Help. Finding max value in array of objects.

Situation:
>
I have an array of objects. I want to find the maximum value of a
given numeric value that exists in each of the objects. Suppose the
array of objects is called "stat" and the numeric value I want to
check is called "member".
>
Obviously, a for loop would do it:
>
var maximum=stat.member[0];
for (i=1;i<stat.length;i++) {
if (stat.member[i]>maximum) {
maximum=stat.member[i];
}
}
>
But is there a better way? I know how to bend Math.max to my will to
have is find the max value of an array. Could it be forced into
servitude to solve this case? Or perhaps there's a relatively clean
way to use sort here?
>
Ideally, I'd want to have a function that takes the name of the
element I want to check.
>
maximum=getMax(stat,"member");
>
Looking for a clear, compact solution. Clever is optional.

timothytoe <timothy...@gmail.comwrites:>>>>>>>>
If you have javascript 1.8 (probably not, but see below):
>
var max = stat.reduce( function(a,b) { return a b ? a : b } );
>
For older versions: there's a definition of Array.prototype.reduce at
>
http://developer.mozilla.org/en/docs...5_Reference:Ob...
>
Joost.

Oh. That's cool. Probably only worth it if I would want to use reduce
in more places than one.

Situation:
>
I have an array of objects. I want to find the maximum value of a
given numeric value that exists in each of the objects. Suppose the
array of objects is called "stat" and the numeric value I want to
check is called "member".
>
Obviously, a for loop would do it:
>
var maximum=stat.member[0];
for (i=1;i<stat.length;i++) {
if (stat.member[i]>maximum) {
maximum=stat.member[i];
}
}

But is there a better way? I know how to bend Math.max to my will to
have is find the max value of an array. Could it be forced into
servitude to solve this case? Or perhaps there's a relatively clean
way to use sort here?
>
Ideally, I'd want to have a function that takes the name of the
element I want to check.
>
maximum=getMax(stat,"member");
>
Looking for a clear, compact solution. Clever is optional.

maximum= array.sort()[array.length-1]
Mick

>
I don't quite get that. You're ignoring "member." I'm not just looking
for the max in an array. I'm looking for the maximum of an element in
an object in an array.
>
For that matter, Joost ignored it too. Neither solution works as is.

>
>
I don't quite get that. You're ignoring "member." I'm not just looking
for the max in an array. I'm looking for the maximum of an element in
an object in an array.
>
For that matter, Joost ignored it too. Neither solution works as is.

timothytoe wrote:>>>
array == stat.member, no?
maximum= stat.member.sort()[stat.member.length-1]
>
or I'm missing something.
Mick

Situation:
>
I have an array of objects. I want to find the maximum value of a
given numeric value that exists in each of the objects. Suppose the
array of objects is called "stat" and the numeric value I want to
check is called "member".

alert(Math.min.apply(null, stat)); // =20

On Feb 6, 11:10 pm, timothytoe <timothy...@gmail.comwrote:
>>>
var stat = [{ member: -1 }, { member: 1 },
{ member: 20 }, { member: 5 }];
>
var toString = Object.prototype.toString;
>
Object.prototype.toString = function () {
return this.member;
>
};
>
alert(Math.min.apply(null, stat)); // =20
>
Object.prototype.toString = toString;

Yes. It makes no difference. The important point is that it is not the
browser(s) that "you" have (you, JD, were replying to the OP, who is
presumed to be an author) have that matters ; it is the browsers that
the author's readers have. AISB.

The distinction between author's system and readers' systems is too
often forgotten.

It would be useful if the FAQ were to list, or explicitly link to, a
chronological table showing when the major browsers were released, and
which versions of JavaScript they came with.

On Feb 6, 10:15 pm, AKS <aksus...@yandex.ruwrote:>
Ah. Good. I get it.

>
>However, I think that iterating through the array is in most cases more
>efficient than sorting it and retrieving the last element. The former
>algorithm has a complexity of O(n) in the Worst Case, even Heapsort has
>O(n * log(n)) then. Meaning it is already slower with 3 or more unsorted items.