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I'm working on school-app. person enter students marks from frontend and I've to store it in my backend. I know my data-structure is quite bad. but this is only way I can comfortly use and fit it in my front end application/website.
codeSandbox link
Full Code:
//This data is already set need to push information in this array.
let student = [{
"detail": {
"name": "Mark",
"surname":"widen"
},
}];
//formatting the query in json.
const keys = Object.keys(query)[0].split(",")
const values = Object.values(query)[0].split(",")
const newObj = {}
for (let i = 0; i < keys.length; i++) {
newObj[keys[i]] = values[i]
}
// I've to push it along with "academic-year". so,
for (let a = 0; a < newObj.length; a++) {
const year = a + "st-Year"
console.log(year) // Expected Output: 1st-Year and 2nd-Year
}
// How to run this both for-loop synchronously way ?? AND
//pushing with "ObtainedMarks" and "year" (Error over here)
student.push({
ObtainedMarks: {
year : [
{ physics: newObj }
],
year : [
{ physics: newObj }
]
}
})
console.log(student) //Here's I want expected Output
Expected Output:
let student = [{
"detail": {
"name": "Mark",
"surname":"widen"
},
ObtainedMarks: {
"1st-Year": [
{ physics: { "marks": "500" } } //Physics subject is default.
],
"2nd-Year": [
{ physics: { "mark": "200" } } //Physics subject is default.
]
}
}];
I want to push returned data in student array. with 1st-Year
and 2nd-Year's for-loop.
You can do the conversion in your for-loop
let student = [{
"detail": {
"name": "Mark",
"surname": "widen"
},
}];
let query = {
"marks,mark": "500,200"
}
const keys = Object.keys(query)[0].split(",");
const values = Object.values(query)[0].split(",");
const marks = {}
for (let i = 0; i < keys.length; i++) {
marks[i === 0 ? `${i+1}st-year` : `${i+1}nd-year`] = [{
physics: {
[keys[i]]: values[i]
}
}];
}
student.push({
obtainedMarks: marks
});
console.log(student);
Alternative way: map through the keys and create an object from entries after manipulating the data.
let student = [{
"detail": {
"name": "Mark",
"surname": "widen"
},
}];
let query = {
"marks,mark": "500,200"
}
const keys = Object.keys(query)[0].split(",");
const values = Object.values(query)[0].split(",");
const marks = Object.fromEntries(keys.map((k, i) => {
return [
i === 0 ? `${i+1}st-year`: `${i+1}nd-year`,
[{ physics: { [k]: values[i] }}]
];
}));
student.push({
obtainedMarks: marks
});
console.log(student);
I got array of objects (tasks). Each task has property named 'category' and 'duration'.
var tasks = [
{
_id : "123",
category : "someCategory",
duration: "3432"
},
{
_id : "113",
category : "someCategory",
duration: "23"
},
{
_id : "124",
category : "someCategory 2",
duration: "1343"
},
{
_id : "2124",
category : "someCategory 2",
duration: "1343"
},
{
_id : "7124",
category : "someCategory 5",
duration: "53"
},
{
_id : "34",
category : "someCategory",
duration: "753"
}
]
I'd like to group tasks by category (unique) and sum duration of each category.
Result should be like:
var categories = ["someCategory", "someCategory 2" ... ]
var duration = [ <summary duration of "someCategory">, <summary duration of "someCategory 2">, ... ]
I have groupBy function which gives me all categories. I can find uniqueCategories using Array.prototype.filter but still I have to sum 'duration'.
var categoryMap = groupBy(tasks, 'category');
var uniqueCategories = categoryMap.get('category').filter((x, i, a) => a.indexOf(x) == i);
function groupBy(list, property) {
var map = new Map();
list.forEach(function(item) {
const key = property;
if(!map.has(key)) {
map.set(key, [item[key]])
} else {
map.get(key).push(item[key])
}
})
return map;
}
Then I create array of { key : value } and sum by key i.e.
[
{
someCategory : 3432
},
{
someCategory : 23
}
.
.
.
]
Finally I achieve my goal but code looks messy and isn't readable at all...
Is there better approach to do it in Javascript?
You could just return one object with category: duration.
var tasks = [{"_id":"123","category":"someCategory","duration":"3432"},{"_id":"113","category":"someCategory","duration":"23"},{"_id":"124","category":"someCategory 2","duration":"1343"},{"_id":"2124","category":"someCategory 2","duration":"1343"},{"_id":"7124","category":"someCategory 5","duration":"53"},{"_id":"34","category":"someCategory","duration":"753"}]
var result = tasks.reduce(function(r, e) {
r[e.category] = (r[e.category] || 0) + +e.duration
return r;
}, {})
console.log(result)
var tasks = [{"_id":"123","category":"someCategory","duration":"3432"},{"_id":"113","category":"someCategory","duration":"23"},{"_id":"124","category":"someCategory 2","duration":"1343"},{"_id":"2124","category":"someCategory 2","duration":"1343"},{"_id":"7124","category":"someCategory 5","duration":"53"},{"_id":"34","category":"someCategory","duration":"753"}]
var arr = [];
tasks.forEach(v => arr.push(v.category));
var newArr = [...new Set(arr)];
var arr2 = [];
newArr.forEach(function(v) {
var obj = {};
obj.category = v;
obj.duration = 0;
arr2.push(obj);
});
arr2.forEach(v => tasks.forEach(c => c.category == v.category ? v.duration += parseInt(c.duration) : v));
console.log(arr2);
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 an 'item' object in JavaScript, and the item can have settings like
color, size, etc.
I need to get all possible combinations in an array.
So lets say we have an item that looks like this:
var newItem = {
name: 'new item',
Settings: [
{name: 'color', values: ['green', 'blue', 'red']},
{name: 'size', values: ['15', '18', '22']},
{name: 'gender',values: ['male', 'female']}
]
};
I need to somehow get this:
[
[{SettingName:'color',value:'green'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}]
]
This can be a good interview question.
See JS Bin for running example.
getAllPermutations(newItem);
function getAllPermutations(item) {
var permutations = [];
getAllPermutations0(item, permutations, []);
console.log(permutations);
}
function getAllPermutations0(item, permutations, array) {
if (array && array.length === item.Settings.length) {
permutations.push(array.slice()); // The slice clone the array
return;
}
var index = array.length;
var setting = item.Settings[index];
for (var i = 0; i < setting.values.length; i++) {
if (index === 0)
array = [];
var currValue = setting.values[i];
array.push({
SettingName: setting.name,
value: currValue
});
getAllPermutations0(item, permutations, array);
array.pop(); // pop the old one first
}
}
Here is a none recursive solution. It takes an empty or existing settings "matrix" and a values array, and return a new matrix as a combination of existing matrix content cloned for each new value, appended with pairs of new value setting items.
[A] -> [1,2] gives [A][1][A][2]
[A][1][A][2] -> [X,Y] gives [A][1][X][A][2][Y][A][2][X][A][1][Y]
and so on
function processSettings(settings, name, values) {
if (settings.length == 0) {
values.forEach(function(value) {
settings.push( [{ SettingName: name, value: value }] )
})
} else {
var oldSettings = JSON.parse(JSON.stringify(settings)), settings = [], temp, i = 0
for (i; i<values.length; i++) {
temp = JSON.parse(JSON.stringify(oldSettings))
temp.forEach(function(setting) {
setting.push( { SettingName: name, value: values[i] } )
settings.push(setting)
})
}
}
return settings
}
You can now create the desired settings literal this way :
var settings = []
for (var i=0; i<newItem.Settings.length; i++) {
var item = newItem.Settings[i]
settings = processSettings(settings, item.name, item.values)
}
demo -> http://jsfiddle.net/b4ck98mf/
The above produces this :
[
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}]
]
You can use Array.prototype.map(), for loop, while loop, Array.prototype.concat(). Iterate gender values; select each of color, size value in succession beginning at index 0 of either; iterating the furthest adjacent array from current gender, increment the index of the closest adjacent array; merge the resulting two gender arrays to form a single array containing all combinations of gender, color, size
var colors = newItem.Settings[0].values;
var sizes = newItem.Settings[1].values;
var gen = newItem.Settings[2].values;
var i = sizes.length;
var res = [].concat.apply([], gen.map(function(value, key) {
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
arr.push([{
SettingName: "gender",
value: value
}, {
SettingName: "size",
value: sizes[curr]
}, {
SettingName: "color",
value: colors[++next]
}])
}
next = -1;
}
return arr
}))
var newItem = {
"name": "new item",
"Settings": [{
"name": "color",
"values": [
"green",
"blue",
"red"
]
}, {
"name": "size",
"values": [
"15",
"18",
"22"
]
}, {
"name": "gender",
"values": [
"male",
"female"
]
}]
}
var colors = newItem.Settings[0].values;
var sizes = newItem.Settings[1].values;
var gen = newItem.Settings[2].values;
var i = sizes.length;
var res = [].concat.apply([], gen.map(function(value, key) {
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
arr.push([{
SettingName: "gender",
value: value
}, {
SettingName: "size",
value: sizes[curr]
}, {
SettingName: "color",
value: colors[++next]
}])
}
next = -1;
}
return arr
}))
document.querySelector("pre").textContent = JSON.stringify(res, null, 2)
<pre></pre>
plnkr http://plnkr.co/edit/C2fOJpfwOrlBwHLQ2izh?p=preview
An approach using Array.prototype.reduce(), Array.prototype.sort(), Object.keys(), for loop, while loop
var newItem = {
name: 'new item',
Settings: [
{
name: 'color',
values: ['green', 'blue', 'red']
},
{
name: 'size',
values: ['15', '18', '22']
},
{
name: 'gender',
values: ['male', 'female']
}
]
};
var props = ["SettingName", "value"];
var settings = newItem.Settings;
function p(settings, props) {
var data = settings.reduce(function(res, setting, index) {
var name = setting.name;
var obj = {};
obj[name] = setting.values;
res.push(obj);
return res.length < index ? res : res.sort(function(a, b) {
return a[Object.keys(a)[0]].length - b[Object.keys(b)[0]].length
})
}, []);
var key = data.splice(0, 1)[0];
return [].concat.apply([], key[Object.keys(key)].map(function(value, index) {
return data.reduce(function(v, k) {
var keys = [v, k].map(function(obj) {
return Object.keys(obj)[0]
});
var i = Math.max.apply(Math, [v[keys[0]].length, k[keys[1]].length]);
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
var a = {};
a[props[0]] = keys[0];
a[props[1]] = v[keys[0]][++next];
var b = {};
b[props[0]] = keys[1];
b[props[1]] = k[keys[1]][next];
var c = {};
c[props[0]] = Object.keys(key)[0];
c[props[1]] = value;
arr.push([a, b, c]);
};
next = -1;
}
return arr
});
}));
}
document.querySelector("pre").textContent = JSON.stringify(
p(settings, props), null, 2
);
<pre></pre>
This question already has answers here:
Most efficient method to groupby on an array of objects
(58 answers)
How to merge two arrays in JavaScript and de-duplicate items
(89 answers)
Closed 7 months ago.
What's the correct way to merge two arrays in Javascript?
I've got two arrays (for example):
var a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
var a2 = [{ id : 1, count : "1"}, {id : 2, count : "2"}]
I want to be able to end up with something like:
var a3 = [{ id : 1, name : "test", count : "1"},
{ id : 2, name : "test2", count : "2"}]
Where the two arrays are being joined based on the 'id' field and extra data is simply being added.
I tried to use _.union to do this, but it simply overwrites the values from the second array into the first one
Short ES6 solution
const a3 = a1.map(t1 => ({...t1, ...a2.find(t2 => t2.id === t1.id)}))
This should do the trick:
var mergedList = _.map(a1, function(item){
return _.extend(item, _.findWhere(a2, { id: item.id }));
});
This assumes that the id of the second object in a1 should be 2 rather than "2"
Assuming IDs are strings and the order does not matter, you can
Create a hash table.
Iterate both arrays and store the data in the hash table, indexed by the ID. If there already is some data with that ID, update it with Object.assign (ES6, can be polyfilled).
Get an array with the values of the hash map.
var hash = Object.create(null);
a1.concat(a2).forEach(function(obj) {
hash[obj.id] = Object.assign(hash[obj.id] || {}, obj);
});
var a3 = Object.keys(hash).map(function(key) {
return hash[key];
});
In ECMAScript6, if the IDs are not necessarily strings, you can use Map:
var hash = new Map();
a1.concat(a2).forEach(function(obj) {
hash.set(obj.id, Object.assign(hash.get(obj.id) || {}, obj))
});
var a3 = Array.from(hash.values());
ES6 simplifies this:
let merge = (obj1, obj2) => ({...obj1, ...obj2});
Note that repeated keys will be merged, and the value of the second object will prevail and the repeated value of the first object will be ignored.
Example:
let obj1 = {id: 1, uniqueObj1Key: "uniqueKeyValueObj1", repeatedKey: "obj1Val"};
let obj2 = {id: 1, uniqueObj2Key: "uniqueKeyValueObj2", repeatedKey: "obj2Val"};
merge(obj1, obj2)
// {id: 1, uniqueObj1Key: "uniqueKeyValueObj1", repeatedKey: "obj2Val", uniqueObj2Key: "uniqueKeyValueObj2"}
merge(obj2, obj1)
// {id: 1, uniqueObj2Key: "uniqueKeyValueObj2", repeatedKey: "obj1Val", uniqueObj1Key: "uniqueKeyValueObj1"}
Complete solution (with Lodash, not Underscore)
var a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
var a2 = [{ id : 1, count : "1"}, {id : 2, count : "2"}]
var merge = (obj1, obj2) => ({...obj1, ...obj2});
_.zipWith(a1, a2, merge)
(2) [{…}, {…}]
0: {id: 1, name: "test", count: "1"}
1: {id: 2, name: "test2", count: "2"}
If you have an array of arrays to merge you can do it like this:
var arrayOfArraysToMerge = [a1, a2, a3, a4]; //a3 and a4 are arrays like a1 and a2 but with different properties and same IDs.
_.zipWith(...arrayOfArraysToMerge, merge)
(2) [{…}, {…}]
0: {id: 1, name: "test", count: "1", extra1: "val1", extra2: 1}
1: {id: 2, name: "test2", count: "2", extra1: "val2", extra2: 2}
reduce version.
var a3 = a1.concat(a2).reduce((acc, x) => {
acc[x.id] = Object.assign(acc[x.id] || {}, x);
return acc;
}, {});
_.values(a3);
I think it's common practice in functional language.
Already there are many great answers, I'll just add another one which is from a real problem I needed to solve yesterday.
I had an array of messages with user ids, and one array of users containing users' names and other details. This is how I managed to add user details to the messages.
var messages = [{userId: 2, content: "Salam"}, {userId: 5, content: "Hello"},{userId: 4, content: "Moi"}];
var users = [{id: 2, name: "Grace"}, {id: 4, name: "Janetta"},{id: 5, name: "Sara"}];
var messagesWithUserNames = messages.map((msg)=> {
var haveEqualId = (user) => user.id === msg.userId
var userWithEqualId= users.find(haveEqualId)
return Object.assign({}, msg, userWithEqualId)
})
console.log(messagesWithUserNames)
Vanilla JS solution
const a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
const a2 = [{ id : 1, count : "1"}, {id : 2, count : "2"}]
const merge = (arr1, arr2) => {
const temp = []
arr1.forEach(x => {
arr2.forEach(y => {
if (x.id === y.id) {
temp.push({ ...x, ...y })
}
})
})
return temp
}
console.log(merge(a1, a2))
The lodash implementaiton:
var merged = _.map(a1, function(item) {
return _.assign(item, _.find(a2, ['id', item.id]));
});
The result:
[
{
"id":1,
"name":"test",
"count":"1"
},
{
"id":2,
"name":"test2",
"count":"2"
}
]
Wanted to add this answer which is derived from #daisihi answer above. Main difference is that this uses the spread operator.
Also, at the end I remove the id because it was not desirable in the first place.
const a3 = [...a1, ...a2].reduce((acc, x) => {
acc[x.id] = {...acc[x.id] || {}, ...x};
return acc;
}, {});
This part was taken from another post. removing a property from a list of objects in an array
const newArray = Object.values(a3).map(({id, ...keepAttrs}) => keepAttrs);
Found other solutions failing for some cases, so writing a better one here
const a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
const a2 = [{ id : 3, count : "3"}, { id : 1, count : "1"}, {id : 2, count : "2"}]
const mergeHelper = new Map(a1.map(x => [x.id, x]));
for (const x of a2) {
if (mergeHelper.has(x.id)) {
const item = mergeHelper.get(x.id);
mergeHelper.set(x.id, {...item, ...x});
} else {
mergeHelper.set(x.id, x);
}
}
const mergedList = [...mergeHelper.values()];
// For sorted array
// const mergedSortedList = [...mergeHelper.values()].sort((a, b) => a.id - b.id);
console.log(mergedList)
Using js Map is way faster than other approaches, helps when array length is huge.
A working TypeScript version:
export default class Merge {
static byKey(a1: any[], a2: any[], key: string) {
const res = a1.concat(a2).reduce((acc, x) => {
acc[x[key]] = Object.assign(acc[x[key]] || {}, x);
return acc;
}, {});
return Object.entries(res).map(pair => {
const [, value] = pair;
return value;
});
}
}
test("Merge", async () => {
const a1 = [{ id: "1", value: "1" }, { id: "2", value: "2" }];
const a2 = [{ id: "2", value: "3" }];
expect(Merge.byKey(a1, a2, "id")).toStrictEqual([
{
id: "1",
value: "1"
},
{ id: "2", value: "3" }
]);
});
try this
var a1 = [{ id : 1, name : "test"}, { id : 2, name : "test2"}]
var a2 = [{ id : 1, count : "1"}, {id : 2, count : "2"}]
let arr3 = a1.map((item, i) => Object.assign({}, item, a2[i]));
console.log(arr3);
How about this?
const mergeArrayObjects = (arr1: any[], arr2: any[], mergeByKey: string): any[] => {
const updatedArr = [];
for (const obj of arr1) {
const arr1ValueInArr2 = arr2.find(
a => a[mergeByKey] === obj[mergeByKey],
);
if (arr1ValueInArr2) {
updatedArr.push(Object.assign(obj, arr1ValueInArr2));
} else {
updatedArr.push(obj);
}
}
const mergeByKeyValuesInArr1 = arr1.map(a => a[mergeByKey]);
const remainingObjInArr2 = arr2.filter(a => !mergeByKeyValuesInArr1.includes(a[mergeByKey]) )
return updatedArr.concat(remainingObjInArr2)
}
You can write a simple object merging function like this
function mergeObject(cake, icing) {
var icedCake = {}, ingredient;
for (ingredient in cake)
icedCake[ingredient] = cake[ingredient];
for (ingredient in icing)
icedCake[ingredient] = icing[ingredient];
return icedCake;
}
Next, you need to do use a double-loop to apply it to your data structre
var i, j, a3 = a1.slice();
for (i = 0; i < a2.length; ++i) // for each item in a2
for (j = 0; i < a3.length; ++i) // look at items in other array
if (a2[i]['id'] === a3[j]['id']) // if matching id
a3[j] = mergeObject(a3[j], a2[i]); // merge
You can also use mergeObject as a simple clone, too, by passing one parameter as an empty object.
const a3 = a1.map(it1 => {
it1.test = a2.find(it2 => it2.id === it1.id).test
return it1
})
If you have exactly the same number of items in both array with same ids you could do something like this.
const mergedArr = arr1.map((item, i) => {
if (item.ID === arr2[i].ID) {
return Object.assign({}, item, arr2[i]);
}
});
function mergeDiffs(Schedulearray1, Schedulearray2) {
var secondArrayIDs = Schedulearray2.map(x=> x.scheduleid);
return Schedulearray1.filter(x=> !secondArrayIDs.includes(x.scheduleid)).concat(Schedulearray2);
}
None of them worked for me. I wrote own:
const formatteddata=data.reduce((a1,a2)=>{
for (let t=0; t<a1.length; t++)
{var id1=a1[t].id
for (let tt=0; tt<a2.length; tt++)
{var id2=a2[tt].id
if(id1==date2)
{a1[t]={...a1[t],...a2[tt]}}
}
}
return a1
})
works with any amount of arrays of objects in arrays, with varying length and not always coinsciding dates