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I have an array containing objects where there is a rpId key in some of the objects. The goal is to separate/move the objects that return undefined to a separate array and remove them out of the first array.
e.g.:
results = [{id: 1}, {id: 2, rpId: 1076}, {id: 3}, {id: 4, rpId: 303}];
goal: results = [{id: 2, rpId: 1076}, {id: 4, rpId: 303}] and stations = [{id: 1}, {id: 3}]
My current approach can be seen below. As visible, I get a wrong array1 because it contains an object with a rpId, plus array2 returns the keys of the object and I'd like to read the entire object, not just the "undefined" of the key.
const array1 = [{id: 1}, {id: 2, rpId: 1076}, {id: 3}, {id: 4, rpId: 303}];
const array2 = [];
const mapping = array1.map((e) => e.rpId);
console.log("mapping",mapping);
mapping.forEach(function(elem, index){
elem === undefined ? array2.push(elem) && array1.splice(index, elem === undefined) && console.log(elem): console.log("defined", elem);
}),
console.log("1", array1); // [{ id: 2, rpId: 1076 }, { id: 3 }]
console.log("2", array2); // [undefined, undefined]
Just check if the rpId property is undefined in each element.
const array1 = [{id: 1}, {id: 2, rpId: 1076}, {id: 3}, {id: 4, rpId: 303}];
const array2 = [];
array1.forEach(function(elem, index){
if(elem.rpId === undefined)
array2.push(elem) && array1.splice(index, 1)
});
console.log(array1);
console.log(array2);
One can also use Array#filter or push elements into two separate arrays based on the condition for better performance.
const array1 = [{id: 1}, {id: 2, rpId: 1076}, {id: 3}, {id: 4, rpId: 303}];
const yes = [], no = [];
array1.forEach(elem=>(elem.rpId!==undefined?yes:no).push(elem));
console.log(yes);
console.log(no);
You can use filter too:
let results = [
{ id: 1 },
{ id: 2, rpId: 1076 },
{ id: 3 },
{ id: 4, rpId: 303 },
];
const stations = results.filter((c) => !c.rpId);
results = results.filter((c) => c.rpId);
console.log("stations", stations);
console.log("results", results);
const GiveItACreativeName = (arr) => {
const result = []
const stations = []
arr.forEach((el) => {
if('rpId' in el) result.push(el);
else stations.push(el);
});
return {result, stations}
}
console.log(
GiveItACreativeName([{id: 1}, {id: 2, rpId: 1076}, {id: 3}, {id: 4, rpId: 303}])
);
My apologies if this has been addressed before, but I couldn't get it to work with anything I found.
Assume I have 2 arrays - arr1, arr2. I want to update the objects in arr1 if the the property id matches in arr1 and arr2. Objects that exist in arr2 but not in arr1 - meaning the property id does not exist in arr1 - should be pushed to arr1.
Example:
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
]
# Expected Outcome
let outcome = [
{id: 0, name: "Mark"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"},
{id: 4, name: "Sara"}
]
You can use reduce and find for this:
const arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
];
const arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
];
arr2.reduce((res, item) => {
const existingItem = res.find(x => x.id === item.id);
if (existingItem) { existingItem.name = item.name; }
else { res.push(item); }
return res;
}, arr1);
console.log(arr1);
You could do this:
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
]
var res = arr1.reduce((acc, elem)=>{
var x = arr2.find(i=>i.id === elem.id);
if(x){
acc.push(x)
}else{
acc.push(elem)
}
return acc
}, []);
console.log(res)
Assuming you want to mutate the objects in arr1 rather than creating new ones, one way to do it would be using for...of to iterate the objects in arr2 and then check if there's already an object with the same id in arr1 using Array.prototype.find():
If there is one, you mutate it with Object.assign.
Otherwise, push the new object to arr1:
const arr1 = [
{ id: 0, name: 'John' },
{ id: 1, name: 'Sara' },
{ id: 2, name: 'Domnic' },
{ id: 3, name: 'Bravo' },
];
const arr2 = [
{ id: 0, name: 'Mark', sometingElse: 123 },
{ id: 2, foo: 'bar' },
{ id: 4, name: 'Sara' },
];
for (const currentElement of arr2) {
let previousElement = arr1.find(el => el.id === currentElement.id);
if (previousElement) {
Object.assign(previousElement, currentElement);
} else {
arr1.push(currentElement);
}
}
console.log(arr1);
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max-height: 100% !important;
}
if you want to try something different you can use foreach and filter to achieve this
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}]
arr1.forEach(x=>{
arr2.forEach(y=>{
if(x.id==y.id){
x.name=y.name
}
})
})
arr2.filter((a)=>{if(!arr1.some(b=>a.id==b.id)) arr1.push(a)})
console.log(arr1)
You should be able to use Array.prototype.find to sort this out!
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
];
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
];
let updateArrayOfObjects = (arr1, arr2) => {
for (let obj of arr2) {
let item = arr1.find(v => v.id === obj.id);
if (item) item.name = obj.name;
else arr1.push({ ...obj });
}
return arr1;
};
console.log(updateArrayOfObjects(arr1, arr2));
I want to remove same object from array by comparing 2 arrays.
Sample Data:
arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = []; // new array with with no same values it should be unique.
arr1.map((val, i)=>{
arr2.map((val2)=>{
if(val.id == val2.id){
console.log('Matched At: '+ i) // do nothing
}else{
newArray.push(val);
}
})
})
console.log(newArray); // e.g: [{id: 2, name: "b"}, {id: 3, name: "c"},];
Array.filter combined with not Array.some.
The trick here is also to not some,..
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
], arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const newArray=arr1.filter(a=>!arr2.some(s=>s.id===a.id));
console.log(newArray);
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As mentioned in comments the question could be interpreted slightly differently. If you also want the unqiue items from arr2, you basically just do it twice and join. IOW: check what not in arr2 is in arr1, and then check what not in arr1 that's in arr2.
eg..
const notIn=(a,b)=>a.filter(f=>!b.some(s=>f.id===s.id));
const newArray=[...notIn(arr1, arr2), ...notIn(arr2, arr1)];
Update 2:
Time complexity, as mentioned by qiAlex there is loops within loops. Although some will short circuit on finding a match, if the dataset gets large things could slow down. This is were Set and Map comes in.
So to fix this using a Set.
const notIn=(a,b)=>a.filter(a=>!b.has(a.id));
const newArray=[
...notIn(arr1, new Set(arr2.map(m=>m.id))),
...notIn(arr2, new Set(arr1.map(m=>m.id)))
];
const isInArray = (arr, id, name) => arr.reduce((result, curr) => ((curr.name === name && curr.id === id) || result), false)
const newArray = arr1.reduce((result, curr) => (isInArray(arr2, curr.id, curr.name) ? result : result.concat(curr)), [])
You can update you code using filter() method, instead of using .map() method like:
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
], arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = []; // new array with with no same values it should be unique.
newArray = arr1.filter(function(a) {
for(var i=0; i < arr2.length; i++){
if(a.id == arr2[i].id) return false;
}
return true;
});
console.log(newArray);
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You check each element in first array whether its id lies in the second array by using Array.prototype.some. If the element is not present then only yield it.
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
const arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const result = arr1.filter(x => !arr2.some(y => y.id === x.id));
console.log(result);
I think a simple comparer can works for getting differences and then concat them.
with this method you dont need to check which array is bigger.
arr1 = [ {id: 1, name: "a"}, {id: 2, name: "b"}, {id: 3, name: "c"}, {id: 4, name: "d"}];
arr2 = [ {id: 1, name: "a"}, {id: 4, name: "d"},];
function localComparer(b){
return function(a){
return b.filter(
function(item){
return item.id == a.id && item.name == a.name
}).length == 0;
}
}
var onlyInArr1 = arr1.filter(localComparer(arr2));
var onlyInArr2 = arr2.filter(localComparer(arr1));
console.log(onlyInArr1.concat(onlyInArr2));
We can filter values by checking whether some element is not contained in current array:
const result = arr1.reduce((a, c) => {
if (!arr2.some(a2 => a2.id === c.id))
a.push(c);
return a;
}, [])
An example:
let arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
let arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const result = arr1.reduce((a, c) => {
if (!arr2.some(a2 => a2.id === c.id))
a.push(c);
return a;
}, [])
console.log(result);
Try this one -
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
const arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const arr3 = [...arr1, ...arr2];
const mySubArray = _.uniq(arr3, 'id');
console.log(mySubArray);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>
So many loops in every answer.
Complexity of the code my answer is 2N,
Idea is:
to merge arrays.
first loop - mark duplicates somehow
second loop - filter duplicates out
arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = [...arr1, ...arr2].reduce((acc, item, index) => {
acc.items.push(item);
if (typeof acc.map[item.id] !== 'undefined') {
acc.items[acc.map[item.id]] = null;
acc.items[index] = null;
}
acc.map[item.id] = index;
return acc
}, {map: {}, items: []}).items.filter(item => !!item)
console.log(newArray);
I have an object like:
const arr = {
[1]: [{id: 1, category: 1}, {id: 2, category: 2}],
[2]: [{id: 3, category: 2}, {id: 4, category: 2}],
[3]: [{id: 5, category: 3}, {id: 6, category: 3}]
}
Is it possible to move for example {id: 2, category: 2} to another place to have:
const arr = {
[1]: [{id: 1, category: 1}],
[2]: [{id: 2, category: 2}, {id: 3, category: 2}, {id: 4, category: 2}],
[3]: [{id: 5, category: 3}, {id: 6, category: 3}]
}
items with category = 2 should be in the array[2], 3 in 3 etc. It's possible to do that with Object.keys and filtering somehow? Or maybe there is a better solution?
You can use Object.keys() to iterate through arr, then using array#reduce you can group your array on the category value.
const arr = {[1]: [{id: 1, category: 1}, {id: 2, category: 2}], [2]: [{id: 3, category: 2}, {id: 4, category: 2}], [3]: [{id: 5, category: 3}, {id: 6, category: 3}]},
result = Object.keys(arr).reduce((r,k) => {
arr[k].forEach(({id, category}) => {
r[category] = r[category] || [];
r[category].push({id, category});
})
return r;
},{});
console.log(result);
You can use array#concat all the values of your object, then using array#reduce group object based on category.
const arr = {[1]: [{id: 1, category: 1}, {id: 2, category: 2}], [2]: [{id: 3, category: 2}, {id: 4, category: 2}], [3]: [{id: 5, category: 3}, {id: 6, category: 3}]},
result = [].concat(...Object.values(arr)).reduce((r,{id, category}) => {
r[category] = r[category] || [];
r[category].push({id, category});
return r;
},{});
console.log(result);
Yes, the Object.keys can be used to traverse the array. Then, using Array.splice and Array.push, you can remove item and push it to proper place. Please check the following code:
const arr = {
[1]: [{id: 1, category: 1}, {id: 2, category: 2}, {id: 7, category: 3}],
[2]: [{id: 3, category: 2}, {id: 4, category: 2}],
[3]: [{id: 5, category: 3}, {id: 6, category: 3}]
};
Object.keys(arr).forEach(function(k){
for(var j = 0; j < arr[k].length;j++) {
if(arr[k][j].category!=k) {
var removedItem = arr[k].splice(arr[k].indexOf(j), 1);
var pushedArr = arr[removedItem[0].category];
if(pushedArr) {
pushedArr.push(removedItem[0]);
j--;
}
}
}
});
console.log(arr);
Is it possible to concat two arrays with objects and let the second array overwrite the first array where they have the same id:
// array 1
[
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
]
// array 2:
[
{id: 1, name: "newFoo"},
{id: 4, name: "y"},
{id: 5, name: "z"}
]
// out:
[
{id: 1, name: "newFoo"}, // overwriten by array 2
{id: 2, name: "bar"}, // not changed (from array 1)
{id: 3, name: "baz"}, // not changed (from array 1)
{id: 4, name: "y"}, // added (from array 2)
{id: 5, name: "z"} // added (from array 2)
]
If it is possible I would like to do this without the use of third party libraries
var a = [
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
];
var b = [
{id: 1, name: "fooboo"},
{id: 4, name: "bar"},
{id: 5, name: "baz"}
];
/* iterate through each of b, if match found in a, extend with that of a. else push into b ...*/
b.forEach(m => {
var item = a.find(n => n.id === m.id);
if(item) { return Object.assign(item, m); }
a.push(m);
});
console.log(a);
You can do
let arr1 = [
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
]
let arr2 = [
{id: 1, name: "newFoo"},
{id: 4, name: "y"},
{id: 5, name: "z"}
]
let result = arr1.concat(arr2).reduce((a, b) => {
a[b.id] = b.name;
return a;
},{})
result = Object.keys(result).map(e => {
return {id : e, name : result[e]};
});
console.log(result);
Explanation
I am using the property of objects that they don't keep duplicate keys, so for an array concated together, I reduce it to an object with id as it's key and name as its value, hence overriding all duplicates. In the next step I converted this back into an array.
Check you my solution. There is no "rewrite", i just use a second array as base and don't write value if it has same id.
let a = [
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
];
let b = [
{id: 1, name: "newFoo"},
{id: 4, name: "y"},
{id: 5, name: "z"}
];
let duplicateId;
a.forEach(aitem => {
duplicateId = false;
b.forEach(bitem => {
if (aitem.id === bitem.id)
duplicateId = true;
});
if (!duplicateId)
b.push(aitem);
});
console.log(b);
Maybe you can use Object.assign and Object.entries to achieve, lets say:
const arr1 = [
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
]
const arr2 = [
{id: 1, name: "newFoo"},
{id: 4, name: "y"},
{id: 5, name: "z"}
]
const obj3 = Object.entries(Object.assign({}, ...arr1, arr2))
.map(([prop, value]) => ({[prop]:value}));
Example:
https://jsfiddle.net/0f75vLka/
Another option would be to convert arrays to map with id as key then merge the objects and then convert it back to array.
var arr1 = [
{id: 1, name: "foo"},
{id: 2, name: "bar"},
{id: 3, name: "baz"}
];
var arr2 = [
{id: 1, name: "newFoo"},
{id: 4, name: "y"},
{id: 5, name: "z"}
];
function arr2map(arr) {
var map = {};
for (var i = 0; i < arr.length; i++) {
var item = arr[i];
map[item.id] = item;
}
return map;
}
function map2arr(map) {
var arr = [];
for (var i in map) {
arr.push(map[i]);
}
return arr;
}
var arr1m = arr2map(arr1);
var arr2m = arr2map(arr2);
var arr3m = map2arr( Object.assign({}, arr1m, arr2m) );
//output
alert(JSON.stringify(arr3m));