Re: Help
- To: mathgroup at smc.vnet.net
- Subject: [mg28228] Re: [mg28196] Help
- From: Tomas Garza <tgarza01 at prodigy.net.mx>
- Date: Thu, 5 Apr 2001 03:00:44 -0400 (EDT)
- References: <200104040813.EAA23835@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I only hope your s vector has no zero elements. If m and s have all
nonnegative elements,
In[1]:=
c[m_, s_] := (n = (m - s)/s; negs = Position[n, x_ /; x < 0];
ReplacePart[n, m, negs, negs])
can handle the job decently. For example, with m 100x150 and s 100x1,
In[2]:=
m = Table[Random[Integer, {0, 9}], {100}, {150}];
s = Table[Random[Integer, {1, 9}], {100}];
In[3]:=
c[m, s]; // Timing
Out[3]=
{12.31 Second, Null}
If you allow for both negative and positive elements, then some
modifications must be made to the above. You should have given us some more
information about the nature of your matrix and vector.
Tomas Garza
Mexico City
----- Original Message -----
From: <Yannis.Paraskevopoulos at ubsw.com>
To: mathgroup at smc.vnet.net
Subject: [mg28228] [mg28196] Help
> Hi all,
>
> I would appreciate your help on the following problem:
>
>
> assume that we have a matrix m with dimensions (RxC) and a vector s
> with dimensions (Rx1). Now what I want is to create another matrix Q
>
> If m[[i,j]]>s[[i]], then Q=(m[[i,j]]-s[[i]])/s[[i]], for {i,1,R},{j,1,C}
>
>
> I'm sure there must be an elegant and fast way to do that.
>
>
> Best Regards
>
> yannis
>
>
>
>
> Visit our website at http://www.ubswarburg.com
>
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- References:
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- From: Yannis.Paraskevopoulos@ubsw.com
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