const censusMembers = Object.freeze([
{
id: 1,
name: 'Bob'
}, {
id: 2,
name: 'Sue'
}, {
id: 3,
name: 'Mary',
household_id: 2
}, {
id: 4,
name: 'Elizabeth',
household_id: 6
}, {
id: 5,
name: 'Tom'
}, {
id: 6,
name: 'Jill'
}, {
id: 7,
name: 'John',
household_id: 6
}
]);
In this array, A dependent can be determined by the presence of a household_id. The household_id is a reference to the ID of the employee that that member is a depended of (ex in the censusMembers list 'Mary' is a dependent of 'Sue')
How to build a function that takes in an id and the array of members(census members) and returns all dependents that belong to the user that has that id.
If the id is of a dependent, or isn't in the censusMember array then the function should return null.
If there are no dependents then the function should return an empty arrray.
for example:
if I give input as id 6
then output shoul be
[
{"id":4,"name":"Elizabeth","household_id":6},
{"id":7,"name":"John","household_id":6}
]
Here is code that seems to do what you would like:
const {curry, find, propEq, has, filter} = R
const householdMembers = curry((census, id) => {
const person = find(propEq('id', id), census);
return person
? has('household_id', person)
? null
: filter(propEq('household_id', id), census)
: null
})
var censusMembers = Object.freeze([
{id: 1, name: 'Bob'},
{id: 2, name: 'Sue' },
{id: 3, name: 'Mary', household_id: 2 },
{id: 4, name: 'Elizabeth', household_id: 6},
{id: 5, name: 'Tom'},
{id: 6, name: 'Jill'},
{id: 7, name: 'John', household_id: 6}
])
const householders = householdMembers(censusMembers)
console.log(householders(6))
//=> [
// {id: 4, name: 'Elizabeth','household_id': 6},
// {id: 7, name: 'John', 'household_id': 6}
// ]
console.log(householders(7)) //=> null
console.log(householders(8)) //=> null
console.log(householders(5)) //=> []
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>
But I would suggest that you might want to rethink this API. The empty array is a perfectly reasonable result when nothing is found. Making it return null for some of these cases makes the output much harder to use. For instance, if you wanted to retrieve the list of names of household members, you could simply write const householderNames = pipe(householders, prop('name')). Or you could do this if your function always returned a list.
Having a single function return multiple types like this is much harder to understand, and much, much harder to maintain. Note how much simpler the following version is, one that always returns a (possibly empty) list:
const members = curry((census, id) => filter(propEq('household_id', id), census))
Related
I have 2 array of objects like
const arrayOne = [{id: 1, name: 'one'}, {id: 2, name: 'two'}, {id: 3, name: 'three'}];
const arrayTwo = [{id: 2, name: 'two'}, {id: 3, name: 'three'}];
Here, I need to compare both these arrays and remove matching objects from arrayOne, which should finally give
this.arrayOne = [{id: 1, name: 'one'}];
I tried like below but it is removing all objects from the array
this.arrayOne = this.arrayOne.filter(o1 => this.arrayTwo.some(o2 => o1.id === o2.id));
What I am doing wrong here? Please suggest. Thanks
const arrayOne = [
{ id: 1, name: "one" },
{ id: 2, name: "two" },
{ id: 3, name: "three" },
];
const arrayTwo = [
{ id: 2, name: "two" },
{ id: 3, name: "three" },
];
const arrayTwoIds = new Set(arrayTwo.map((el) => el.id));
const arrayOneFiltered = arrayOne.filter((el) => !arrayTwoIds.has(el.id));
console.log(arrayOneFiltered);
// [ { id: 1, name: 'one' } ]
Depending on the size of the array, creating a set can improve performance, as you do not need to loop over arrayTwo arrayOne.length times but only once. After that, you can look up the existence of an id in arrayTwo in constant time.
Yet, as pointed out in another answer, this is not necessary if the arrays are small (like in your example). In this case, you could also use this one-liner:
arrayOne = arrayOne.filter((elOne) => !arrayTwo.some((elTwo) => elOne.id === elTwo.id));
Here, arrayOne would need to be mutable, i.e. defined with let.
You can find it by comparing it with id.
And arrayOne must be a let.
let arrayOne = [{id: 1, name: 'one'}, {id: 2, name: 'two'}, {id: 3, name: 'three'}];
const arrayTwo = [{id: 2, name: 'two'}, {id: 3, name: 'three'}];
arrayOne = arrayOne.filter(one => !arrayTwo.find(two => one.id == two.id));
console.log(arrayOne);
const arrayOne = [{
id: 1,
name: 'one'
}, {
id: 2,
name: 'two'
}, {
id: 3,
name: 'three'
}];
const arrayTwo = [{
id: 2,
name: 'two'
}, {
id: 3,
name: 'three'
}];
const arrayTwoId = arrayTwo.map(el => (el.id)); // extract id from arrayTwo
const result = arrayOne.filter(el => !arrayTwoId.includes(el.id));
console.log(result);
Extract all the ids from the arrayTwo.
filter those objects who do not match the array of ids of arrayTwo.
Your way is correct. but you miss the not operation (!) before arrayTwo.some.
So the correct way is this:
const arrayOne = [
{id: 1, name: 'one'},
{id: 2, name: 'two'},
{id: 3, name: 'three'}
];
const arrayTwo = [
{id: 2, name: 'two'},
{id: 3, name: 'three'}
];
// Shared Items between arrayOne and arrayTwo (this what you done)
const sharedObjects = arrayOne.filter(o1 => arrayTwo.some(o2 => o1.id === o2.id));
console.log(sharedObjects);
// arrayOne - arrayTwo (this is what you want)
const arrayOneUniqueObjects = arrayOne.filter(o1 => !arrayTwo.some(o2 => o1.id === o2.id));
console.log(arrayOneUniqueObjects);
Also, You can find more details here:
bobbyhadz.com/blog/javascript-get-difference-between-two-arrays-of-objects
I have an array of objects I need to sort based on another array objects. This is the given array that needs to be sorted:
const arr1 = [
{
id: 21,
name: 'Joey',
vehicle: 'car'
},
{
id: 6,
name: 'Kevin'
vehicle: 'car'
},
{
id: 10,
name: 'Luis'
vehicle: 'van'
}
]
And this is the array that is in the proper order:
const arr2 = [
{
id: 6,
name: 'Kevin'
},
{
id: 21,
name: 'Joey'
},
{
id: 10,
name: 'Luis'
}
]
There is no specific order to arr2 its just the data that comes back from my db. I basically just need to put the ids in arr1 in the same order as thet are in arr2.
Ive tried using findIndex and sort but I am very confused
that ?
const arr1 =
[ { id: 21, name: 'Joey', vehicle: 'car' }
, { id: 6, name: 'Kevin', vehicle: 'car' }
, { id: 10, name: 'Luis', vehicle: 'van' }
]
const arr2 =
[ { id: 6, name: 'Kevin' }
, { id: 21, name: 'Joey' }
, { id: 10, name: 'Luis' }
]
// Arr1 ordered..
const arr1_ord = arr2.map(a2=> arr1.find(x=>x.id===a2.id))
console.log( arr1_ord )
.as-console-wrapper {max-height: 100%!important;top:0}
Also, if there are only 2 items in arr2 I still want those elements that are missing to be at the end of the sorted arr1. Would this solve that?
I add add another case : arr2 element doesn't have a arr1 same id
const arr1 =
[ { id: 21, name: 'Joey', vehicle: 'car' }
, { id: 6, name: 'Kevin', vehicle: 'car' }
, { id: 12, name: 'George', vehicle: 'carriage' } // not in arr2
, { id: 10, name: 'Luis', vehicle: 'van' }
]
const arr2 =
[ { id: 6, name: 'Kevin' }
, { id: 21, name: 'Joey' }
, { id: 88, name: 'none' } // not in arr1
, { id: 10, name: 'Luis' }
]
// Arr1 ordered..
const arr1_ord = arr2.reduce((res, {id},i,{[i+1]:next})=>
{
let a1 = arr1.find(x=>x.id===id)
if (a1) res.push(a1) // if exist in arr1
if (next) return res
else return [...res, arr1.filter(r=>!res.some(z=>z.id===r.id))] // add the missing elements
},[])
console.log( arr1_ord )
.as-console-wrapper {max-height: 100%!important;top:0}
Array.prototype.sort can take in a custom comparison function to sort however you'd like it to.
The function takes in two arguments (firstValue, secondValue)
If that function returns a positive value, then secondValue goes before the firstValue, if it's 0 or negative, then firstValue is sorted before secondValue. First/second value are from the array you are sorting. In your case, you are sorting based on a different array, but that is fine since you can do that in your function.
You can do:
arr1.sort((firstValue, secondValue) => {
return findIdInArr2(firstValue.id) - findIdInArr2(secondValue.id);
});
Where you would have to define findIdInArr2 to find the index in arr2, and you can use Array.prototype.findIndex to solve that problem. Where findIndex similarly takes in a function to find the index of something in an array.
function findIdInArr2(id) {
const foundIndex = arr2.findIndex(function(obj) {
return obj.id === id;
});
// If not found, this will be -1. But that would sort all
// unfound objects at the beginning of the array.
// To place these objects at the end of the array, it needs to
// return a number higher than the rest. So return infinity.
return (foundIndex === -1) ? Infinity : foundIndex;
}
note: findIndex is not available in IE, so you'd need to add a polyfill if you are planning on supporting IE.
I am having the array of objects like below
Array 1:
[{id: 1, name: 'Golden', isEdited: true}, {id: 2, name: 'Pearl'}]
Array 2:
[{id: 1, name: 'Golden'}, {id: 2, name: 'Pearlblue'}, , {id: 3, name: 'Orange'}]
Now i would like to merge the two arrays if the object contains isEdited flag means then that object should not be updated.
Expected result should be
[{id: 1, name: 'Golden', isEdited: true}, {id: 2, name: 'Pearlblue'}, {id: 3, name: 'Orange'}]
I have tried with the below approach
b.map((battr) => {
return {
...a[battr.id],
...battr
}
})
But it returns the output as
[{id: 1, name: 'Golden'}, {id: 2, name: 'Pearlblue'}, {id: 3, name: 'Orange'}]
Be carefull, your ids don't match your array indexes.
Hence in your attempt a[battr.id] should be replaced by something like a.find(a => a.id === battr.id)
Your example is confusing since in both arrays the name is 'Golden' for id=1, but assuming you want only the properties from array A if isEdited=true, your solution could be :
b.map((b_attr) => {
var matchingA = a.find(a_attr => a_attr.id === b_attr.id);
if (matchingA && matchingA.isEdited)
return matchingA;
else
return { ...matchingA, ...b_attr }
})
I need to find a simplest way for setting order to array of objects.
For example, there is an array:
var array = [
{id: 1, name: "Matt"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"},
{id: 4, name: "Bruce"}
];
and I have provided
var order = [1,4,2,3];
which refers to object id property of array items.
Now I need to reorder array so it should be like:
var array = [
{id: 1, name: "Matt"},
{id: 4, name: "Bruce"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"}
]
Use Array#sort method for sorting and inside custom sort function use Array#indexOf method to get index.
var array = [{
id: 1,
name: "Matt"
}, {
id: 2,
name: "Jack"
}, {
id: 3,
name: "Morgan"
}, {
id: 4,
name: "Bruce"
}];
var order = [1, 4, 2, 3];
array.sort(function(a, b) {
// sort based on the index in order array
return order.indexOf(a.id) - order.indexOf(b.id);
})
console.log(array);
You can also use reduce() on [1,4,2,3] array to return object where keys will be elements and values will be index of each element and then sort by that object.
var array = [
{id: 1, name: "Matt"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"},
{id: 4, name: "Bruce"}
];
var s = [1,4,2,3].reduce((r, e, i) => {return r[e] = i, r}, {});
var result = array.sort(function(a, b) {
return s[a.id] - s[b.id];
});
console.log(result)
I guess anything that involves sort can not be more efficient than an O(2n) solution. So i would like to do this job with two reduces as follows;
var arr = [{id: 1, name: "Matt"}, {id: 2, name: "Jack"}, {id: 3, name: "Morgan"}, {id: 4, name: "Bruce"}],
order = [1,4,2,3],
lut = order.reduce((t,e,i) => (t[e] = i,t),{}),
result = arr.reduce((res,obj) => (res[lut[obj.id]] = obj, res) ,[]);
console.log(result);
I have 2 arrays of objects, they each have an id in common. I need a property from objects of array 2 added to objects array 1, if they have matching ids.
Array 1:
[
{
id: 1,
name: tom,
age: 24
},
{
id: 2,
name: tim,
age: 25
},
{
id: 3,
name: jack,
age: 24
},
]
Array 2:
[
{
id: 1,
gender: male,
eyeColour: blue,
weight: 150
},
{
id: 2,
gender: male,
eyeColour: green,
weight: 175
},
{
id: 3,
gender: male,
eyeColour: hazel,
weight: 200
},
]
Desired Outcome:
[
{
id: 1,
name: tom,
age: 24,
eyeColour: blue,
},
{
id: 2,
name: tim,
age: 25,
eyeColour: green,
},
{
id: 3,
name: jack,
age: 24,
eyeColour: hazel,
},
]
I tried using lodash _.merge function but then I end up with all properties into one array, when I only want eyeColour added.
Lodash remains a highly useful bag of utilities, but with the advent of ES6 some of its use cases are less compelling.
For each object (person) in the first array, find the object in the second array with matching ID (see function findPerson). Then merge the two.
function update(array1, array2) {
var findPerson = id => array2.find(person => person.id === id);
array1.forEach(person => Object.assign(person, findPerson(person.id));
}
For non-ES6 environments, rewrite arrow functions using traditional syntax. If Array#find is not available, write your own or use some equivalent. For Object.assign, if you prefer use your own equivalent such as _.extend.
This will merge all properties from array2 into array1. To only merge eyeColour:
function update(array1, array2) {
var findPerson = id => array2.find(person => person.id === id);
array1.forEach(person => {
var person2 = findPerson(person.id));
var {eyeColour} = person2;
Object.assign(person, {eyeColour});
});
}
Just noticed Paul answered while I was working on my answer but I'll add my very similar code anyway:
var getEyeColour = function (el) { return _.pick(el, 'eyeColour'); }
var out = _.merge(arr1, _.map(arr2, getEyeColour));
DEMO
You can use pick to get only the properties you want before merging:
var result = _.merge( arr1, _.map( arr2, function( obj ) {
return _.pick( obj, 'id', 'eyeColour' );
}));
A solution in plain Javascript
This is a more generic solution for merging two arrays which have different properties to union in one object with a common key and some properties to add.
var array1 = [{ id: 1, name: 'tom', age: 24 }, { id: 2, name: 'tim', age: 25 }, { id: 3, name: 'jack', age: 24 }, ],
array2 = [{ id: 1, gender: 'male', eyeColour: 'blue', weight: 150 }, { id: 2, gender: 'male', eyeColour: 'green', weight: 175 }, { id: 3, gender: 'male', eyeColour: 'hazel', weight: 200 }, ];
function merge(a, b, id, keys) {
var array = [], object = {};
function m(c) {
if (!object[c[id]]) {
object[c[id]] = {};
object[c[id]][id] = c[id];
array.push(object[c[id]]);
}
keys.forEach(function (k) {
if (k in c) {
object[c[id]][k] = c[k];
}
});
}
a.forEach(m);
b.forEach(m);
return array;
}
document.write('<pre>' + JSON.stringify(merge(array1, array2, 'id', ['name', 'age', 'eyeColour']), 0, 4) + '</pre>');
I was looking for the same, but I want to match Id before merging
And in my case, second array may have different number of items, finally I came with this:
var out = arr1.map(x => {
return { ...x, eyeColour: arr2.find(y => x.id === y.id)?.eyeColour }
});