       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

>

```

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