Reorder JSON by value, showing a,b,c,a,b,c - json

Question

I have some JSON that is in the below format. I want to have a script that rearranges this so that it's ordered by preference, but with alternating values. For instance, showing a,b,c,a,b,c. Can anyone help with this?
[
{
"name" : "Tim",
"preference" : "b"
},
{
"name" : "Tom",
"preference" : "b"
},
{
"name" : "Steve",
"preference" : "a"
},
{
"name" : "Rick",
"preference" : "a"
},
{
"name" : "Nile",
"preference" : "c"
},
{
"name" : "James",
"preference" : "c"
}
]

Answer 1

Underscore provides utilities suited to this task.
First, group your input by the preference field:
var groups = _.groupBy(input, 'preference');
Then convert this into an array of arrays:
var arrays = _.values(groups);
Then, "zip" the groups:
var result = _.zip.apply(null, arrays);
In one line:
var result = _.zip.apply(null, _.values(_.groupBy(input, 'preference')));
Non-Underscore version
If you can't/don't want to use Underscore, then you'll have to write your own versions of groupBy and zip:
function groupBy(array, prop) {
var result = {};
for (var i = 0; i < array.length; i++) {
var entry = array[i];
var val = entry[prop];
if (!result[val]) result[val] = [];
result[val].push(entry);
}
return result;
}
Either this or Underscore's _.groupBy will transform your input into
{
b: [ { name: 'Tim', preference: 'b' }, ... ],
a: [ { name: 'Rick', preference: 'a' }, ... ]
}
To get an array of the arrays:
function values(obj) {
return Object.keys(obj) . sort() . map(function(key) { return obj[key]; });
}
This will result in
[
[ { name: 'Tim', preference: 'b' }, ... ],
[ { name: 'Rick', preference: 'a' }, ... ]
]
Then for zip:
function zip(arrays) {
var result = [];
var n = 0;
var more = true;
var array;
while (more) {
more = false;
for (var i = 0; i < arrays.length; i++) {
array = arrays[i];
if (n < array.length) {
more = true;
result.push(array[n]);
}
}
n++;
}
return result;
}
Note: this implementation of zip takes an array of arrays as a parameter, unlike Underscore's version, which takes the arrays as individual parameters.
Then
zip(values(groupBy(input, 'preference')))