Re: Problem with the 'if' command
- To: mathgroup at smc.vnet.net
- Subject: [mg96618] Re: Problem with the 'if' command
- From: dh <dh at metrohm.com>
- Date: Tue, 17 Feb 2009 06:27:15 -0500 (EST)
- References: <gncme2$fid$1@smc.vnet.net>
Hi,
your condition does not check for Even/Odd. Use EvenQ or OddQ for this
purpose.
hope this helps, Daniel
mathandpi wrote:
> Hi everyone,
> I'm a new Mathematica user so I may be missing something fairly obvious, but I'm having trouble with the 'if' command.
> I'm writing a function that is supposed to return the median of a list (I know such a function already exists, but I need to create one myself).
>
> What I have is:
>
> newMedian[list_] :=
> If[Length[list]/2 == 0,
> 1/2*(list[[(Length[list]/2) + 1]] + list[[(Length[list])/2]])
> , list[[(Length[list] + 1)/2]]]
>
> if the list has an off number of members (condition is false), it evaluates as expected. If it's even, however,
>
> newMedian[{1, 2, 3, 4}] returns:
> Part::pspec: Part specification 5/2 is neither an integer nor a list of integers.
>
> BUT
>
> list={1,2,3,4};
>
> 1/2*(list[[(Length[list]/2) + 1]] + list[[(Length[list])/2]])
> returns 5/2, as expected so that code is right.
>
> It seems likes its actually evaluating the false part of the code (trying to find the 5/2'ith element in a list), even though the condition is true.
>
> What's going on here?
>
> Thanks
>