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Re: Tips for writing correct, non trivial Mathematica Libraries
Posted:
Jan 20, 2012 1:58 AM


On 19 Jan 2012, at 14:14, Nehal Patel wrote:
> On Jan 19, 2012, at 6:39 AM, Andrzej Kozlowski wrote: > >> >> On 19 Jan 2012, at 11:12, Bill Rowe wrote: >> >>>> >>>> geom[list_] := Apply[Times, list]^(1/Length[list]) >>> >>>> So, this does a bad thing for geom[ x+ y] (returns (Sqrt[x y]) >>> >>> What were you expecting here? This looks correct to me >> >> Clearly he sees what actually is a powerful feature of Mathematica as a "bad thing". The feature is that a function whose arguments are not restricted by pattern maching can be put to a more general use. I think it's great and not "bad". If you want to restrict the function to work on lists you can do this: >> > > Hi  thanks for your comment. Actually I think the pattern matching is the coolest thing. my point is that when combined with how Mathematica treats + (and in general, Flat and Orderless operators) it becomes a lot harder to reason about whether a function handles all edge cases correctly. So for instance nobody would expect the geometric mean of x + y to be Sqrt[x y], it should be (x + y)^1 = x+ y
No, it shouldn't be. There is no such thing as a mean of a number or an algebraic expression. What you would expect is
In[7]:= geom[{x + y}]
Out[7]= x + y
which is indeed what you get.
> > Separately, GeometricMean does do something reasonable for GeometricMean[x + y] (give it a try), and I wish it were easier to see what syntax it uses.
It is very easy to do that, it in fact:
Clear[geom]
geom::nonlist = "The argument `1` is neither a nonempty vector nor a nonempty matrix";
geom[ls_] /;If[Head[ls] === List, True, Message[geom::nonlist, x]; False] := Apply[Times, ls]^(1/Length[ls])
geom[{x + y}]
x + y
geom[x + y]
The argument x is neither a nonempty vector nor a nonempty matrix
geom[x + y]
Andrzej Kozlowski



