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

Tomas Garza
Mexico City

----- Original Message -----
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
>
>
>
>
>
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```

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