Related
I'm trying to convert an array of object contains paths with item to tree of data so I wrote a function path loop on the path:
From this array:
[
{ userName: "1", tags: ["A;B"] },
{ userName: "2", tags: ["A;B"] },
{ userName: "3", tags: ["A;"] },
{ userName: "4", tags: ["A;B;C"] },
{ userName: "5", tags: ["A;B"] },
{ userName: "6", tags: ["A;B;C;D"] }
]
to this structure:
[{
name: "A",
families: [{
name: "B",
families: [{
name: "C",
families: [{
name: "D",
families: [],
items: ["6"]
}],
items: ["4"]
}],
items: ["1", "2", "5"]
}],
items: ["3"]
}]
function convertListToTree(associationList) {
let tree = [];
for (let i = 0; i < associationList.length; i++) {
let path = associationList[i].tags[0].split(';');
let assetName = associationList[i].userName;
let currentLevel = tree;
for (let j = 0; j < path.length; j++) {
let familyName = path[j];
let existingPath = findWhere(currentLevel, 'name', familyName);
if (existingPath) {
if (j === path.length - 1) {
existingPath.items.push(assetName);
}
currentLevel = existingPath.families;
} else {
let assets = [];
if (j === path.length - 1) {
assets.push(assetName)
}
let newPart = {
name: familyName,
families: [],
items: assets,
};
currentLevel.push(newPart);
currentLevel = newPart.families;
}
}
}
return tree;
}
function findWhere(array, key, value) {
let t = 0;
while (t < array.length && array[t][key] !== value) {
t++;
}
if (t < array.length) {
return array[t]
} else {
return false;
}
}
But I have some issue here that the expected output is not like I want
[
{
"name": "A",
"families": [
{
"name": "B",
"families": [
{
"name": "C",
"families": [
{
"name": "D",
"families": [],
"items": [
"6"
]
}
],
"items": [
"4"
]
}
],
"items": [
"1",
"2",
"5"
]
},
{
"name": "",
"families": [],
"items": [
"3"
]
}
],
"items": []
}
]
Can someone please help me to fix that
You should be able to use recursion to achieve this, using getFamilies and getUsers functions called at each level:
const allTags = ["A", "B", "C", "D"];
let a = [ { "userName": "1", "tags": ["A;B"] }, { "userName": "2", "tags": ["A;B"] }, { "userName": "3", "tags": ["A;"] }, { "userName": "4", "tags": ["A;B;C"] }, { "userName": "5", "tags": ["A;B"] }, { "userName": "6", "tags": ["A;B;C;D"] } ];
// This function assumes order is not important, if it is, remove the sort() calls.
function arraysEqual(a1, a2) {
return a1.length === a2.length && a1.sort().every(function(value, index) { return value === a2.sort()[index]});
}
function getUserNames(tags, arr) {
return arr.filter(v => arraysEqual(v.tags[0].split(';').filter(a => a),tags)).map(({userName}) => userName);
}
function getFamilies(tags) {
if (tags.length >= allTags.length) return [];
const name = allTags[tags.length];
const path = [...tags, name];
return [{ name, families: getFamilies(path), items: getUserNames(path, a)}];
}
let res = getFamilies([]);
console.log('Result:', JSON.stringify(res, null, 4));
The idea here is to iterate the data (the reduce loop), and whenever a node is missing from the Map (nodesMap), use createBranch to recursively create the node, create the parent (if needed...), and then assign the node to the parent, and so on. The last step is to get a unique list of root paths (A in your data), and extract them from the Map (tree) to an array.
const createBranch = ([name, ...tagsList], nodesMap, node) => {
if(!nodesMap.has(name)) { // create node if not in the Map
const node = { name, families: [], items: [] };
nodesMap.set(name, node);
// if not root of branch create the parent...
if(tagsList.length) createBranch(tagsList, nodesMap, node);
};
// if a parent assign the child to the parent's families
if(node) nodesMap.get(name).families.push(node);
};
const createTree = data => {
const tree = data.reduce((nodesMap, { userName: item, tags: [tags] }) => {
const tagsList = tags.match(/[^;]+/g).reverse(); // get all nodes in branch and reverse
const name = tagsList[0]; // get the leaf
if(!nodesMap.has(name)) createBranch(tagsList, nodesMap); // if the leaf doesn't exist create the entire branch
nodesMap.get(name).items.push(item); // assign the item to the leaf's items
return nodesMap;
}, new Map());
// get a list of uniqnue roots
const roots = [...new Set(data.map(({ tags: [tags] }) => tags.split(';')[0]))];
return roots.map(root => tree.get(root)); // get an array of root nodes
}
const data = [{"userName":"1","tags":["A;B"]},{"userName":"2","tags":["A;B"]},{"userName":"3","tags":["A;"]},{"userName":"4","tags":["A;B;C"]},{"userName":"5","tags":["A;B"]},{"userName":"6","tags":["A;B;C;D"]}];
const result = createTree(data);
console.log(result);
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Allow me to make two small changes, and ramda's mergeDeepWithKey will do most of the work for you.
Changes, before we start:
Make tags an array rather than an array containing one string (i.e. tags[0].split(";"))
Allow families to be a dictionary-like object rather than an array (if you ever need your array format, it's Object.values(dict))
Solution:
Transform every entry to a path of the desired format using reduce
Merge all paths with custom logic:
When merging name entries, don't change the name
When merging items entries, concatenate
const inp = [
{ userName: "1", tags: ["A","B"] },
{ userName: "2", tags: ["A","B"] },
{ userName: "3", tags: ["A"] },
{ userName: "4", tags: ["A","B","C"] },
{ userName: "5", tags: ["A","B"] },
{ userName: "6", tags: ["A","B","C","D"] }
];
// Transform an input element to a nested path of the right format
const Path = ({ userName, tags }) => tags
.slice(0, -1)
.reduceRight(
(families, name) => ({ name, families: { [families.name]: families },
items: []
}),
({ name: last(tags), families: {}, items: [userName] })
);
// When merging path entries, use this custom logic
const mergePathEntry = (k, v1, v2) =>
k === "name" ? v1 :
k === "items" ? v1.concat(v2) :
null;
const result = inp
.map(Path)
// Watch out for inp.length < 2
.reduce(
mergeDeepWithKey(mergePathEntry)
)
console.log(JSON.stringify(result, null, 2));
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.min.js"></script>
<script>const { mergeDeepWithKey, last } = R;</script>
I've taken the following sample from a different question. And I am able to identify the object. But I also need to find our the position of that object. For example:
var arr = [{
Id: 1,
Categories: [{
Id: 1
},
{
Id: 2
},
]
},
{
Id: 2,
Categories: [{
Id: 100
},
{
Id: 200
},
]
}
]
If I want to find the object by the Id of the Categories, I can use the following:
var matches = [];
var needle = 100; // what to look for
arr.forEach(function(e) {
matches = matches.concat(e.Categories.filter(function(c) {
return (c.Id === needle);
}));
});
However, I also need to know the position of the object in the array. For example, if we are looking for object with Id = 100, then the above code will find the object, but how do I find that it's the second object in the main array, and the first object in the Categories array?
Thanks!
Well, if every object is unique (only in one of the categories), you can simply iterate over everything.
var arr = [{
Id: 1,
Categories: [{Id: 1},{Id: 2}]
},
{
Id: 2,
Categories: [{Id: 100},{Id: 200}]
}
];
var needle = 100;
var i = 0;
var j = 0;
arr.forEach(function(c) {
c.Categories.forEach(function(e) {
if(e.Id === needle) {
console.log("Entry is in position " + i + " of the categories and in position " + j + " in its category.");
}
j++;
});
j = 0;
i++;
});
function findInArray(needle /*object*/, haystack /*array of object*/){
let out = [];
for(let i = 0; i < haystack.lenght; i++) {
if(haystack[i].property == needle.property) {
out = {pos: i, obj: haystack[i]};
}
}
return out;
}
if you need the position and have to filter over an property of the object you can use a simple for loop. in this sample your result is an array of new object because there can be more mathches than 1 on the value of the property.
i hope it helps
Iterate over the array and set index in object where match found
var categoryGroups = [{
Id : 1,
Categories : [{
Id : 1
}, {
Id : 2
},
]
}, {
Id : 2,
Categories : [{
Id : 100
}, {
Id : 200
},
]
}
]
var filterVal = [];
var needle = 100;
for (var i = 0; i < categoryGroups.length; i++) {
var subCategory = categoryGroups[i]['Categories'];
for (var j = 0; j < subCategory.length; j++) {
if (subCategory[j]['Id'] == findId) {
filterVal.push({
catIndex : i,
subCatIndex : j,
id : needle
});
}
}
}
console.log(filterVal);
Here is solution using reduce:
var arr = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 }, ] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }, ] } ]
const findPositions = (id) => arr.reduce((r,c,i) => {
let indx = c.Categories.findIndex(({Id}) => Id == id)
return indx >=0 ? {mainIndex: i, categoryIndex: indx} : r
}, {})
console.log(findPositions(100)) // {mainIndex: 1, categoryIndex: 0}
console.log(findPositions(1)) // {mainIndex: 0, categoryIndex: 0}
console.log(findPositions(200)) // {mainIndex: 1, categoryIndex: 1}
console.log(findPositions(0)) // {}
Beside the given answers with fixt depth searh, you could take an recursive approach by checking the Categories property for nested structures.
function getPath(array, target) {
var path;
array.some(({ Id, Categories = [] }) => {
var temp;
if (Id === target) {
path = [Id];
return true;
}
temp = getPath(Categories, target);
if (temp) {
path = [Id, ...temp];
return true;
}
});
return path;
}
var array = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 },] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }] }];
console.log(getPath(array, 100));
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I have the code below which I expect to map the result from the nested array and return a single array having both id's but I get 2 arrays instead. Can someone please guide me on what I'm doing wrongly?
arrayVal = [{
sources: {
data: [{
id: 1
}]
}
},
{
sources: {
data: [{
id: 2
}]
}
}
]
for (let sub of arrayVal) {
let result = sub.sources.data.map(x => (x.id))
console.log(result)
}
Right now, you're calling map for each element in arrayVal, so you get two arrays. Use reduce instead, to transform an array of objects into another array that's not necessarily one-to-one with the input elements:
const arrayVal=[{sources:{data:[{id:1}]}},{sources:{data:[{id:2}]}}];
const result = arrayVal.reduce((a, { sources: { data } }) => (
[...a, ...data.map(({ id }) => id)]
), []);
console.log(result)
Try following
var arrayVal = [{sources: {data: [{id: 1}]}},{sources: {data: [{id: 2}]}}];
// Create an array on sources.data and merge it into 1 collection (array)
var result = arrayVal.reduce((a, c) => [...a, ...c.sources.data.map(({id}) => id)], []);
console.log(result);
For reference, Array.reduce
Also, you can improve your code as follows
var arrayVal = [{sources: {data: [{id: 1}]}},{sources: {data: [{id: 2}]}}];
let result = [];
for (let sub of arrayVal) {
result.push(sub.sources.data.map(x => (x.id)));
}
console.log([].concat(...result))
arrayVal = [{
sources: {
data: [{
id: 1
}]
}
},
{
sources: {
data: [{
id: 2
}]
}
}
]
let result = [];
for (let sub of arrayVal) {
result = result.concat(sub.sources.data.map(x => (x.id)))
}
console.log(result)
I think concat is what you were missing here, Hope this is what you trying to achieve
Try this
arrayVal = [{
sources: {
data: [{
id: 1
}]
}
},
{
sources: {
data: [{
id: 2
}]
}
}
]
let result = arrayVal.map((x) => x.sources.data[0].id)
console.log(result)
You can do something like this:
arrayVal = [{
sources: {
data: [{
id: 1
}]
}
},
{
sources: {
data: [{
id: 2
}]
}
}
]
var flat = arrayVal.reduce(function(prev,curr,cI){
prev.push(curr.sources.data.map(x => (x.id))[0]);
return prev; // ********* Important ******
}, []);
I have an array like bellow each index contains different set of objects,I want to create an uniformal data where object missing in each index will with Value:0 ,
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
]
];
how can I get an array like bellow using above above array
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
{axis:"Sending Money",value:0,id:6},
{axis:"Other",value:0,id:7},
],
[
{axis:"Email",value:0,id:1},
{axis:"Social Networks",value:0,id:2},
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
]
];
There are two functions:
getAllEntries that find all objects and stores them into a variable accEntries. Then accEntries is used to search for all occurrences in a sub-array of d. This whole process is done in checkArray.
checkArray is used to fetch all found and not-found entries in d. Both Arrays (found and not-found) are then used to build a new sub-array that contains either found entries with certain values and/or not-found entries with values of 0.
Hope this helps:
var d = [
[
{
axis: 'Email',
value: 59,
id: 1
},
{
axis: 'Social Networks',
value: 56,
id: 2
},
],
[
{
axis: 'Sending Money',
value: 18,
id: 6
},
{
axis: 'Other',
value: 15,
id: 7
},
]
];
function getAllEntries(array) {
var uniqueEntries = [];
array.forEach(function (subarray) {
subarray.forEach(function (obj) {
if (uniqueEntries.indexOf(obj) === - 1) uniqueEntries.push(obj);
});
});
return uniqueEntries;
}
function checkArray(array, acceptedEntries) {
var result = [];
array.forEach(function (subArray) {
var subResult = [];
var foundEntries = [];
subArray.forEach(function (obj) {
if (foundEntries.indexOf(obj.axis) === - 1) foundEntries.push(obj.axis);
});
var notFound = acceptedEntries.filter(function (accepted) {
return foundEntries.indexOf(accepted.axis) === - 1;
});
foundEntries.forEach(function (found) {
subArray.forEach(function (obj) {
if (obj.axis === found) subResult.push(obj);
});
});
notFound.forEach(function (notfound, index) {
subResult.push({
axis: notfound.axis,
value: 0,
id: notfound.id
});
});
result.push(subResult);
});
return result;
}
var accEntries = getAllEntries(d);
var result = checkArray(d, accEntries);
console.log(result);
You can loop over the array to find all the unique objects and then again loop over to push the values that are not present comparing with the array of objects of unique keys.
You can use ES6 syntax to find if an object with an attribute is present like uniKeys.findIndex(obj => obj.axis === val.axis); and the to push with a zero value use the spread syntax like d[index].push({...val, value: 0});
Below is the snippet for the implementation
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
{axis:"Social Networks",value:89,id:2},
]
];
var uniKeys = [];
$.each(d, function(index, item) {
$.each(item, function(idx, val){
const pos = uniKeys.findIndex(obj => obj.axis === val.axis);
if(pos == - 1) {
uniKeys.push(val);
}
})
})
$.each(d, function(index, item) {
var temp = [];
$.each(uniKeys, function(idx, val){
const pos = item.findIndex(obj => obj.axis === val.axis);
if(pos == - 1) {
d[index].push({...val, value: 0});
}
})
})
console.log(d);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
How about a short shallowCopy function (Object.assign is not available in IE) and otherwise less than 10 new lines of code?
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2}
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7}
]
];
var newD_0 = [shallowCopy(d[0][0]), shallowCopy(d[0][1]), shallowCopy(d[1][0]), shallowCopy(d[1][1])];
var newD_1 = [shallowCopy(d[0][0]), shallowCopy(d[0][1]), shallowCopy(d[1][0]), shallowCopy(d[1][1])];
newD_0[2].id = 0;
newD_0[3].id = 0;
newD_1[0].id = 0;
newD_1[1].id = 0;
d = [newD_0, newD_1];
function shallowCopy(obj) {
var copy = {};
for (var key in obj) {
if (obj.hasOwnProperty(key)) {
copy[key] = obj[key];
}
}
return copy;
}
console.log(JSON.stringify(d));
RESULT:
[
[
{
"axis":"Email",
"value":59,
"id":1
},
{
"axis":"Social Networks",
"value":56,
"id":2
},
{
"axis":"Sending Money",
"value":18,
"id":0
},
{
"axis":"Other",
"value":15,
"id":0
}
],
[
{
"axis":"Email",
"value":59,
"id":0
},
{
"axis":"Social Networks",
"value":56,
"id":0
},
{
"axis":"Sending Money",
"value":18,
"id":6
},
{
"axis":"Other",
"value":15,
"id":7
}
]
]
I have to merge 2 arrays with key value as follows:
array1 = [
{id:"123", data:[{id:"234",data:"hello"},{id:"345",data:"there"},{id:"xyz", data:"yo"}]},
{id:"456", data:[{id:"34",data:"test"},{id:"45",data:"test2"},{id:"yz", data:"test3"}]},
{id:"789", data:[{id:"23",data:"aaa"},{id:"34",data:"bbb"},{id:"xy", data:"ccc"}]}]
with
array2 = [
{id:"456", data:[{id:"45",data:"changed"},{id:"yz", data:"data"}]},
{id:"789", data:[{id:"456",data:"appended data"}]},
{id:"890", data:[{id:"456",data:"new data"}]}]
to produce something like
merged = [
{id:"123", data:[{id:"234",data:"hello"},{id:"345",data:"there"},{id:"xyz", data:"yo"}]},
{id:"456", data:[{id:"34",data:"test"},{id:"45",data:"changed"},{id:"yz", data:"data"}]},
{id:"789", data:[{id:"23",data:"aaa"},{id:"34",data:"bbb"},{id:"xy", data:"ccc"},{id:"456",data:"appended data"}]},
{id:"890", data:[{id:"456",data:"new data"}]}]
I've been trying this out for quite some time and can't get a solution that meets the scenario. Most of the solutions just do blind merging, not based on the id value. Tried using lodash mergeWith but didn't get the output needed. A Ramda solution is also acceptable.
Thanks,
This links could be helpful to you merge two arrays.
In this code snippet, i have tried to find the common objects between set1 and set2,if there are any i'm finding the unique properties and changing their content and also non existant properties in object2 and pushing it to object1
Check the following snippet.
var arr1 = [{
id: "123",
data: [{
id: "234",
data: "hello"
}, {
id: "345",
data: "there"
}, {
id: "xyz",
data: "yo"
}]
}, {
id: "456",
data: [{
id: "34",
data: "test"
}, {
id: "45",
data: "test2"
}, {
id: "yz",
data: "test3"
}]
}, {
id: "789",
data: [{
id: "23",
data: "aaa"
}, {
id: "34",
data: "bbb"
}, {
id: "xy",
data: "ccc"
}]
}]
var arr2 = [{
id: "456",
data: [{
id: "45",
data: "changed"
}, {
id: "yz",
data: "data"
}]
}, {
id: "789",
data: [{
id: "456",
data: "appended data"
}]
}, {
id: "890",
data: [{
id: "456",
data: "new data"
}]
}]
var arr3 = [];
for (var i in arr1) {
var shared = false;
for (var j in arr2)
if (arr2[j].id == arr1[i].id) {
shared = true;
// arr1[i].data.concat(arr2[j].data);
var set1 = pushproperties(arr1[i].data, arr2[j].data);
arr1[i].data = set1;
arr3.push(arr1[i]);
break;
}
if (!shared) {
arr3.push(arr1[i]);
arr3.push(arr2[j]);
}
}
function pushproperties(set1, set2) {
var filtered = false;
set2.forEach(function(item) {
filtered = set1.every(function(element) {
return element.id != item.id;
});
if (filtered) {
set1.push(item);
}
});
set1.forEach(function(item) {
set2.forEach(function(element) {
if (item.id == element.id) {
item.data = element.data;
}
});
});
return set1;
}
console.log(arr3);
Hope this helps
This a function the merges 2 arrays recursively using Array.prototype.reduce(). If it encounters items with the same id, and they have a data prop, which is an array, it merges them using the logic. If data is not an array, it's overridden by the last item instead.
function mergeArraysDeep(arr1, arr2) {
var unique = arr1.concat(arr2).reduce(function(hash, item) {
var current = hash[item.id];
if(!current) {
hash[item.id] = item;
} else if (Array.isArray(current.data)) {
current.data = mergeArraysDeep(current.data, item.data);
} else {
current.data = item.data;
}
return hash;
}, {});
return Object.keys(unique).map(function(key) {
return unique[key];
});
}
var array1 = [
{id:"123", data:[{id:"234",data:"hello"},{id:"345",data:"there"},{id:"xyz", data:"yo"}]},
{id:"456", data:[{id:"34",data:"test"},{id:"45",data:"test2"},{id:"yz", data:"test3"}]},
{id:"789", data:[{id:"23",data:"aaa"},{id:"34",data:"bbb"},{id:"xy", data:"ccc"}]}
];
var array2 = [
{id:"456", data:[{id:"45",data:"changed"},{id:"yz", data:"data"}]},
{id:"789", data:[{id:"456",data:"appended data"}]},
{id:"890", data:[{id:"456",data:"new data"}]}
];
var result = mergeArraysDeep(array1, array2)
console.log(result);
ES6 version that uses Map, Map.prototype.values(), and array spread:
const mergeArraysDeep = (arr1, arr2) => {
return [...arr1.concat(arr2).reduce((hash, item) => {
const current = hash.get(item.id);
if(!current) {
hash.set(item.id, item);
} else if (Array.isArray(current.data)) {
current.data = mergeArraysDeep(current.data, item.data);
} else {
current.data = item.data;
}
return hash;
}, new Map()).values()];
}
const array1 = [
{id:"123", data:[{id:"234",data:"hello"},{id:"345",data:"there"},{id:"xyz", data:"yo"}]},
{id:"456", data:[{id:"34",data:"test"},{id:"45",data:"test2"},{id:"yz", data:"test3"}]},
{id:"789", data:[{id:"23",data:"aaa"},{id:"34",data:"bbb"},{id:"xy", data:"ccc"}]}
];
const array2 = [
{id:"456", data:[{id:"45",data:"changed"},{id:"yz", data:"data"}]},
{id:"789", data:[{id:"456",data:"appended data"}]},
{id:"890", data:[{id:"456",data:"new data"}]}
];
const result = mergeArraysDeep(array1, array2)
console.log(result);
Finally this is what worked for me. Thanks to #Geeky for showing the way:
function mergeArrays(arr1, arr2) {
var arr3, arrIdx = [];
if (!arr1 || arr1.length ==0) return arr2
for (var i in arr1) {
var shared = false;
for (var j in arr2)
if (arr2[j].id == arr1[i].id) {
shared = true;
joined = _.mergeWith({},arr1[i],arr2[j], function (a,b) {
if (_.isArray(a)) return b.concat(a)})
arr3.push(joined);
break;
}
if (!shared) {
arr3.push(arr1[i]);
}
}
for (var k in arr2) {
if (arrIdx[k] !=k) arr3.push(arr2[k])
}
return arr3
}