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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