correspond with a condition. After that i created a new array what contains data from first. This array contains a value a which should overrides the first array.
const arr = [{
a: 1,
b: 2,
name: 'one'
},
{
a: 1,
b: 7,
name: 'two'
}
]
const b = [1, 2]
const res = arr.filter(i => b.includes(i.b))
const newArr = {
...res,
a: 'newdata'
}
console.log(newArr)
In console i got:
{
"0": {
"a": 1,
"b": 2,
"name": "one"
},
"a": "newdata"
}
But the expected output is:
{
"a": "newdata"
"b": 2,
"name": "one"
},
Why i get this and how to get the expected result?
After you filter the values you're interested in, you could use Array.map(..) to go through all the items and change their a property, here is an example:
const arr = [{
a: 1,
b: 2,
name: 'one'
},
{
a: 1,
b: 7,
name: 'two'
}
]
const b = [1, 2]
const res = arr.filter(i => b.includes(i.b)).map(i => ({...i, a: 'newdata'}));
console.log(res)
You simply spread an array and you get the indices from the array/object as key and the item/object as value.
Instead, you could take a function for a new object with additional key/value and map the filtered result by using the function.
const
arr = [{ a: 1, b: 2, name: 'one' }, { a: 1, b: 7, name: 'two' }],
b = [1, 2, 7],
res = arr.filter(i => b.includes(i.b)),
newObject = o => ({ ...o, a: 'newdata' }),
newArr = res.map(newObject);
console.log(newArr);
I am trying to merge two arrays of objects without using the unionBy method from lodash.
Currently I have the following code working perfectly:
var array1 = [
{ a: 1, b: 'first'},
{ a: 2, b: 'second'}
];
var array2 = [
{ a: 3, b: 'third'},
{ a: 1, b: 'fourth'}
];
var array3 = __.unionBy(array2, array1, 'a');
This outputs:
[
{
"a": 3,
"b": "third"
},
{
"a": 1,
"b": "fourth"
},
{
"a": 2,
"b": "second"
}
]
This is the desired result but I can't use unionBy in my current work environment, so I'm looking for a result that uses either native JS or other lodash methods 3.6.0 or lower.
Concat and use Array#filter with a helper object to remove duplicates:
var array1 = [{"a":1,"b":"first"},{"a":2,"b":"second"}];
var array2 = [{"a":3,"b":"third"},{"a":1,"b":"fourth"}];
var result = array2.concat(array1).filter(function(o) {
return this[o.a] ? false : this[o.a] = true;
}, {});
console.log(result);
If ES6 is an option you can use a Set instead of the helper object:
const array1 = [{"a":1,"b":"first"},{"a":2,"b":"second"}];
const array2 = [{"a":3,"b":"third"},{"a":1,"b":"fourth"}];
const result = array2.concat(array1).filter(function(o) {
return this.has(o.a) ? false : this.add(o.a);
}, new Set());
console.log(result);
If you want to use an arrow function, you can't use the thisArg of Array.filter() to bind the Set as the this of the function (you can't bind this to arrow functions). You can use a closure instead (attribute for the method goes to #NinaScholz).
const array1 = [{"a":1,"b":"first"},{"a":2,"b":"second"}];
const array2 = [{"a":3,"b":"third"},{"a":1,"b":"fourth"}];
const result = [...array2, ...array1]
.filter((set => // store the set and return the actual callback
o => set.has(o.a) ? false : set.add(o.a)
)(new Set()) // use an IIFE to create a Set and store it set
);
console.log(result);
You could take a Set for filtering to get unique values.
var array1 = [{ a: 1, b: 'first' }, { a: 2, b: 'second' }],
array2 = [{ a: 3, b: 'third' }, { a: 1, b: 'fourth' }],
s = new Set,
array3 = array2.map(o => (s.add(o.a), o)).concat(array1.filter(o => !s.has(o.a)));
console.log(array3);
You can use an ES6 Map for this. Construct it with the data, keyed by the a property value, and then take the values out of the Map again:
var array1 = [{"a":1,"b":"first"},{"a":2,"b":"second"}],
array2 = [{"a":3,"b":"third"},{"a":1,"b":"fourth"}];
var result = [...new Map([...array1,...array2].map( o => [o.a, o] )).values()];
console.log(result);
You can merge the 2 arrays and then filter the ones with same property a:
var array1 = [{ a: 1, b: 'first'},{ a: 2, b: 'second'}],
array2 = [{ a: 3, b: 'third'},{ a: 1, b: 'fourth'}],
array3 = [...array2, ...array1].filter((item, pos, arr) =>
arr.findIndex(item2 => item.a == item2.a) == pos);
console.log(array3)
If you want to still be able to specify the property by which to union you can implement you own function like this:
var array1 = [{ a: 1, b: 'first'},{ a: 2, b: 'second'}],
array2 = [{ a: 3, b: 'third'},{ a: 1, b: 'fourth'}],
array3 = unionBy(array1, array2, 'a');
function unionBy(array1, array2, prop){
return [...array2, ...array1].filter((item, pos, arr) =>
arr.findIndex(item2 => item[prop] == item2[prop]) == pos);
}
console.log(array3);
Note: One advantage of my answer over some of the answers is that it preserves the order like in lodash which may or may not be important.
ES5 using Array.filter and Array.find
var array1 = [{ a: 1, b: "first" }, { a: 2, b: "second" }];
var array2 = [{ a: 3, b: "third" }, { a: 1, b: "fourth" }];
function merge(a, b, prop) {
var reduced = a.filter(function(itemA) {
return !b.find(function(itemB) {
return itemA[prop] === itemB[prop];
});
});
return reduced.concat(b);
}
console.log(merge(array1, array2, "a"));
ES6 arrow functions
var array1 = [{ a: 1, b: "first" }, { a: 2, b: "second" }];
var array2 = [{ a: 3, b: "third" }, { a: 1, b: "fourth" }];
function merge(a, b, prop) {
const reduced = a.filter(
itemA => !b.find(itemB => itemA[prop] === itemB[prop])
);
return reduced.concat(b);
}
console.log(merge(array1, array2, "a"));
Another ES6 one line experiment
var array1 = [{ a: 1, b: "first" }, { a: 2, b: "second" }];
var array2 = [{ a: 3, b: "third" }, { a: 1, b: "fourth" }];
const merge = (a, b, p) => a.filter( aa => ! b.find ( bb => aa[p] === bb[p]) ).concat(b);
console.log(merge(array1, array2, "a"));
You could use ES6 find and reduce function smartly!
var array1 = [{"a":1,"b":"first"},{"a":2,"b":"second"}];
var array2 = [{"a":3,"b":"third"},{"a":1,"b":"fourth"}];
var res = array1.concat(array2).reduce((aggr, el)=>{
if(!aggr.find(inst=>inst.a==el.a))
return [...aggr, el];
else
return aggr
},[])
console.log(res);
given this array var array = [{a:1, b: 2, c: 3}, {a: 2, b: 4, c: 4}].
I want to iterate through each item in the array and omit the key c and it's value using lodash.
here is the method I tried
_.each(array, function (obj) {
return _.omit(obj, ['a']);
});
Expected Output should be // [{b: 2, c: 3}, {b: 4, c: 4} ]
but lodash returns the original array// [{a:1, b: 2, c: 3}, {a: 2, b: 4, c: 4}]
Instead of forEach() to return new modified array you should use map(), also first parameter is array.
var array = [{a:1, b: 2, c: 3}, {a: 2, b: 4, c: 4}]
var result = _.map(array, function (obj) {
return _.omit(obj, ['a']);
});
console.log(JSON.stringify(result, 0, 4))
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.15.0/lodash.min.js"></script>
A probably-not-the-lo-dash-iest way of handling it (assuming you actually want to omit a):
_.each(lnkn, function (obj) {
var returnObj = {};
for (prop in obj) {
if (obj.hasOwnProperty(prop) {
if (prop !== 'a') {
returnObj[prop] = obj[prop];
}
}
}
return returnObj;
});
_.forEach is supposed to return the original array. You re looking for _.map.
const array = [{a:1, b: 2, c: 3}, {a: 2, b: 4, c: 4}];
const result = _.map(array, obj => _.omit(obj, 'a'));
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>
Or you can modify the objects in place:
const array = [{a:1, b: 2, c: 3}, {a: 2, b: 4, c: 4}];
_.forEach(array, obj => {
delete obj.a;
});
console.log(array);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>