I am trying to sort my 2D array according to column 3.
when I sort it with sort function, all the array members become duplicates of one member of the original array.
so for example;
my original array:
[12, AAA, eee, 5]
[58, BBB, zzz, 3]
[28, CCC, ddd, 6]
[18, DDD, fff, 9]
I want it to become :
[18, DDD, fff, 9]
[28, CCC, ddd, 6]
[12, AAA, eee, 5]
[58, BBB, zzz, 3]
I use the code :
function sortByColumn(a, colIndex){
a.sort(sortFunctionq);
function sortFunctionq(a, b) {
if (a[colIndex] === b[colIndex]) {
return 0;
}
else {
return (a[colIndex] > b[colIndex]) ? -1 : 1;
}
}
return a;
}
var sorted_a = new Array(15);
sorted_a = sortByColumn(arr, 3);
now the array becomes:
[18, DDD, fff, 9]
[18, DDD, fff, 9]
[18, DDD, fff, 9]
[18, DDD, fff, 9]
I am using javascript on Samsung Gear watch. maybe it does not support the "sort" function correctly.
is there a way to sort 2D array without using sort function ?
final code is:
var sorted_a = new Array(15);
sorted_a = sortByColumn(arrx, 3);
arrx= sorted_a;
function bubbleSort(a, fCompare) {
if( a.length < 2) {return a;}
for( var length = a.length-1; length; --length) {
var noSwaps = true;
var temp;
for(var c=0; c<length; ++c) {
if( fCompare( a[c], a[c+1]) > 0) {
temp = a[c+1];
a[c+1] = a[c];
a[c] = temp;
noSwaps = false;
}
}
if( noSwaps) {break;}
}
}
function sortByColumn(a, colIndex){
function sortFunctionq(a, b) {
if (a[colIndex] === b[colIndex]) {
return 0;
}
else {
return (a[colIndex] > b[colIndex]) ? -1 : 1;
}
}
//return bubbleSort(a, sortFunctionq);
return bubbleSort(a.slice(), sortFunctionq);
}
but now nothing is available in the array.
for those who ask:
if I remove the sort function and use arrx as it is, I can reach 2D array elements with arrx[1][1]
but with the code above arrx[1][1] returns null.
I changed it a bit and now it seems to work.
But now I need to remove duplicates as well. How can I do that ?
current code:
var arrx = new Array(50);
for (var j = 0; j<50; j++){
arrx[j] = arr[j].split("|+");
}
var arry = new Array(50);
arry = bubbleSort(arrx);
function bubbleSort(a) {
for( var r = 49; r >= 0; --r) {
var noSwaps = true;
var temp = new Array(50);
for(var c=0; c<r; ++c) {
if (a[c][3] < a[c+1][3]) {
temp = a[c+1];
a[c+1] = a[c];
a[c] = temp;
noSwaps = false;
}
}
if( noSwaps) {break;}
}
return a;
}
The direct answer to you question is "yes, it is possible to sort an array without using the array's sort method". A simple example using bubble sort:
function bubbleSort(a, fCompare) {
if( a.length < 2)
return a;
for( var length = a.length-1; length; --length) {
var noSwaps = true;
var temp;
for( i=0; i<length; ++i) {
if( fCompare( a[i], a[i+1]) > 0) {
temp = a[i+1];
a[i+1] = a[i];
a[i] = temp;
noSwaps = false;
}
}
if( noSwaps)
break;
}
}
function sortByColumn(a, colIndex){
function sortFunctionq(a, b) {
if (a[colIndex] === b[colIndex]) {
return 0;
}
else {
return (a[colIndex] > b[colIndex]) ? -1 : 1;
}
}
return bubbleSort(a, sortFunctionq);
}
var a = [
[12, 'AAA', 'eee', 5],
[58, 'BBB', 'zzz', 3],
[28, 'CCC', 'ddd', 6],
[18, 'DDD', 'fff', 9],
];
var sortedA = sortByColumn(a,2) // updates a in-place, as well
console.log( JSON.stringify(sortedA))
However
Note that both the sort method of an array and the bubbleSort above change the order of elements in the array being sorted, without creating a shallow copy of the array. While bubbleSort may show the Samsung JS engine has a problem, more than likely it will not and produce the same result.
Because sorting sorts the array in place, you may wish to check if creating a shallow copy of it before sorting solves the problem. EG by replacing the return statement in the example with
return a.slice().sort(functionq) // OR
return bubbleSort(a.slice(), functionq)
Debugging notes:
JavaScript arrays are objects. The value of an object variable is a reference of some kind used by the JavaScript engine to access object properties. The reference could be a memory pointer or some other value used by the engine to access object data. When you assign an object to a variable, its existing content is overwritten. If you assign the same object value to two variables, they hold the same refernce value and refer to the same set of object data.
var arry = new Array(50);
arry = bubbleSort(arrx);
unnecessarily creates a new Array, because the new Array value is overwritten in the second line. It can be simplified as
var arry = bubbleSort( arrx).
Note that JavaScript arrays can grow and shrink and do not have some pre-allocated length.
Both the bubble sort code and the inbuilt array sort method (inherited by an array instance from the Array.prototype object, and documented on MDN under Array.prototype.sort) sort the array in place and return an object reference to the array being sorted. After
arry = bubbleSort(arrx); // OR
arry = arrx.sort(sortFunction)
the value of arry is the same as arrx. If you want to make a copy of the array that is immune to arrx modificatioms of first dimension values, make a shallow copy of the input array before sorting:
arry = bubbleSort(arrx.slice());
If you want to make a copy that is immune to modifications to either dimension value then make a shallow copy of both dimensions' arrays as for example:
arry = bubbleSort( arrx.map(element => element.slice())
This creates new arrays from both dimensions of arrx before sorting.
If you are still getting duplicate entries after this you will need to find out where in the code the duplicates are being assigned.
Tip
Check there are no typos in conditional tests that use = (the assignment operator) instead of == or === operators to test for equality. This is a good way of inadvertently assigning values to something that was not meant to be changed.
All you need to change is to return the sorted array.
const data = [
[12, 'AAA', 'eee', 5],
[58, 'BBB', 'zzz', 3],
[28, 'CCC', 'ddd', 6],
[18, 'DDD', 'fff', 9]
];
function sortByColumn(a, colIndex){
function sortFunctionq(a, b) {
if (a[colIndex] === b[colIndex]) {
return 0;
}
else {
return (a[colIndex] > b[colIndex]) ? -1 : 1;
}
}
return a.sort(sortFunctionq);
//^^^^^^^^^^^^^^^^^^^^^^^^^^^
}
const result = sortByColumn(data, 3);
console.log(result);
Related
I have 10 different arrays. Each array has different numbers.
array1 = [1,2,3,4,5]
array2 = [6,7,8,9,10]
...
array 10 = [51,52,53,54]
let's say I pass in 7. Then I want to know which array it is from and want to return array number. So in this case it is going to be 2.
Should I write a switch statement for each array? Appreciate it in javascript.
try:
var arrays = [array1, array2, ..., array10];
for(var i=0; i<arrays.length; ++i) {
if (arrays[i].indexOf(value) != -1) {
console.log('found in array' + (i+1));
}
}
You cannot directly retrieve the name of array.The reason is this variable is only storing a reference to the object.
Instead you can have a key inside the same array which represent its name. Then indexOf can be used to find the array which contain the number , & if it is so, then get the array name
var array1 = [1,2,3,4,5];
array1.name ="array1";
var array2 = [6,7,8,9,10];
array2.name ="array2";
var array10 = [51,52,53,54]
array10.name ="array10";
var parArray = [array1,array2,array10]
function _getArrayName(number){
for(var o=0;o<parArray.length;o++){
var _tem = parArray[o];
if(parArray[o].indexOf(number) !==-1){
console.log(parArray[o].name);
}
}
}
_getArrayName(6) //prints array2
jsfiddle
One fast method should be using hash tables or as i would like to call LUT. Accordingly this job boils down to a single liner as follows;
var arrs = {
arr1 : [1,2,3,4,5],
arr2 : [6,7,8,9,10],
arr3 : [12,14,16,17],
arr4 : [21,23,24,25,27,20],
arr5 : [31,34,35,39],
arr6 : [45,46,44],
arr7 : [58,59],
arr8 : [66,67,69,61],
arr9 : [72,73,75,79,71],
arr0 : [81,85,98,99,90,80]
},
lut = Object.keys(arrs).reduce((p,c) => {arrs[c].forEach(n => p[n]=c); return p},{}),
findar = n => lut[n];
document.write("<pre>" + findar(12) + "</pre>");
One way to do this is have the arrays in an object and iterate over the keys/values. This method doesn't presume the arrays (and therefore their names) are in sequential order.
Note: this will always return a the first match from the function and terminate the search.
var obj = {
array1: [1, 2, 3, 4, 5],
array2: [6, 7, 8, 9, 10],
array3: [51, 52, 53, 54],
array4: [51, 52, 53, 54, 7]
}
function finder(obj, test) {
var keys = Object.keys(obj);
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (obj[key].indexOf(test) > -1) {
return key.match(/\d+/)[0];
}
}
return false;
}
finder(obj, 7); // '2'
DEMO
If you want to find all instances of a value in all arrays the function needs to be altered slightly.
function finder(obj, test) {
var keys = Object.keys(obj);
var out = [];
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (obj[key].indexOf(test) > -1) {
out.push(key.match(/\d+/)[0]);
}
}
return out;
}
finder(obj, 7); // ['2', '4']
DEMO
I am trying to get this function to split an array into subsets. each subset is to have numbers that are equal to the previous or within 1 from the previous number.
The example I have below should return two subsets but it returns {0, 1, 2, 3} instead. Any idea on what I am doing wrong? Also, is there a better way to dynamically create an array for each new subset? Thanks
function max_tickets() {
var arr = [4, 13, 2, 3];
var myarr = arr.sort(function(a, b){return a-b});
for(var i = 0; i<myarr.length; i++){
var iplus = i+1;
if(i === i || i === iplus){
newArr= [];
newArr.push(i);
}else if (i !== i || i !== iplus){
arr2 =[];
arr2.push(i);
}
}
}
What you are trying to do is usually called "partitioning". The generic version of the problem is to partition an array into sub-arrays using some "rule", or predicate, or condition, which specifies which partition a particular element is supposed to go into, or specifies that it should go into a new partition.
The pseudo code for doing this would be:
To partition an array:
Initialize the resulting array
For each element in the array
If that element starts a new chunk
Create a new empty chunk in the resulting array
Add the element to the most recent chunk
Return the result
This can be expressed in JS quite straightforwardly as
function partition(array, fn) {
return array.reduce((result, elt, i, a) => {
if (!i || !fn(elt, i, a)) result.push([]);
result[result.length - 1].push(elt);
return result;
}, []);
}
Now we need to write the function saying when a new partition should start:
// Is the element within one of the previous element?
function close(e, i, a) {
return Math.abs(e - a[i-1]) > 1;
}
We can now partition the array with
partition([[4, 13, 2, 3], close)
This should work.
function max_tickets() {
var arr = [4, 13, 2, 3];
var myarr = arr.sort(function (a, b) { return a - b });
arrSubsets = [];
arr1 = [];
for (var i = 0; i < myarr.length; i++) {
if (myarr[i - 1] === undefined) {
arr1.push(myarr[i]);
continue;
}
if (myarr[i] - myarr[i - 1] <= 1) {
arr1.push(myarr[i]);
}
else {
arrSubsets.push(arr1);
arr1 = [];
arr1.push(myarr[i]);
}
}
if (arr1.length > 0)
arrSubsets.push(arr1);
}
max_tickets();
Based on your questions:
Any idea on what I am doing wrong?.
Inside of your loop you are using i as if it is the value of the array, but the loop goes from 0 to the value of myarr.length in your particular case 4, so that makes the value of i to be 0, 1, 2, 3.
As you can see you are using the values of the index to compare, instead of using the values of the array in order to use the values of the array you must specify the arrayname[index], in your case myarr[i] that will give you the values: 4, 13, 2, 3.
Also, is there a better way to dynamically create an array for each new subset?
Yes you can create an array inside of another array dynamically inside of a loop:
var b = [];
for(var i = 0; i < 10; i++){
b.push(['I am' + i, i]);
}
As you can see in the previous example I'm creating an array inside of the b array so once the loop finishes the b array will have 10 arrays inside of it with 2 elements each.
I want to 'reduce' the array to only max values for each x (or index 0) value in a JavaScript multidimensional array.
My Array is as follows:
var mulitple = [["1/2013", 1],
["1/2013", 5],
["1/2013", 7],
["1/2013", 6],
["1/2013", 5],
["2/2013", 7],
["2/2013", 10],
["2/2013", 10],
["3/2013", 7],
["3/2013", 10],
["3/2013", 10],
["4/2013", 1],
["4/2013", 5],
["4/2013", 7],
["4/2013", 6],
["4/2013", 5],
["5/2013", 7]];
So the final result should be as follows:
[["1/2013", 7],
["2/2013", 10],
["3/2013", 10],
["4/2013", 7],
["5/2013", 7]];
How can I achieve this in JavaScript.
EDIT:
Aww man who voted my question down.
Anyhow, this is what I have come up with.
var max = 0;
var newarray = [];
for (var i = 1; i < mulitple.length; i++) {
if (mulitple[i - 1][0] == mulitple[i][0]) {
if (mulitple[i - 1][1] > max) {
max = mulitple[i - 1][1];
}
}
else {
if (mulitple[i - 1][1] > max) {
max = mulitple[i - 1][1];
}
newarray.push([mulitple[i - 1][0], max]);
max = 0;
}
}
newarray.push([mulitple[mulitple.length - 1][0], max]);
The problem that I am having is that I can't get that last value (for the lone record) to get in the array. This was my result after I ran the code above.
[["1/2013", 7], ["2/2013", 10], ["3/2013", 10], ["4/2013", 7], ["5/2013", 0]]
This assumes that original array is already sorted. If not, you will have to write additional function to sort out.
function findMaximums(data) {
var out = [], maximums = {}, order = new Set;
data.reduce(function(acc, pair) {
if (
// Accumulator has value and it's lower than current
(acc[pair[0]] && acc[pair[0]][1] < pair[1]) ||
// Accumulator doesn't have value
!acc[pair[0]]
) {
acc[pair[0]] = pair; // Store maximum in accumulator
order.add(pair[0]) // Store order in set
}
return acc;
}, maximums);
order.forEach(function(key) {
out.push(maximums[key]); // Populate out with maximums by order
});
return out;
}
findMaximums(multiple);
/*[
[
"1/2013",
7
],
[
"2/2013",
10
],
[
"3/2013",
10
],
[
"4/2013",
7
],
[
"5/2013",
7
]
]*/
Update 1: same, but without Set.
function findMaximums(data) {
var order = [];
var maximums = data.reduce(function(acc, pair) {
if (
// Accumulator has value and it's lower than current
(acc[pair[0]] && acc[pair[0]][2] < pair[1]) ||
// Accumulator doesn't have value
!acc[pair[0]]
) {
// Store maximum
acc[pair[0]] = pair;
// Store order
if (order.indexOf(pair[0]) === -1) {
order.push(pair[0])
}
}
return acc;
}, {});
return order.map(function(key) {
return maximums[key]; // Populate out with maximums by order
});
}
Update 2: Shorter version.
function findMaximums(data) {
return data.filter(function(p1, i1) {
return !data.some(function(p2, i2) {
return p1[0] === p2[0] && ( (p1[1] < p2[1]) || (p1[1] === p2[1] && i1 > i2) );
});
});
}
In this version I let pair to remain in output if there are no other pairs in input data that:
Have same month.
Have bigger value.
or
Have same value, but occur earlier than tested pair. This prevents duplicates.
Read here more about used Array methods: filter, some.
Assuming the array as defined in the original question, which is sorted to have each grouping together...
Completely untested code:
var reduced = [];
var groupName = '';
var groupMax;
var groupIndex;
var l = multiple.length; // Grab the array length only once
for (var i = 0; i < l; i++){
// Current Group Name doesn't match last Group Name
if (multiple[i][0] !== groupName) {
// Last Group Name is not empty (it's not the first Group)
if (groupName !== '') {
// Assume groupIndex has been set and grab the item
reduced.push(multiple[groupIndex]);
}
// Grab the new Group Name and set the initial Max and Index
groupName = multiple[i][0];
groupMax = multiple[i][1];
groupIndex = i;
}
// Current Value is bigger than last captured Group Max
if (multiple[i][1] > groupMax) {
// Grab the new Group Max and the current Index
groupMax = multiple[i][1];
groupIndex = i;
}
}
// Grab the last index, since there's no new group after the last item
reduced.push(multiple[groupIndex]);
There could be some syntax or logic errors. I didn't actually run this code, but I think the concept is correct.
Here's a tested version using a map to collect all the unique values, then the output is sorted by month/year and is independent of the order of the input data. This also works in all browsers (IE6+).
Working demo: http://jsfiddle.net/jfriend00/dk1tc73s/
function findLargest(list) {
var map = {}, output = [], item, key, val, current;
for (var i = 0; i < list.length; i++) {
item = list[i];
key = item[0];
val = item[1];
current = map[key];
if (current) {
// this date is in the map, so make sure to save the largest
// value for this date
if (val > current) {
map[key] = val;
}
} else {
// this date is not yet in the map, so add it
map[key] = val;
}
}
// map contains all the largest values, output it to an array
// the map is not in a guaranteed order
for (var key in map) {
output.push([key, map[key]])
}
// actually parse to numbers in sort so the comparison
// works properly on number strings of different lengths
function parseDate(str) {
var split = str.split("/");
return {m: +split[0], y: +split[1]};
}
// now sort the output
output.sort(function(t1, t2) {
var diffYear, a, b;
a = parseDate(t1[0]);
b = parseDate(t2[0]);
diffYear = a.y - b.y;
if (diffYear !== 0) {
return diffYear;
} else {
return a.m - b.m;
}
});
return output;
}
for hours i've been trying to figure out how to sort 2 array dependently.
Let's say I have 2 arrays.
First one:
array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
and the second one:
array2 = [3, 7, 1];
I sort the first one with array1.sort(); and it becomes [aaaaaa, cccccc, zzzzzz]
now what I want is that the second one becomes [7, 1, 3]
I think it's quite simple but i'm trying to implement this in something a little more complex, im new and i keep mixing up things.
Thanks
I would "zip" them into one array of objects, then sort that with a custom sort callback, then "unzip" them back into the two arrays you wanted:
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'],
array2 = [3, 7, 1],
zipped = [],
i;
for(i=0; i<array1.length; ++i) {
zipped.push({
array1elem: array1[i],
array2elem: array2[i]
});
}
zipped.sort(function(left, right) {
var leftArray1elem = left.array1elem,
rightArray1elem = right.array1elem;
return leftArray1elem === rightArray1elem ? 0 : (leftArray1elem < rightArray1elem ? -1 : 1);
});
array1 = [];
array2 = [];
for(i=0; i<zipped.length; ++i) {
array1.push(zipped[i].array1elem);
array2.push(zipped[i].array2elem);
}
alert('Sorted arrays:\n\narray1: ' + array1 + '\n\narray2: ' + array2);
Here's a working fiddle.
Here's a simple function that will do the trick:
function sortTogether(array1, array2) {
var merged = [];
for(var i=0; i<array1.length; i++) { merged.push({'a1': array1[i], 'a2': array2[i]}); }
merged.sort(function(o1, o2) { return ((o1.a1 < o2.a1) ? -1 : ((o1.a1 == o2.a1) ? 0 : 1)); });
for(var i=0; i<merged.length; i++) { array1[i] = merged[i].a1; array2[i] = merged[i].a2; }
}
Usage demo (fiddle here):
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
var array2 = [3, 7, 1];
console.log('Before..: ',array1,array2);
sortTogether(array1, array2); // simply call the function
console.log('After...: ',array1,array2);
Output:
Before..: ["zzzzz", "aaaaaa", "ccccc"] [3, 7, 1]
After...: ["aaaaaa", "ccccc", "zzzzz"] [7, 1, 3]
Instead of two arrays of primitive types (strings, numbers) you can make an array of objects where one property of the object is string (containing "aaaaa", "cccccc", "zzzzzz") and another is number (7,1,3). This way you will have one array only, which you can sort by any property and the other property will remain in sync.
It just so happens I had some old code lying around that might do the trick:
function arrVirtualSortGetIndices(array,fnCompare){
var index=array.map(function(e,i,a){return i;});
fnCompare=fnCompare || defaultStringCompare;
var idxCompare=function (aa,bb){return fnCompare(array[aa],array[bb]);};
index.sort(idxCompare);
return index;
function defaultStringCompare(aa,bb){
if(aa<bb)return -1;
if(bb<aa)return 1;
return 0;
}
function defaultNumericalCompare(aa,bb){
return aa-bb;
}
}
function arrReorderByIndices(array,indices){
return array.map(
function(el,ix,ar){
return ar[indices[ix]];
}
);
}
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
var array2 = [3, 7, 1];
var indices=arrVirtualSortGetIndices(array1);
var array2sorted=arrReorderByIndices(array2,indices);
array2sorted;
/*
7,1,3
*/
Sorry, I don't do 'fors'. At least not when I don't have to.
And fiddle.
Also, an alternative fiddle that sorts the results when given an array of objects like this:
given:
var list = [
{str:'zzzzz',value:3},
{str:'aaaaa',value:7},
{str:'ccccc',value:1}
];
outputs:
[
{str: "aaaaa", value: 7},
{str: "ccccc", value: 1},
{str: "zzzzz", value: 3}
]
Assumption:
The arrays are the same length (this is implied by your question)
the contents can be compared with > and < (true in your example, but I wanted to make it clear that it was assumed here)
So then we can use an insertion sort.
var value,len = array1.length;
for (i=0; i < len; i++) {
value = array1[i];
for (j=i-1; j > -1 && array1[j] > value; j--) {
array1[j+1] = array1[j];
array2[j+1] = array2[j];
}
items[j+1] = value;
}
Using a solution found here to find the new indices after sorting an array, you can apply those indices to array2 like so.
function sortWithIndices(toSort) {
for (var i = 0; i < toSort.length; i++) {
toSort[i] = [toSort[i], i];
}
toSort.sort(function(left, right) {
return left[0] < right[0] ? -1 : 1;
});
toSort.sortIndices = [];
for (var j = 0; j < toSort.length; j++) {
toSort.sortIndices.push(toSort[j][2]);
toSort[j] = toSort[j][0];
}
return toSort;
}
var array1 = ['zzzz', 'aaaa', 'cccc'];
var array2 = [3, 7, 1];
// calculate the indices of array1 after sorting. (attached to array1.sortIndices)
sortWithIndices(array1);
// the final array after applying the sorted indices from array1 to array2
var final = [];
// apply sorted indices to array2
for(var i = 0; i < array1.sortIndices.length; i++)
final[i] = array2[array1.sortIndices[i]];
// output results
alert(final.join(","));
JSFiddle Demo
I wanted to change the rows into columns of an array.
[
[1],
[1,2],
[1,2,3],
[4,2,3],
[4,5,3],
[4,5,6]
]
to
[
[1,1,1,4,4,4],
[2,2,2,5,5],
[3,3,3,6]
]
I tried
var res = [];
for(i in this.fields) {
for(j in this.fields[i].value) {
if(i === 0) res[j] = [];
res[j][i] = this.fields[i].value[j];
}
}
this gives me empty set.
Create this function:
function transpose(arr) {
return Object.keys(arr[0]).map(function (c) {
return arr.map(function (r) {
return r[c];
});
});
}
and then:
var transposedArray = transpose(originalArray);
What you're asking looks a little weird because you have different lengths and you're ignoring undefined values, but it is still achievable.
Don't use for..in loops for Array, use a normal for. Also, you'll need to know how many items you'll have in your new parent Array, which is the max of the lengths of the original child Arrays.
var arrR = [ // will refer to "down" and "across" as in this literal
[1],
[1, 2],
[1, 2, 3],
[4, 2, 3],
[4, 5, 3],
[4, 5, 6]
];
function r2c(arr) {
var arrC = [], // next get the longest sub-array length
x = Math.max.apply(Math, arr.map(function (e) {return e.length;})),
y = arr.length,
i, j;
for (i = 0; i < x; ++i) { // this is the loop "down"
arrC[i] = [];
for (j = 0; j < y; ++j) // and this is the loop "across"
if (i in arr[j])
arrC[i].push(arr[j][i]);
}
return arrC;
}
var arrC = r2c(arrR);
/* [
[1, 1, 1, 4, 4, 4],
[2, 2, 2, 5, 5],
[3, 3, 3, 6]
] */
You should still consider if you're happy with [[1], [1, 2], [1]] becoming [[1, 1, 1], [2]], which I would consider unexpected (the position of 2 is completely lost), but seems to be what you intend.
Similar to Pauls but doesn't need to get the max length first:
function transpose(arr) {
// Loop over arrays as long as one has values
// Arrays should be contiguous, may fail if sparse
for (var result = [], i=0, more; more; i++) {
more = false;
// Get the ith element of each array (if there is one)
for (var j=0, jLen=arr.length; j<jLen; j++) {
// Don't add missing members
if (arr[j].hasOwnProperty(i)) {
// Add array for result if not already there
result[i] = result[i] || [];
// Do transpose
result[i][j] = arr[j][i];
// Only keep going while there is data
more = true;
}
}
}
return result;
}
BTW, a fixed version of your original function is:
function transpose2(fields) {
// Make sure the result array is initialised
var res = [];
// Don't forget to keep counters local - declare them
// I've removed *this* as it's a plain function, use it if
// it's an instance method
for(var i in fields) {
// Values are read directly, there is no "value" accessor
for(var j in fields[i]) {
// Don't rely on order of enumeration - may not start at 0
if(!res[j]) res[j] = [];
// Do the transpose
res[j][i] = fields[i][j];
}
}
return res;
}
But as noted above, for..in is not liked for arrays, particularly as there are many libraries that extend built-ins like Array.prototype so you will traverse those properties too. But if you're cool with that, this is a good way to deal with sparse arrays. You can add a hasOwnProperty test to avoid inherited enumerables.
Note also that the order of enumeration isn't necessarily from '0' or in any particular order, hence changed way of initialising res[j].