Sort a Javascript Array by frequency and then filter repeats
Compute the frequency of each item first.
{
apples: 1,
oranges: 4,
bananas: 2
}
Then create an array from this frequency object which will also remove the duplicates.
["apples", "oranges", "bananas"]
Now sort this array in descending order using the frequency map we created earlier.
function compareFrequency(a, b) {
return frequency[b] - frequency[a];
}
array.sort(compareFrequency);
Here's the entire source (using the newly introduced Array functions in ECMA 5) and combining the de-duplication and frequency map generation steps,
function sortByFrequency(array) {
var frequency = {};
array.forEach(function(value) { frequency[value] = 0; });
var uniques = array.filter(function(value) {
return ++frequency[value] == 1;
});
return uniques.sort(function(a, b) {
return frequency[b] - frequency[a];
});
}
Same as above using the regular array iteration.
function sortByFrequencyAndRemoveDuplicates(array) {
var frequency = {}, value;
// compute frequencies of each value
for(var i = 0; i < array.length; i++) {
value = array[i];
if(value in frequency) {
frequency[value]++;
}
else {
frequency[value] = 1;
}
}
// make array from the frequency object to de-duplicate
var uniques = [];
for(value in frequency) {
uniques.push(value);
}
// sort the uniques array in descending order by frequency
function compareFrequency(a, b) {
return frequency[b] - frequency[a];
}
return uniques.sort(compareFrequency);
}
// returns most frequent to least frequent
Array.prototype.byCount= function(){
var itm, a= [], L= this.length, o= {};
for(var i= 0; i<L; i++){
itm= this[i];
if(!itm) continue;
if(o[itm]== undefined) o[itm]= 1;
else ++o[itm];
}
for(var p in o) a[a.length]= p;
return a.sort(function(a, b){
return o[b]-o[a];
});
}
//test
var A= ["apples","oranges","oranges","oranges","bananas","bananas","oranges"];
A.byCount()
/* returned value: (Array) oranges,bananas,apples */