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I have got a . (dot) separated string, from which I want to create nested JSON object. The length of the string is not fixed. For example,
var string = 'a.b.c.d';
Then my JSON object should be as following:
a: {
b: {
c:{
d: {
//Some properties here.
}
}
}
}
I've tried following code:
var packageName = "a.b.c.d"
var splitted = packageName.split('.');
var json = {};
for(var i=0;i<splitted.length-1;i++){
json[splitted[i]] = splitted[i+1];
}
But this returns
{
a: 'b',
b: 'c',
c: 'd'
}
But this is not what I want. I've also searched on google and found similar questions, but no solutions answer my problem. For example this.
A good use case for reduce
packageName = "a.b.c.d";
initProps = {hi: 'there'};
obj = packageName.split('.').reduceRight((o, x) => ({[x]: o}), initProps);
console.log(JSON.stringify(obj))
If you find loops easier to work with, a loop could be written concisely as
result = {};
ptr = result;
for (let prop of packageName.split('.'))
ptr = ptr[prop] = {};
You need to create a new object each time and attribute it to the last object created. And it goes until splitted.length, not splitted.length - 1, because you're using <, not <=.
var packageName = "a.b.c.d";
var splitted = packageName.split('.');
var json = {};
var current = json;
for (var i = 0; i < splitted.length; i++) {
current[splitted[i]] = {};
current = current[splitted[i]];
}
console.log(json);
You may use the last splittted part as property for some payload.
I suggest to keep the object reference and use a temporary variable for aceessing an creating a new property, if necessary.
Please avoid the use of JSON for not stringified objects.
var packageName = "a.b.c.d",
splitted = packageName.split('.'),
result = {},
temp = result,
i;
for (i = 0; i < splitted.length - 1; i++) {
temp[splitted[i]] = temp[splitted[i]] || {};
temp = temp[splitted[i]];
}
temp[splitted[i]] = { some: 'data' };
console.log(result);
Case: We have 'n' number of arrays stored in an array (Array of Arrays). Now that each child array in this parent array can have elements that may or may not be present in other child arrays. Output - I need to create an array which has the all the elements present in all the child arrays excluding the duplicates.
I do not want to concatenate all the arrays into a single array and use unique method to filter out. I need to create unique array then and there during iteration.
Ex:
var a[] = [1,2,3,4,5];
var b[] = [1,2,7,8];
var c[] = [1,2,3,4,5,6,7,8];
var d[] = [9,10,11,12];
var arr[] = [a,b,c,d]
Output must be [1,2,3,4,5,6,7,8,9,10,11,12]
P.S: I can concat the arrays and use jquery unique function to resolve this, but i need a solution in javascript alone. Thanks
You can use array#reduce to flatten your array and then use Set to get distinct values and use array#from to get back array from Set.
var a = [1,2,3,4,5];
var b = [1,2,7,8];
var c = [1,2,3,4,5,6,7,8];
var d = [9,10,11,12];
var arr = [a,b,c,d]
var result = Array.from(new Set(arr.reduce((r,a) => r.concat(a))));
console.log(result);
Try using .filter when adding each array to the final one, filtering out the duplicates:
a.filter(function(item) {
return !finalArray.contains(item));
});
Answer using Sets:
var a = [1,2,3,4,5];
var b = [1,2,7,8];
var c = [1,2,3,4,5,6,7,8];
var d = [9,10,11,12];
var concat = a.concat(b).concat(c).concat(d);
var union = new Set(concat);
//console.log(union);
ES6 Answer:
let a = new Set([1,2,3,4,5]);
let b = new Set([1,2,7,8]);
let c = new Set([1,2,3,4,5,6,7,8]);
let d = new Set([9,10,11,12]);
let arr = new Set([...a,...b,...c,...d]);
//Result in arr.
Whats going on???
From https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set:
The Set object lets you store unique values of any type, whether
primitive values or object references.
So when we initialise Sets passing arrays to the constructor we basically ensure that there are no duplicate values.
Then in the last line, we concat all the Sets we initialised prior into a final set.
The ... notation converts the Set into an array, and when we pass the 4 arrays to the constructor of the Set they get concatenated and a Set of their unique values is created.
Here is a functional alternative written in ES5.
var flatten = function(list) {
return list.reduce(function(acc, next) {
return acc.concat(Array.isArray(next) ? flatten(next) : next);
}, []);
};
var unique = function(list) {
return list.filter(function(element, index) {
return list.indexOf(element) === index;
})
}
var a = [1,2,3,4,5];
var b = [1,2,7,8];
var c = [1,2,3,4,5,6,7,8];
var d = [9,10,11,12];
var arr = [a,b,c,d];
var result = unique(flatten(arr));
console.log(result);
If you support ES6, arrow function can make that code even shorter.
Here is a solution that uses a plain object for resolving duplicates, and only uses basic ES3 JavaScript. Runs in IE 5.5 and higher, and with O(n) time complexity.
function uniques(arr) {
var obj = {}, result = [];
for (var i = 0; i < arr.length; i++) {
obj[arr[i]] = true;
}
for (var prop in obj) {
if (obj.hasOwnProperty(prop)) result.push(+prop);
}
return result;
}
// Example use
var a = [1,2,3,4,5],
b = [1,2,7,8],
c = [1,2,3,4,5,6,7,8],
d = [9,10,11,12];
var result = uniques(a.concat(b, c, d));
console.log('Result: ' + result);
As an object can only have a unique set of properties (no duplicates), the use of all array values as properties in an object will give you an object with a property for each unique value. This happens in the first loop. NB: the value given to those properties is not relevant; I have used true.
Then the result is just the conversion of those properties back to array values. This happens in the second loop.
var a = [1,2,3,4,5];
var b = [1,2,7,8];
var c = [1,2,3,4,5,6,7,8];
var d = [9,10,11,12];
var result = a.concat(b,c,d);
function remvDup(result){
var tmp = [];
for(var i = 0; i < result.length; i++){
if(tmp.indexOf(result[i]) == -1){
tmp.push(result[i]);
}
}
return tmp;
}
console.log(remvDup(result));
Becuase the OP mentioned that he cannot use 'Set' as it is not supported on the targeted browsers, I would recommand using the 'union' function from the lodash library.
See union's documentation here
My goal is to create an array like this:
[{"str":"a","number":1},{"str":"a","number":2},{"str":"b","number":1},{"str":"b","number":2}]
so I wrote this javascript
abc = ["a","b"]
num = [1,2]
arr = []
a = {}
for (var i in abc)
{
str = abc[i]
a.str = str;
for(var x in num)
{
number = num[x]
a.number = number
console.log(a)
arr.push(a)
}
}
the console log looks fine, but the array looks like this:
[{"str":"b","number":2},{"str":"b","number":2},{"str":"b","number":2},{"str":"b","number":2}]
Can anyone could explain this?
This is happening because you are actually working with a reference to the same object, thus modifying the same over and over.
To fix it you must declare a new object in every iteration you want to use a different one.
Try something like this:
var abc = ["a", "b"];
var num = [1, 2];
var arr = [];
for (var i in abc) {
for (var x in num) {
var a = {};
a.str = abc[i];
a.number = num[x];
arr.push(a);
}
}
console.log(arr);
Also, don't forget to declare your variables with var or let and end your statements with ;.
As said in the comments, you’ve pushed your a object to arr many times, instead of adding four separate objects. To fix this issue, you could declare a in the for (var x in num) loop, every time as a new object (using const or let). But I’ve simplified it further, see the code below.
To iterate through JavaScript arrays, you should use .forEach method.
let abc = ['a', 'b'];
let num = [1, 2];
let arr = [];
abc.forEach(letter => {
num.forEach(number => {
arr.push({number: number, str: letter});
});
});
abc = ["a","b"]
num = [1,2]
arr = []
for (var i in abc)
{
for(var x in num)
{
a = {} ---------------- Reset "a"
str = abc[i] --------------------- 1
a.str = str; --------------------- 2
number = num[x]
a.number = number
console.log(a)
arr.push(a)
}
}
console.log(arr)
// Move 1 and 2 inside the second loop
Using map :
let tempArray = abc.map((e,i) => { return num.map((ee,ii) => { return {"str": e, "number": ee }; } ) });
$.merge(tempArray[0], tempArray[1]);
I want to filter out values for an array by passing another array to the filter function.
x = [1,2,3];
y = [2,3];
var n = x.filter(filterByArray);
function filterByArray(element, index, array, myOtherArray){
// some other code
});
What is the best way to pass "y" to the "myOtherArray" prototype in the function?
You can use the second parameter of .filter(callback[, thisArg]) to set its this value to something "useful" such as your second array
function filterByArray(element, index, array) {
return this.lookup.indexOf(element) > -1; // this.lookup == y
};
var x = [1,2,3],
y = [2,3];
var result = x.filter(filterByArray, {lookup: y});
console.log(result);
fiddle
You cannot change the signature of the callback but you can have a separate class that takes the other array as parameter:
function MyFilter(otherArray) {
this.otherArray = otherArray;
}
MyFilter.prototype.filterByArray = function(element, index, array) {
// you can use this.otherArray here
};
and then:
x = [1,2,3];
y = [2,3];
var myFilter = new MyFilter(y);
var n = x.filter(myFilter.filterByArray);
I have an array of items as follows in Javascript:
var users = Array();
users[562] = 'testuser3';
users[16] = 'testuser6';
users[834] = 'testuser1';
users[823] = 'testuser4';
users[23] = 'testuser2';
users[917] = 'testuser5';
I need to sort that array to get the following output:
users[834] = 'testuser1';
users[23] = 'testuser2';
users[562] = 'testuser3';
users[823] = 'testuser4';
users[917] = 'testuser5';
users[16] = 'testuser6';
Notice how it is sorted by the value of the array and the value-to-index association is maintained after the array is sorted (that is critical). I have looked for a solution to this, tried making it, but have hit a wall.
By the way, I am aware that this is technically not an array since that would mean the indices are always iterating 0 through n where n+1 is the counting number proceeding n. However you define it, the requirement for the project is still the same. Also, if it makes a difference, I am NOT using jquery.
The order of the elements of an array is defined by the index. So even if you specify the values in a different order, the values will always be stored in the order of their indices and undefined indices are undefined:
> var arr = [];
> arr[2] = 2;
> arr[0] = 0;
> arr
[0, undefined, 2]
Now if you want to store the pair of index and value, you will need a different data structure, maybe an array of array like this:
var arr = [
[562, 'testuser3'],
[16, 'testuser6'],
[834, 'testuser1'],
[823, 'testuser4'],
[23, 'testuser2'],
[917, 'testuser5']
];
This can be sorted with this comparison function:
function cmp(a, b) {
return a[1].localeCompare(b[1]);
}
arr.sort(cmp);
The result is this array:
[
[834, 'testuser1'],
[23, 'testuser2'],
[562, 'testuser3'],
[823, 'testuser4'],
[917, 'testuser5'],
[16, 'testuser6']
]
If I understand the question correctly, you're using arrays in a way they are not intended to be used. In fact, the initialization style
// Don't do this!
var array = new Array();
array[0] = 'value';
array[1] = 'value';
array[2] = 'value';
teaches wrong things about the nature and purpose of arrays. An array is an ordered list of items, indexed from zero up. The right way to create an array is with an array literal:
var array = [
'value',
'value',
'value'
]
The indexes are implied based on the order the items are specified. Creating an array and setting users[562] = 'testuser3' implies that there are at least 562 other users in the list, and that you have a reason for only knowing the 563rd at this time.
In your case, the index is data, and is does not represent the order of the items in the set. What you're looking for is a map or dictionary, represented in JavaScript by a plain object:
var users = {
562: 'testuser3',
16: 'testuser6',
834: 'testuser1',
823: 'testuser4',
23: 'testuser2',
917: 'testuser5'
}
Now your set does not have an order, but does have meaningful keys. From here, you can follow galambalazs's advice to create an array of the object's keys:
var userOrder;
if (typeof Object.keys === 'function') {
userOrder = Object.keys(users);
} else {
for (var key in users) {
userOrder.push(key);
}
}
…then sort it:
userOrder.sort(function(a, b){
return users[a].localeCompare(users[b]);
});
Here's a demo
You can't order arrays like this in Javascript. Your best bet is to make a map for order.
order = new Array();
order[0] = 562;
order[1] = 16;
order[2] = 834;
order[3] = 823;
order[4] = 23;
order[5] = 917;
In this way, you can have any order you want independently of the keys in the original array.
To sort your array use a custom sorting function.
order.sort( function(a, b) {
if ( users[a] < users[b] ) return -1;
else if ( users[a] > users[b] ) return 1;
else return 0;
});
for ( var i = 0; i < order.length; i++ ) {
// users[ order[i] ]
}
[Demo]
Using the ideas from the comments, I came up with the following solution. The naturalSort function is something I found on google and I modified it to sort a multidimensional array. Basically, I made the users array a multidimensional array with the first index being the user id and the second index being the user name. So:
users[0][0] = 72;
users[0][1] = 'testuser4';
users[1][0] = 91;
users[1][1] = 'testuser2';
users[2][0] = 12;
users[2][1] = 'testuser8';
users[3][0] = 3;
users[3][1] = 'testuser1';
users[4][0] = 18;
users[4][1] = 'testuser7';
users[5][0] = 47;
users[5][1] = 'testuser3';
users[6][0] = 16;
users[6][1] = 'testuser6';
users[7][0] = 20;
users[7][1] = 'testuser5';
I then sorted the array to get the following output:
users_sorted[0][0] = 3;
users_sorted[0][1] = 'testuser1';
users_sorted[1][0] = 91;
users_sorted[1][1] = 'testuser2';
users_sorted[2][0] = 47;
users_sorted[2][1] = 'testuser3';
users_sorted[3][0] = 72;
users_sorted[3][1] = 'testuser4';
users_sorted[4][0] = 20;
users_sorted[4][1] = 'testuser5';
users_sorted[5][0] = 16;
users_sorted[5][1] = 'testuser6';
users_sorted[6][0] = 18;
users_sorted[6][1] = 'testuser7';
users_sorted[7][0] = 12;
users_sorted[7][1] = 'testuser8';
The code to do this is below:
function naturalSort(a, b) // Function to natural-case insensitive sort multidimensional arrays by second index
{
// setup temp-scope variables for comparison evauluation
var re = /(-?[0-9\.]+)/g,
x = a[1].toString().toLowerCase() || '',
y = b[1].toString().toLowerCase() || '',
nC = String.fromCharCode(0),
xN = x.replace( re, nC + '$1' + nC ).split(nC),
yN = y.replace( re, nC + '$1' + nC ).split(nC),
xD = (new Date(x)).getTime(),
yD = xD ? (new Date(y)).getTime() : null;
// natural sorting of dates
if ( yD )
if ( xD < yD ) return -1;
else if ( xD > yD ) return 1;
// natural sorting through split numeric strings and default strings
for( var cLoc = 0, numS = Math.max(xN.length, yN.length); cLoc < numS; cLoc++ ) {
oFxNcL = parseFloat(xN[cLoc]) || xN[cLoc];
oFyNcL = parseFloat(yN[cLoc]) || yN[cLoc];
if (oFxNcL < oFyNcL) return -1;
else if (oFxNcL > oFyNcL) return 1;
}
return 0;
}
// Set values for index
var users = Array();
var temp = Array();
users.push(Array('72', 'testuser4'));
users.push(Array('91', 'testuser2'));
users.push(Array('12', 'testuser8'));
users.push(Array('3', 'testuser1'));
users.push(Array('18', 'testuser7'));
users.push(Array('47', 'testuser3'));
users.push(Array('16', 'testuser6'));
users.push(Array('20', 'testuser5'));
// Sort the array
var users_sorted = Array();
users_sorted = users.sort(naturalSort);
I'd use map once to make a new array of users,
then a second time to return the string you want from the new array.
var users= [];
users[562]= 'testuser3';
users[16]= 'testuser6';
users[834]= 'testuser1';
users[823]= 'testuser4';
users[23]= 'testuser2';
users[917]= 'testuser5';
var u2= [];
users.map(function(itm, i){
if(itm){
var n= parseInt(itm.substring(8), 10);
u2[n]= i;
}
});
u2.map(function(itm, i){
return 'users['+itm+']= testuser'+i;
}).join('\n');
/*returned value: (String)
users[834]= testuser1
users[23]= testuser2
users[562]= testuser3
users[823]= testuser4
users[917]= testuser5
users[16]= testuser6
*/
If you want to avoid any gaps. use a simple filter on the output-
u2.map(function(itm, i){
return 'users['+itm+']= testuser'+i;
}).filter(function(itm){return itm}).join('\n');
Sparse arrays usually spell trouble. You're better off saving key-value pairs in an array as objects (this technique is also valid JSON):
users = [{
"562": "testuser3"
},{
"16": "testuser6"
}, {
"834": "testuser1"
}, {
"823": "testuser4"
}, {
"23": "testuser2"
}, {
"917": "testuser5"
}];
As suggested, you can use a for loop to map the sorting function onto the array.
Array.prototype.sort() takes an optional custom comparison function -- so if you dump all of your users into an array in this manner [ [562, "testuser3"], [16, "testuser6"] ... etc.]
Then sort this array with the following function:
function(comparatorA, comparatorB) {
var userA = comparatorA[1], userB = comparatorB[1]
if (userA > userB) return 1;
if (userA < userB) return -1;
if (userA === userB) return 0;
}
Then rebuild your users object. (Which will loose you your sorting.) Or, keep the data in the newly sorted array of arrays, if that will work for your application.
A oneliner with array of array as a result:
For sorting by Key.
let usersMap = users.map((item, i) => [i, item]).sort((a, b) => a[0] - b[0]);
For sorting by Value. (works with primitive types)
let usersMap = users.map((item, i) => [i, item]).sort((a, b) => a[1] - b[1]);