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How can I get an array with all the unique values based on a property name?
In my case my object looks like this and I want an array with the unique documentID's.
const file = {
invoice: {
invoiceID: 1,
documentID: 5
},
reminders: [
{
reminderID: 1,
documentID: 1
},
{
reminderID: 2,
documentID: 1
}
]
}
The result should be an array [5, 1] //The unique documentID's are 5 and 1
It doesn't seem like possible to add a property name to the Object.values() function.
You can use Set to get unique documentID.
const file = {
invoice: {
invoiceID: 1,
documentID: 5
},
reminders: [
{
reminderID: 1,
documentID: 1
},
{
reminderID: 2,
documentID: 1
}
],
payments: {
documentID : 5
}
};
var keys = Object.keys(file).map(key=>file[key].map ? file[key].map(i=>i.documentID) : file[key].documentID)
var keysFlattened= [].concat.apply([], keys);
var unique = new Set(keysFlattened);
console.log(Array.from(unique));
I use something like this that does what you want I think
const keepUniqueBy = key => (array, item) => {
if (array.find(i => item[key] === i[key])) {
return array;
} else {
return [ ...array, item ];
}
};
Then you can simply: const unique = reminders.reduce(keepUniqueBy('documentID'))
NB: It's probably low performing, but for small arrays it doesn't matter.
I am basically trying to convert a flat json file to tree view. Here the parent child relationship required for tree view is mentained by links key using source and target.
Here is the sample raw input:
{
"nodes" : [
{
name: "bz_db",
index: 0
},
{
name: "mysql",
index: 1
},
{
name: "postgres",
index: 2
},
{
name: "it-infra",
index: 3
},
{
name: "user-count",
index: 4
}
],
links: [
{
source: 0, target: 1
},
{
source: 0, target: 3
},
{
source: 1, target: 3
},
{
source: 3, target: 4
}
]
}
As you can see the link field maintains this relation ship, and finally I want my data in this format:
{
name: "bz_db",
children: [
{
name: "mysql",
children: [
{
name: "it-infra",
children: [
{
name: "user_count",
children: []
}
]
}
]
},
{
name: "it-infra",
children: [{
name: "user_count",
children: []
}
]
}
]
}
I tried to solve this, but it worked for 1 level (to show immediate child of a selected root element.
var findObjectByKeyValue = function(arrayOfObject, key, value){
return _.find(arrayOfObject, function(o){ return o[key] == value})
}
var rootObject = findObjectByKeyValue(sample_raw_input.nodes, 'name', 'bz_db');
var treeObject = {
name: rootObject.name,
index: rootObject.index,
root: true,
children: []
};
angular.forEach(dependencyData.links, function(eachLink){
if(treeObject.index == eachLink.source){
var rawChildObject = findObjectByKeyValue(dependencyData.nodes, 'index', eachLink.target);
var childObject = {};
childObject.index = rawChildObject.index;
childObject.name = rawChildObject.name;
childObject.children = [];
treeObject.children.push(childObject);
}
});
But the above code returns me only first level of depndencies, but i want hierarchical relationship.
I know i can use recursion here. But I am not so comfortable with it.
Josh's answer uses a sequence of map->filter->map->find calls, each of which iterate thru a collection of data. This loop of loop of loop of loops results in a stunning amount of computational complexity as the number of nodes in your collection increases.
You can dramatically simplify the creation of the tree by using a single reduce pass on each nodes and links. Map can also perform look-ups in logarithmic time, compared to Array's find which requires linear time (slower). When you consider this operation is called for each element of the input, it's clear to see a significant difference in time.
const makeTree = (nodes = [], links = []) =>
links.reduce
( (t, l) =>
t.set ( l.source
, MutableNode.push ( t.get (l.source)
, t.get (l.target)
)
)
, nodes.reduce
( (t, n) => t.set (n.index, MutableNode (n.name))
, new Map
)
)
.get (0)
Lastly, we provide the MutableNode interface we relied upon
const MutableNode = (name, children = []) =>
({ name, children })
MutableNode.push = (node, child) =>
(node.children.push (child), node)
Below is a full program demonstration. JSON.stringify is used only to display the result
const MutableNode = (name, children = []) =>
({ name, children })
MutableNode.push = (node, child) =>
(node.children.push (child), node)
const makeTree = (nodes = [], links = []) =>
links.reduce
( (t, l) =>
t.set ( l.source
, MutableNode.push ( t.get (l.source)
, t.get (l.target)
)
)
, nodes.reduce
( (t, n) => t.set (n.index, MutableNode (n.name))
, new Map
)
)
.get (0)
const data =
{ nodes:
[ { name: "bz_db", index: 0 }
, { name: "mysql", index: 1 }
, { name: "postgres", index: 2 }
, { name: "it-infra", index: 3 }
, { name: "user-count", index: 4 }
]
, links:
[ { source: 0, target: 1 }
, { source: 0, target: 3 }
, { source: 1, target: 3 }
, { source: 3, target: 4 }
]
}
const tree =
makeTree (data.nodes, data.links)
console.log (JSON.stringify (tree, null, 2))
You can rely on tracking an object reference and do this without any recursion. Using Object.assign, map the list of nodes to its children:
// Assuming that input is in `input`
const nodes = input.nodes.reduce((a, node) => {
a[node.index] = { ...node, index: undefined };
return a;
}, []);
// organize the links by their source
const links = input.links.reduce((a, link) => {
return a.set((a.get(link.source) || []).concat(nodes[link.target]);
}, new Map());
// Apply side effect of updating node children
nodes.forEach(node => Object.assign(node, {
children: links.get(node.index),
}));
So I'm taking the list of nodes, and assigning to each (to mutate the node itself -- keep in mind this is a side-effect) a new array. Those children are all the links that link this node, and we Array#map them to convert their target ID into the actual node we want.
just share sample, a little different from yours.
But it give you a hint with recursive function.
jsFiddle flat array json transform to recursive tree json
function getNestedChildren(arr, parent) {
var out = []
for(var i in arr) {
if(arr[i].parent == parent) {
var children = getNestedChildren(arr, arr[i].id)
if(children.length) {
arr[i].children = children
}
out.push(arr[i])
}
}
return out
}
var flat = [
{id: 1, title: 'hello', parent: 0},
{id: 2, title: 'hello', parent: 0},
{id: 3, title: 'hello', parent: 1},
{id: 4, title: 'hello', parent: 3},
{id: 5, title: 'hello', parent: 4},
{id: 6, title: 'hello', parent: 4},
{id: 7, title: 'hello', parent: 3},
{id: 8, title: 'hello', parent: 2}
]
var nested = getNestedChildren(flat, 0)
console.log(nested)
I am using Lodash in my Angular project and I was wondering if there is a better way to write the following code:
$scope.new_arr = _.map(arr1, function(item){
return _.assign(item, {new_id: _.find(arr2, {id: item.id})});
});
$scope.new_arr = _.filter($scope.new_arr, function (item) {
return item.new_id !== undefined;
});
I am trying to combine values from one array to same objects in other array, and I want to ignore the objects that not appear in both arrays (it is something like join or left outer join in the sql language).
Here is a fiddle with an example of this code: Click me!
i think is better to use chaining
$scope.new_arr = _.chain(arr1)
.map(function(item) {
return _.merge(
{}, // to avoid mutations
item,
{new_id: _.find(arr2, {id: item.id})}
);
})
.filter('new_id')
.value();
https://jsfiddle.net/3xjdqsjs/6/
try this:
$scope.getItemById = (array, id) => {
return array.find(item => item.id == id);
};
$scope.mergeArrays = () => {
let items_with_ids = arr1.filter(item => !_.isNil($scope.getItemById(arr2,item.id)));
return items_with_ids.map(item => _.assign(item, {new_id: $scope.getItemById(arr2,item.id)}));
};
The answers provided here are all runtime of O(n^2), because they first run an outer loop on the first array, with an inner loop on the second array. You can instead run this in O(n). First, create a hashmap of all the ids in arr2 in a single loop; this will allow us an order 1 lookup. In the second loop on arr1, check this hashmap to determine if those items exist with O(n). Total Complexity is n + n = 2n, which is just O(n).
// provision some test arrays
var arr1 = [
{
id: 2
},
{
id: 4
},
{
id: 6
}
]
var arr2 = [
{
id: 3
},
{
id: 4
},
{
id: 5
},
{
id: 6
}
]
// First, we create a map of the ids of arr2 with the items. Complexity: O(n)
var mapIdsToArr2Items = _.reduce(arr2, function(accumulator, item) {
accumulator[item.id] = item;
return accumulator;
}, {});
// Next, we use reduce (instead of a _.map followed by a _.filter for slightly more performance.
// This is because with reduce, we loop once, whereas with map and filter,
// we loop twice). Complexity: O(n)
var combinedArr = _.reduce(arr1, function(accumulator, item) {
// Complexity: O(1)
if (mapIdsToArr2Items[item.id]) {
// There's a match/intersection! Arr1's item matches an item in arr 2. Include it
accumulator.push(item);
}
return accumulator;
}, []);
console.log(combinedArr)
You could first make a Map with arr1 and then map the items of arr2 with the properties of arr1.
var arr1 = [{ id: 1, title: 'z' }, { id: 2, title: 'y' }, { id: 3, title: 'x' }, { id: 4, title: 'w' }, { id: 5, title: 'v' }],
arr2 = [{ id: 2, name: 'b' }, { id: 3, name: 'c' }, { id: 4, name: 'd' }, { id: 5, name: 'e' }],
map = new Map(arr1.map(a => [a.id, a])),
result = arr2.map(a => Object.assign({}, a, map.get(a.id)));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
See jsfiddle here: https://jsfiddle.net/remenyLx/2/
I have data that contains objects that each have an array of images. I want only the first image of each object.
var data1 = [
{
id: 1,
images: [
{ name: '1a' },
{ name: '1b' }
]
},
{
id: 2,
images: [
{ name: '2a' },
{ name: '2b' }
]
},
{
id: 3
},
{
id: 4,
images: []
}
];
var filtered = [];
var b = data1.forEach((element, index, array) => {
if(element.images && element.images.length)
filtered.push(element.images[0].name);
});
console.log(filtered);
The output needs to be flat:
['1a', '2a']
How can I make this prettier?
I'm not too familiar with JS map, reduce and filter and I think those would make my code more sensible; the forEach feels unnecessary.
First you can filter out elements without proper images property and then map it to new array:
const filtered = data1
.filter(e => e.images && e.images.length)
.map(e => e.images[0].name)
To do this in one loop you can use reduce function:
const filtered = data1.reduce((r, e) => {
if (e.images && e.images.length) {
r.push(e.images[0].name)
}
return r
}, [])
You can use reduce() to return this result.
var data1 = [{
id: 1,
images: [{
name: '1a'
}, {
name: '1b'
}]
}, {
id: 2,
images: [{
name: '2a'
}, {
name: '2b'
}]
}, {
id: 3
}, {
id: 4,
images: []
}];
var result = data1.reduce(function(r, e) {
if (e.hasOwnProperty('images') && e.images.length) r.push(e.images[0].name);
return r;
}, [])
console.log(result);
All answers are creating NEW arrays before projecting the final result : (filter and map creates a new array each) so basically it's creating twice.
Another approach is only to yield expected values :
Using iterator functions
function* foo(g)
{
for (let i = 0; i < g.length; i++)
{
if (g[i]['images'] && g[i]["images"].length)
yield g[i]['images'][0]["name"];
}
}
var iterator = foo(data1) ;
var result = iterator.next();
while (!result.done)
{
console.log(result.value)
result = iterator.next();
}
This will not create any additional array and only return the expected values !
However if you must return an array , rather than to do something with the actual values , then use other solutions suggested here.
https://jsfiddle.net/remenyLx/7/
I have an array of objects and I want to get a new array from it that is unique based only on a single property, is there a simple way to achieve this?
Eg.
[ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
Would result in 2 objects with name = bill removed once.
Use the uniq function
var destArray = _.uniq(sourceArray, function(x){
return x.name;
});
or single-line version
var destArray = _.uniq(sourceArray, x => x.name);
From the docs:
Produces a duplicate-free version of the array, using === to test object equality. If you know in advance that the array is sorted, passing true for isSorted will run a much faster algorithm. If you want to compute unique items based on a transformation, pass an iterator function.
In the above example, the function uses the objects name in order to determine uniqueness.
If you prefer to do things yourself without Lodash, and without getting verbose, try this uniq filter with optional uniq by property:
const uniqFilterAccordingToProp = function (prop) {
if (prop)
return (ele, i, arr) => arr.map(ele => ele[prop]).indexOf(ele[prop]) === i
else
return (ele, i, arr) => arr.indexOf(ele) === i
}
Then, use it like this:
const obj = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
obj.filter(uniqFilterAccordingToProp('abc'))
Or for plain arrays, just omit the parameter, while remembering to invoke:
[1,1,2].filter(uniqFilterAccordingToProp())
If you want to check all the properties then
lodash 4 comes with _.uniqWith(sourceArray, _.isEqual)
A better and quick approach
var table = [
{
a:1,
b:2
},
{
a:2,
b:3
},
{
a:1,
b:4
}
];
let result = [...new Set(table.map(item => item.a))];
document.write(JSON.stringify(result));
Found here
You can use the _.uniqBy function
var array = [ { id: 1, name: 'bob' }, { id: 2, name: 'bill' }, { id: 1, name: 'bill' },{ id: 2, name: 'bill' } ];
var filteredArray = _.uniqBy(array,function(x){ return x.id && x.name;});
console.log(filteredArray)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>
In the above example, filtering is based on the uniqueness of combination of properties id & name.
if you have multiple properties for an object.
then to find unique array of objects based on specific properties, you could follow this method of combining properties inside _.uniqBy() method.
I was looking for a solution which didn't require a library, and put this together, so I thought I'd add it here. It may not be ideal, or working in all situations, but it's doing what I require, so could potentially help someone else:
const uniqueBy = (items, reducer, dupeCheck = [], currentResults = []) => {
if (!items || items.length === 0) return currentResults;
const thisValue = reducer(items[0]);
const resultsToPass = dupeCheck.indexOf(thisValue) === -1 ?
[...currentResults, items[0]] : currentResults;
return uniqueBy(
items.slice(1),
reducer,
[...dupeCheck, thisValue],
resultsToPass,
);
}
const testData = [
{text: 'hello', image: 'yes'},
{text: 'he'},
{text: 'hello'},
{text: 'hell'},
{text: 'hello'},
{text: 'hellop'},
];
const results = uniqueBy(
testData,
item => {
return item.text
},
)
console.dir(results)
In case you need pure JavaScript solution:
var uniqueProperties = {};
var notUniqueArray = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ];
for(var object in notUniqueArray){
uniqueProperties[notUniqueArray[object]['name']] = notUniqueArray[object]['id'];
}
var uniqiueArray = [];
for(var uniqueName in uniqueProperties){
uniqiueArray.push(
{id:uniqueProperties[uniqueName],name:uniqueName});
}
//uniqiueArray
unique array by id property with ES6:
arr.filter((a, i) => arr.findIndex(b => b.id === a.id) === i); // unique by id
replace b.id === a.id with the relevant comparison for your case