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Re: Conditions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg29252] Re: Conditions
  • From: bghiggins at ucdavis.edu (Brian Higgins)
  • Date: Fri, 8 Jun 2001 04:15:37 -0400 (EDT)
  • References: <9fna8p$bih$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Yannis, Try this

myMatrix = Table[Random[Integer, {0, 10}], {10}, {7}];
nRow = First[Dimensions[myMatrix]];
nCol = Last[Dimensions[myMatrix]];
Table[If[Length[Cases[myMatrix[[j]], x_ /; x == 0]] > 0, v[j] = 0, 
    v[j] = myMatrix[[j, nCol]]], {j, 1, nRow}]


Cheers,

Brian




Yannis.Paraskevopoulos at ubsw.com wrote in message news:<9fna8p$bih$1 at smc.vnet.net>...
> Hi again,
> 
> I've been puzzled by the following problem which I  have to admit it's 
> not complicated but yet...
> 
> assume that we have a matrix m with dimensions (RxC) then create a 
> vector v, which has value 0 when there is a zero at any point across 
> m's row. Otherwise v's value would be equal to the last value of the 
> corresponding m row.
> 
> 
> I tried for example:
> 
> 
> I initiated a vector v={a,b}. I had to that as 
> 
> Table[If[m[[i,j]]==0,v[[i]]=0,v[[i]]=g[[i,5]]],{i,2},{j,5}];
> v
> 
> Out:= {1,1} when I expected {0,1}
> 
> 
> I also tried
> 
> 
> v:=0 /; g/.x_:>x==0
> v:=Table[g[[i,5]],{i,2}]
> 
> with same results.
> 
> Its obvious to me that I do something stupid but I can't figure it out. 
> Any help would be very much appreciated.
> 
> Best regards
> 
> 
> yannis
> 
> 
> Visit our website at http://www.ubswarburg.com
> 
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