I have an array of objects. I want to find the "maximum" of this array based on a function that returns whichever object is bigger when given 2 objects.
function comparison(first, second) {
// ... arbitrary comparison based on properties...
return first; // or second
}
var a = [obj1, obj2, obj3];
var maxObj = ????(comparison);
What do I fill in here? What's elegant and short?
Something like this should be quicker than sort (depending on the data):
/*
values: array of values to test.
fn: function that takes two arguements and returns true if the first is bigger.
*/
var maximum = function(values, fn) {
var currentValue, maxValue = values.pop();
while(values.length)
maxValue = fn(maxValue, currentValue = values.pop()) ? maxValue : currentValue;
return maxValue;
}
Examples: http://jsfiddle.net/SaBJ4/2/
Even better, use Array.reduce:
var a = ['abc', 'defg', 'highlkasd', 'ac', 'asdh'];
a.reduce(function(a, b) { return a.length > b.length ? a : b; }); // highlkasd
reduce but you'll need to special-case empty arrays and arrays with only one element (depending on the behavior of comparison), using an explicit initial value can help.What's wrong with the obvious approach?
for(var i = 0, max; i < a.length; ++i)
max = typeof max == 'undefined' ? a[i] : comparison(a[i], max);
Wrap that up however you like.
Or you can take advantage of the fact that a = []; x = a[0] leaves you with undefined in x and do it RobG's way:
for(var i = 1, max = a[0]; i < a.length; ++i)
max = comparison(a[i], max);
That nicely avoids a bunch of typeof operators and comparisons that you really don't need.
max as a[0] and i as 1 and avoid the ternary expression?
[obj,obj,obj].sort(comparison)
// aka
var sorted = [obj,obj,obj].sort(function(a,b){
// return 1/0/-1
});
Then either pop the top or bottom element off (however you're sorting) to get the "max" object.
