Allow obtaining the mean of a distribution

[haykam821]

Fixes #284

[codecov]

Codecov Report

Attention: Patch coverage is 91.66667% with 3 lines in your changes are missing coverage. Please review.

Project coverage is 82.98%. Comparing base (5e04cb9) to head (4580c3a).
Report is 7 commits behind head on main.

Files	Patch %	Lines
core/src/num/dist.rs	86.66%	2 Missing ⚠️
core/src/error.rs	0.00%	1 Missing ⚠️

Additional details and impacted files

@@            Coverage Diff             @@
##             main     #285      +/-   ##
==========================================
+ Coverage   82.89%   82.98%   +0.09%     
==========================================
  Files          52       52              
  Lines       14947    15000      +53     
==========================================
+ Hits        12390    12448      +58     
+ Misses       2557     2552       -5     

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[haykam821]

I'm not sure how the 'there must be at least one part in a dist' error can be produced; I assume it's a failsafe prevented by parsing (e.g. d0 is invalid). I assume I can ignore the missing coverage as a result.

[bgkillas]

probably too annoying to implement for the little performance difference in most scenarios(aspecially because of how poor the performance of dice is to begin with) but to get the mean for dice it can just be (dice.sum()+dice.len())/2 via some simple logic about how the distribution is in the center always, and wouldn't mode/median be the same as mean? unless there are non dice distributions then it makes sense

[haykam821]

I realize that I implemented the mean calculation incorrectly; the current formula should work for all currently supported distributions, but a skewed distribution will have an incorrect mean due to the formula not accounting for probabilities.

The correct calculation is more like the following, with no division at the end:

result = Exact::new(k, true)
	.mul(&Exact::new(Complex::from(Real::from(v)), true), int)?
	.add(result, int)?;

The big rational to real to complex to exact conversion is a bit verbose, so if there's a more direct way I can do this, I can use that method instead.

[printfn]

Thanks for the PR! It is actually already possible to generate skewed distributions with fend, for example:

> d6 / d2
0.5:  8.33%  ###############
  1: 16.67%  ##############################
1.5:  8.33%  ###############
  2: 16.67%  ##############################
2.5:  8.33%  ###############
  3: 16.67%  ##############################
  4:  8.33%  ###############
  5:  8.33%  ###############
  6:  8.33%  ###############
> mean(d6 / d2)
approx. 2.8333333333
> 0.5*1/12 + 1*2/12 + 1.5*1/12 + 2*2/12 + 2.5*1/12 + 3*2/12 + 4/12 + 5/12 + 6/12
2.625

Would you be able to fix this bug and add a test case for it?

I'm not sure how the 'there must be at least one part in a dist' error can be produced; I assume it's a failsafe prevented by parsing (e.g. d0 is invalid). I assume I can ignore the missing coverage as a result.

The code coverage checks are just to remind people to write the occasional test. There's no need to worry about small areas of missed coverage like that.

The big rational to real to complex to exact conversion is a bit verbose, so if there's a more direct way I can do this, I can use that method instead.

You're right, there's no shorter way to do that conversion atm unfortunately.

[haykam821]

Thank you! I had a feeling there was something I missed that would allow for skewed distributions. In that case, I've committed the fix along with test cases.