Help with some cstimer thing

[Discomantis]

Is there any way to show average of best X of Y on cstimer?

Such as your average of your best 2 solves in 3 solves?

 

[Davian]

@qq280833822

 

[qq280833822]

At present, there is no such feature, can you explain what is this value for?

 

[Discomantis]

At present, there is no such feature, can you explain what is this value for?

I forgot to mention this at the start, but Y has to be greater than X.




So, let's say you want to compare how well your first 50 solves of an ao100 were compared to your second half. Or maybe you want to see your best progress towards a session personal best ao12 after 10 solves. Then you could use this feature.


Here is why it works:

I'll be using the first example I mentioned above for why it works. (The one with an ao100.)

If you wanted to see how well you did for the first half of that ao100, you need to assume that the other half would have been done perfectly.

So, if you did perfectly for the rest of the ao100, your counting times would be:

Your 45 best current session times
A time of 0.01 x 45



Now if you did that for the second half ao50 as well , you would have the same exact thing. In order to compare the two using this metric, you can just compare the average of your best 45, since the 0.01's will change the average by the exact same amount. Whichever has the lower number was the better one.

When this is crunched down into one sentence, you would say it as your average of your best 50 of 100, or once shortened would look something like this: aob50o100.


(Average of best 50 of 100)






Another way that this could be used but in a slightly different but likely more useful way is that you could see who is winning in a cstimer online battle. You could just implement the average that you care about that you haven't been able to calculate (because your number of total solves it too few) and behind the scenes the calculation would do this:

Average of best X of Y

In which Y is the number of solves already done

And X is Y minus ((the average you care about times 0.2) rounded up to the nearest whole number)

or in simpler terms, X is Y minus the number of solves are removed at either side for the average that you care about.




It's a bit complicated, poorly worded, and long, but I tried to explain it as best as I could. I'm not a coder, so I don't know how hard this will be to implement. Thanks for reading if you've read this far.

 

[qq280833822]

When calculating the average, the current strategy is to remove 5% of solves at each end (best/worst) by default. Do you mean to implement another average calculation strategy, keeping the best 5% and then removing the worst X solves?

 

[Discomantis]

When calculating the average, the current strategy is to remove 5% of solves at each end (best/worst) by default. Do you mean to implement another average calculation strategy, keeping the best 5% and then removing the worst X solves?

Yes. Sorry for not making that more clear.


It is useful so you can find out the average of what your counting times would be if you do the rest of the average perfectly.

Using that information, you can compare results or find out how well you have done in your first half of an ao100 compared to your second half.

You could also use it to see your X best solves in the session (Not best as in your best times, but best as in terms of your best potential of getting a really good average.)


You could also use it to see how well you are doing compared to a friend if you are doing some sort of 1v1 battle.

Sorry for not responding right away. It took me a while to find a clear way to phrase this.