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Re: set option flag in function definition
- To: mathgroup at smc.vnet.net
- Subject: [mg128160] Re: set option flag in function definition
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Wed, 19 Sep 2012 05:00:19 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
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- References: <20120918074139.41A2D6853@smc.vnet.net>
On Sep 18, 2012, at 3:41 AM, Joug Raw <jougraw at gmail.com> wrote:
> ...I want to have a function that executes certain operation when I chose the
> corresponding option flag.
>
> For example,
>
> a function called "f"
>
> when I call it like f[arg1,arg2,arg3, optionX]
> it executes command
> arg1/arg2+arg3
>
> If I call it like f[arg1,arg2,arg3,optionY]
> it executes command
> arg1*arg2-arg3
>
> If I call it like f[arg1,arg2,arg3,optionSomethingElse]
> it returns me
> "No appropriate option flag was found. Please indicate correct options"
>
> ...I am wondering if I can do
> the same in Mathematica easily?
Yes, it can be done very easily in Mathematica:
f[a_, b_, c_, method -> "one"] := a/b+c
f[a_, b_, c_, method -> "two"] := a b-c
f[a_, b_, c_, method -> _] := "No relevant value for 'method' found"
f[a_, b_, c_, opt_ -> _] := "I don't understand this option"
(I shortened argument names; the option name "method" I used is quite arbitrary.)
The key to the above is that Mathematica always uses a more specific rule before it uses a more general one.
If, though, you want a single expression for defining the function f, then yes that's also possible but more complicated; you cold use, e.g., FilterRules.
---
Murray Eisenberg
murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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