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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]);
What's the best way to convert this array of comma separated values
[ 'com--test,LFutx9mQbTTyRo4A9Re5ksjdnfsI4cKN4q2,on',
'com--fxtrimester,SEzlksdfMpW3FxkSbzL7eo5MmqkPczCl2,on',
'com--fxtrimester,LFutx9mQbTTyRoldksfns4A9Re5I4cKN4q2,on' ]
Into three arrays?
[ 'com--test', [ LFutx9mQbTTyRo4A9Re5Ilsdf4cKN4q2', ['on',
'com--fxtrimester', SEzMpW3FxkSbzL7eo5MmlkdfqkPczCl2', 'on',
'com--fxtrimester' ] LFutksdfx9mQbTTyRo4A9Re5I4cKN4q2 ] 'on']
I was trying something like:
var indexToSplit = unique.indexOf(',');
var status = unique.slice(3, indexToSplit - 1);
var use = unique.slice(2, indexToSplit - 2);
var pros = unique.slice(0, indexToSplit - 3);
console.log(pros);
But I figured that is wrong ... any help is appreciated!
You will have to loop over array and use string.split to get seperate parts.
Once you have seperate parts, you can push them to necessary array;
var d = [ 'com--test,LFutx9mQbTTyRo4A9Re5ksjdnfsI4cKN4q2,on',
'com--fxtrimester,SEzlksdfMpW3FxkSbzL7eo5MmqkPczCl2,on',
'com--fxtrimester,LFutx9mQbTTyRoldksfns4A9Re5I4cKN4q2,on'];
var result = [[],[],[]];
var len = 3;
d.forEach(function(str, i){
var _tmp = str.split(',');
for (var i = 0; i<len; i++){
result[i].push(_tmp[i])
}
})
console.log(result)
A little generic way.
Loop over data and split each string using comma(,)
Loop over split values and check if necessary array exists.
If not, initialise array, but you cannot do p[i] = [] as this will push to first value. You will have to also initialise all previous values. For this, you can use new Array(length). By default, if length is greater than 0, all indexes will be initialise to undefined.
Now push to necessary array. Position will be maintained.
var d = ['com--test,LFutx9mQbTTyRo4A9Re5ksjdnfsI4cKN4q2,on',
'com--fxtrimester,SEzlksdfMpW3FxkSbzL7eo5MmqkPczCl2,on',
'com--fxtrimester,LFutx9mQbTTyRoldksfns4A9Re5I4cKN4q2,on,test'
];
var result = d.reduce(function(p, c, index) {
var _tmp = c.split(',');
for (var i = 0; i < _tmp.length; i++) {
// Check if position not defined.
if (p[i] === undefined)
// Initialize array and add default (undefined) to all elements before current element
p[i] = new Array(index);
p[i].push(_tmp[i])
}
return p;
}, [])
console.log(result)
With map this becomes:
for positions X out of 0, 1 and 2:
convert each item in the list into an array, and choose the Xth item
var start = [ 'com--test,LFutx9mQbTTyRo4A9Re5ksjdnfsI4cKN4q2,on',
'com--fxtrimester,SEzlksdfMpW3FxkSbzL7eo5MmqkPczCl2,on',
'com--fxtrimester,LFutx9mQbTTyRoldksfns4A9Re5I4cKN4q2,on' ]
var out = [0,1,2].map(i =>
start.map(x => x.split(',')[i]) )
console.log(out)
Since your question does not ask for a more general case, i am safely assuming it for 3 array. We can use forEach function on array below code can be one amongst the possible solutions
var arr1 = [];
var arr2 = [];
var arr3 = [];
var x = ['com--test,LFutx9mQbTTyRo4A9Re5ksjdnfsI4cKN4q2,on', 'com--fxtrimester,SEzlksdfMpW3FxkSbzL7eo5MmqkPczCl2,on', 'com--fxtrimester,LFutx9mQbTTyRoldksfns4A9Re5I4cKN4q2,on']
x.forEach(function(data) {
var dataArray = data.split(',');
arr1.push(dataArray[0]);
arr2.push(dataArray[1]);
arr3.push(dataArray[2]);
});
console.log(arr1)
console.log(arr2)
console.log(arr3)
I have a question of JS arrays.
Example:
var fullArr = [1,2,3,4];
var partArr = [2,3];
var newArr = [];
We have a main array fullArr and a partial array partarr. I want to create a function/filter, which is looking for existing items in fullArr and not in partArr.
In this example above newArr must be equal to [1,4].
I've tried doing something like this, but it's not working properly.
for (var k in fullArray) { // [1,2,3,4]
for (var j in selectedArray) { // [1,4]
if (fullArray[k] == selectedArray[j]) {
newArray.splice(selectedArray[j] - 1, 1); // must be [2,3]
break;
}
}
}
What is a good way of making this? Thanks.
Here's one
var newArr = fullArr.filter(function(f) { // The filter() method creates a new array with all elements that pass the test implemented by the provided function.
return partArr.indexOf(f) == -1; // The indexOf() method returns the first index at which a given element can be found in the array, or -1 if it is not present.
})
to impress the girls, you can also
var newArr = fullArr.filter(function(f) {
return !~partArr.indexOf(f);
})
Here is the code for your requirement.
var fullArr = [1,2,3,4];
var partArr = [2,3];
var newArr = [];
for(var i=0;i<fullArr.length;i++){
if(partArr.indexOf(fullArr[i]) == -1)
newArr.push(fullArr[i]);
};
Here is the working Link
Hope it works :)
In fact, you want a common part between arrays. Obviously you can choose splice or indexOf to have O(n * m) or even O(m * n^2) performance. It's obviously suboptimal for any array larger than few elements
Or you can use objects as hash maps to find differences in (in worst case) O(n + m log m):
var fullArr = [1,2,3,4];
var partArr = [2,3];
var temporaryObject = Object.create(null);
partArr.forEach(el=>temporaryObject[el] = true); // temporaryObject after this operation is {"2": true, "3": true}
var newArr = fullArr.filter(el=>temporaryObject[el]);
In this example I have used ES6 feature called "arrow functions". It translates to following ES5 code:
var partArr = [2, 3];
var temporaryObject = Object.create(null);
partArr.forEach(function (el) {
temporaryObject[el] = true;
}); // temporaryObject after this operation is {"2": true, "3": true}
var newArr = fullArr.filter(function (el) {
return temporaryObject[el];
});
You can use the filter() function that works on arrays:
var newArr = fullArr.filter(function(val, i, arr) {
return partArr.indexOf(val) === -1;
});
This will return a new array containing the values of every iteration that returns true.
Should you ever need to do this on an object in the future a great way is to first convert the object keys to an array and then run the filter:
Object.keys(myObj).function(val, i, arr) {
return partArr.indexOf(val) === -1;
});
Here are few other approaches:
var fullArr = [1,2,3,4];
var partArr = [2,3];
var newArr = [];
1.
fullArr.map(function(element){
if(partArr.indexOf(element) === -1) newArr.push(element);
})
console.log(newArr);
2.
for(i in fullArr){
if(partArr.indexOf(fullArr[i]) === -1) newArr.push(fullArr[i]);
}
console.log(newArr);
3.
fullArr.forEach(function(element){
if(partArr.indexOf(element) === -1) newArr.push(element);
})
console.log(newArr);
I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?
You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})
This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.
a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.
Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.
The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4
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]);