I'm iterating over array couples and I need to sort one by the order of the other.
Say I have these two arrays:
aLinks = [4,5,6]
bLinks = [1,2,3,4,5,6]
I need to return:
aLinks = [4,5,6]
bLinks = [4,5,6,1,2,3]
meaning that i need to have the items that match first array first and than the rest,
sorted by order if possible.
I'm working with d3 so I'm using forEach to go through the link sets and save the order of aLinks.
I don't know how to apply this order to bLinks
var linkOrder = [];
linkSets.forEach(function(set, i) {
linkOrder = [];
set.aLinks.forEach(function(link,i){
linkOrder.push(link.path);
})
});
You can do it like:
Take out the matching items from second array into a temp array
Sort the temp array
Sort the second array containing only items that did not match
Concatenate the second array into the temp array
Code - With the fix provided by User: basilikum
var first = [4,5,6];
var second = [1,7,3,4,6,5,6];
var temp = [], i = 0, p = -1;
// numerical comparator
function compare(a, b) { return a - b; }
// take out matching items from second array into a temp array
for(i=0; i<first.length; i++) {
while ((p = second.indexOf(first[i])) !== -1) {
temp.push(first[i]);
second.splice(p, 1);
}
}
// sort both arrays
temp.sort(compare);
second.sort(compare);
// concat
temp = temp.concat(second);
console.log(temp);
Working Demo: http://jsfiddle.net/kHhFQ/
You end up with A + sort(A-B) - so you just need to compute the difference between the 2 arrays. Using some underscore convenience methods for example:
var A = [4,5,6];
var B = [1,2,3,4,5,6];
var diff = _.difference(A,B);
var result = _.flattern(A, diff.sort());
iterate the first array, removing the values from the second array and then appending them to the start of the array to get the right order :
var arr1 = [4,5,6];
var arr2 = [1,2,3,4,6,5];
arr1.sort(function(a,b) {return a-b;});
for (i=arr1.length; i--;) {
arr2.splice(arr2.indexOf(arr1[i]), 1);
arr2.unshift( arr1[i] );
}
FIDDLE
Related
how to push more than one element at one index of a array in javascript?
like i have
arr1["2018-05-20","2018-05-21"];
arr2[5,4];
i want resulted 4th array to be like:
arr4[["2018-05-20",5],["2018-05-21",4]];
tried pushing like this:
arr1.push("2018-05-20","2018-05-21");
arr1.push(5,4);
and then finally as:
arr4.push(arr1);
But the result is not as expected. Please someone help.
Actually i want to use this in zingChart as :
Options Data
Create an options object, and add a values array of arrays.
Calendar Values
In each array, provide the calendar dates with corresponding number values in the following format.
options: {
values: [
['YYYY-MM-DD', val1],
['YYYY-MM-DD', val2],
...,
['YYYY-MM-DD', valN]
]
}
Your question is not correct at all, since you cannot push more than one element at the same index of an array. Your result is a multidimensional array:
[["2018-05-20",5],["2018-05-21",4]]
You have to create a multidimensional array collecting all your data (arrAll)
Then you create another multidimensional array (arrNew) re-arranging previous data
Try the following:
// Your Arrays
var arr1 = ["2018-05-20","2018-05-21"];
var arr2 = [5, 4];
//var arr3 = [100, 20];
var arrAll = [arr1, arr2];
//var arrAll = [arr1, arr2, arr3];
// New Array definition
var arrNew = new Array;
for (var j = 0; j < arr1.length; j++) {
var arrTemp = new Array
for (var i = 0; i < arrAll.length; i++) {
arrTemp[i] = arrAll[i][j];
if (i === arrAll.length - 1) {
arrNew.push(arrTemp)
}
}
}
//New Array
Logger.log(arrNew)
Assuming the you want a multidimensional array, you can put all the input variables into an array. Use reduce and forEach to group the array based on index.
let arr1 = ["2018-05-20","2018-05-21"];
let arr2 = [5,4];
let arr4 = [arr1, arr2].reduce((c, v) => {
v.forEach((o, i) => {
c[i] = c[i] || [];
c[i].push(o);
});
return c;
}, []);
console.log(arr4);
I've searched SO for a way to do this but most questions only support two arrays (I need a solution for multiple arrays).
I don't want to compare exact objects, I want to compare objects by their ID, as their other parameters may differ.
So here's the example data:
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}, etc.]
data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}, etc.]
data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}, etc.]
...
I'd like to return all objects whose ID appears in all arrays, and I don't mind which of the set of matched objects is returned.
So, from the data above, I'd expect to return any one of the following:
{'id':'22','name':'andrew'}
{'id':'22','name':'mary'}
{'id':'22','name':'john'}
Thanks
First, you really need an array of arrays - using a numeric suffix is not extensible:
let data = [ data1, data2, ... ];
Since you've confirmed that the IDs are unique within each sub array, you can simplify the problem by merging the arrays, and then finding out which elements occur n times, where n is the original number of sub arrays:
let flattened = data.reduce((a, b) => a.concat(b), []);
let counts = flattened.reduce(
(map, { id }) => map.set(id, (map.get(id) || 0) + 1), new Map()
);
and then you can pick out those objects that did appear n times, in this simple version they'll all come from the first sub array:
let found = data[0].filter(({ id }) => counts.get(id) === data.length);
Picking an arbitrary (unique) match from each sub array would be somewhat difficult, although picking just one row of data and picking the items from that would be relatively easy. Either would satisfy the constraint from the question.
If you want the unique object by Name
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'mary'}]
data2 = [{'id':'26','name':'mary'},{'id':'85','name':'bill'}]
data3 = [{'id':'29','name':'sophie'},{'id':'22','name':'john'}]
flattened = [ ...data1, ...data2, ...data3 ];
counts = flattened.reduce(
(map, { name }) => map.set(name, (map.get(name) || 0) + 1), new Map()
);
names = []
found = flattened.filter(({ name }) => {
if ((counts.get(name) > 1) && (!names.includes(name))) {
names.push(name);
return true
}
return false
});
its too many loops but , if u can find the common id which is present in all the arrays then it would make your finding easier i think .you can have one array value as reference to find the common id
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}];
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}];
var data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}];
var arrays = [data1, data2, data3];
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);
There's mention you you need n arrays, but also, given that you can:
put all the arrays into an array called data
you can:
combine your arrays
get a list of duplicated IDs (via sort by ID)
make that list unique (unique list of IDs)
find entries in the combined list that match the unique IDs
where the count of those items match the original number of arrays
Sample code:
// Original data
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}]
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}]
var data3 = [{'id':'13','name':'steve'},{'id':'22','name':'john'}]
var arraycount = 3;
// Combine data into a single array
// This might be done by .pushing to an array of arrays and then using .length
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort?v=control
var data = [].concat(data1).concat(data2).concat(data3);
//console.log(data)
// Sort array by ID
// http://stackoverflow.com/questions/840781/easiest-way-to-find-duplicate-values-in-a-javascript-array
var sorted_arr = data.slice().sort(function(a, b) {
return a.id - b.id;
});
//console.log(sorted_arr)
// Find duplicate IDs
var duplicate_arr = [];
for (var i = 0; i < data.length - 1; i++) {
if (sorted_arr[i + 1].id == sorted_arr[i].id) {
duplicate_arr.push(sorted_arr[i].id);
}
}
// Find unique IDs
// http://stackoverflow.com/questions/1960473/unique-values-in-an-array
var unique = duplicate_arr.filter(function(value, index, self) {
return self.indexOf(value) === index;
});
//console.log(unique);
// Get values back from data
//https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter?v=control
var matches = [];
for (var i = 0; i < unique.length; ++i) {
var id = unique[i];
matches.push(data.filter(function(e) {
return e.id == id;
}))
}
//console.log(matches)
// for data set this will be 13 and 22
// Where they match all the arrays
var result = matches.filter(function(value, index, self) {
return value.length == arraycount;
})
//console.log("Result:")
console.log(result)
Note: There's very likely to be more efficient methods.. I've left this in the hope part of it might help someone
var arr1 = ["558", "s1", "10"];
var arr2 = ["55", "s1", "103"];
var arr3 = ["55", "s1", "104"];
var arr = [arr1, arr2, arr3];
console.log(arr.reduce((p, c) => p.filter(e => c.includes(e))));
// output ["s1"]
I am trying to build an array that should look like this :
[
[{"name":"Mercury","index":0}],
[{"name":"Mercury","index":1},{"name":"Venus","index":1}],
[{"name":"Mercury","index":2},{"name":"Venus","index":2},{"name":"Earth","index":2}],
...
]
Each element is the concatenation of the previous and a new object, and all the indexes get updated to the latest value (e.g. Mercury's index is 0, then 1, etc.).
I have tried to build this array using the following code :
var b = [];
var buffer = [];
var names = ["Mercury","Venus","Earth"]
for (k=0;k<3;k++){
// This array is necessary because with real data there are multiple elements for each k
var a = [{"name":names[k],"index":0}];
buffer = buffer.concat(a);
// This is where the index of all the elements currently in the
// buffer (should) get(s) updated to the current k
for (n=0;n<buffer.length;n++){
buffer[n].index = k;
}
// Add the buffer to the final array
b.push(buffer);
}
console.log(b);
The final array (b) printed out to the console has the right number of objects in each element, but all the indexes everywhere are equal to the last value of k (2).
I don't understand why this is happening, and don't know how to fix it.
This is happening because every object in the inner array is actually the exact same object as the one stored in the previous outer array's entries - you're only storing references to the object, not copies. When you update the index in the object you're updating it everywhere.
To resolve this, you need to create new objects in each inner iteration, or use an object copying function such as ES6's Object.assign, jQuery's $.extend or Underscore's _.clone.
Here's a version that uses the first approach, and also uses two nested .map calls to produce both the inner (variable length) arrays and the outer array:
var names = ["Mercury","Venus","Earth"];
var b = names.map(function(_, index, a) {
return a.slice(0, index + 1).map(function(name) {
return {name: name, index: index};
});
});
or in ES6:
var names = ["Mercury","Venus","Earth"];
var b = names.map((_, index, a) => a.slice(0, index + 1).map(name => ({name, index})));
Try this:
var names = ["Mercury","Venus","Earth"];
var result = [];
for (var i=0; i<names.length; i++){
var _temp = [];
for(var j=0; j<=i; j++){
_temp.push({
name: names[j],
index:i
});
}
result.push(_temp);
}
console.log(result)
try this simple script:
var b = [];
var names = ["Mercury","Venus","Earth"];
for(var pos = 0; pos < names.length; pos++) {
var current = [];
for(var x = 0; x < pos+1; x++) {
current.push({"name": names[x], "index": pos});
}
b.push(current);
}
I have a problem sorting an array. I am not the smartest concerning these sort algorithms.
The array should have following structure:
var arr = [
[week, IssuesPriority1, IssuesPriority2, IssuesPriority3],
[week, IssuesPriority1, IssuesPriority2, IssuesPriority3],
[week, IssuesPriority1, IssuesPriority2, IssuesPriority3],
...
];
So for each week there is a number of issues for the priority very high, high, medium.
The string that needs to be parsed in this structure is following:
var string =
"26|3|1,27|6|1,28|7|1,29|2|1,30|2|1,31|2|1,32|2|1,33|3|1,
35|1|1,34|2|1,36|0|1,37|0|1,38|1|1,26|11|2,27|10|2,28|9|2,
29|13|2,30|10|2,31|8|2,32|10|2,33|12|2,34|14|2,35|11|2,
36|11|2,37|12|2,38|14|2,27|17|3,26|13|3,29|26|3,28|21|3,30|25|3,
31|20|3,34|30|3,32|18|3,33|25|3,35|33|3,36|28|3,38|28|3,37|27|3";
var arr = string.split(",");
for(var i = 0; i < arr.length; i++){
var currentArr = arr[i].split("|");
var week = currentArr[0];
var issues = currentArr[1];
var priority = currentArr[2];
}
I have a lack of ideas sorting it in the desired way. Can you help me?
I don't think you want any sorting at all. You are looking for grouping!
var arr = string.split(",");
var weeks = {};
for (var i = 0; i < arr.length; i++) {
var currentArr = arr[i].split("|");
var week = currentArr[0];
var issue = currentArr[1];
var priority = currentArr[2];
if (!(week in weeks))
weeks[week] = {1:[], 2:[], 3:[]};
// if the number of issues levels were unknown,
// you'd start with an empty object instead
// and create the arrays dynamically in a manner similar to the weeks
weeks[week][priority].push(issue);
}
return Object.keys(weeks).map(function(week) {
return [week, weeks[week][1], weeks[week][2], weeks[week][3]];
});
(to get the result ordered by week number, add a sort(function(a,b){return a-b}) before the .map() call)
In your situation I would recommand to put the values in the array first. In the second step I would sort the array using the sort method.
function getSortedArrayByString(myString) {
var arraySplittedString, i, tmpValueArray, tmpInnerArray, resultingArray;
arraySplittedString = myString.split(",");
resultingArray = [];
for(i = 0; i < arraySplittedString.length; i++){
// tmpArray has the format of [<week>, <IssuesPriority1>, <IssuesPriority2>]
tmpValueArray = arraySplittedString[i].split("|");
// Push it in the new array.
resultingArray.push(tmpValueArray);
}
// Sort array by weeks ascending.
resultingArray.sort( function (a, b) {
return a[0] - b[0];
});
return resultingArray;
}
Running fiddle.
If you also want to sort by the count of issues, you simply can customize the inner sort function.
With this solution all values are saved as strings. You can convert them by using the parseInt function.
I have thousands of legacy code that stores array information in a non array.
For example:
container.object1 = someobject;
container.object2 = someotherobject;
container.object3 = anotherone;
What I want to have is:
container.objects[1], container.objects[2], container.objects[3] etc.
The 'object' part of the name is constant. The number part is the position it should be in the array.
How do I do this?
Assuming that object1, object2, etc... are sequential (like an array), then you can just iterate through the container object and find all the sequential objectN properties that exist and add them to an array and stop the loop when one is missing.
container.objects = []; // init empty array
var i = 1;
while (container["object" + i]) {
container.objects.push(container["object" + i]);
i++;
}
If you want the first item object1 to be in the [1] spot instead of the more typical [0] spot in the array, then you need to put an empty object into the array's zeroth slot to start with since your example doesn't have an object0 item.
container.objects = [{}]; // init array with first item empty as an empty object
var i = 1;
while (container["object" + i]) {
container.objects.push(container["object" + i]);
i++;
}
An alternate way to do this is by using keys.
var unsorted = objectwithobjects;
var keys = Object.keys(unsorted);
var items = [];
for (var j=0; j < keys.length; j++) {
items[j] = unsorted[keys[j]];
}
You can add an if-statement to check if a key contains 'object' and only add an element to your entry in that case (if 'objectwithobjects' contains other keys you don't want).
That is pretty easy:
var c = { objects: [] };
for (var o in container) {
var n = o.match(/^object(\d+)$/);
if (n) c.objects[n[1]] = container[o];
}
Now c is your new container object, where c.object[1] == container.object1