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MathGroup Archive 2005

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Re: Re: random matrix from row and column sums

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53854] Re: [mg53831] Re: random matrix from row and column sums
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 30 Jan 2005 03:18:15 -0500 (EST)
  • References: <csinpt$njk$1@smc.vnet.net><200501200847.DAA04066@smc.vnet.net> <csqqs1$1mh$1@smc.vnet.net> <200501291102.GAA24672@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

"Random" covers a lot of territory, of course. Choosing overall solutions EQUALLY OFTEN probably requires listing them all, first, and that's what my first solution did.

Bobby

On Sat, 29 Jan 2005 06:02:39 -0500 (EST), Valeri Astanoff <astanoff at yahoo.fr> wrote:

> Bob,
>
> Your solution is by far faster than mine, but what kind of random is
> it?
>
> Consider this example :
> [...]
> In[9]:=
> bob=Table[randomFill[{10,10,10,10,10},{10,10,10,10,10}],{100}]//Flatten;
>
> In[10]:=
> BinCounts[bob,{-1,10}]
> Out[10]=
> {1147,355,224,191,134,117,105,68,70,53,36}
>
> In[11]:=
> val=Table[rand[{10,10,10,10,10},{10,10,10,10,10}],{100}]//Flatten;
>
> In[12]:=
> BinCounts[val,{-1,10}]
> Out[12]=
> {348,613,740,445,226,102,24,2,0,0,0}
>
> In[13]:=
> rep=N[{Mean[#],Mode[#],Median[#],StandardDeviation[#],Skewness[#]}]&;
>
> In[14]:=
> rep[bob]
> Out[14]=
> {2.,0.,1.,2.64809,1.33292}
>
> In[15]:=
> rep[val]
> Out[15]=
> {2.,2.,2.,1.37228,0.525969}
>
> With your fast but too much skewed solution 0 is three times
> more likely to occur than 1 and five times more likely to occur than 2,
>
> which is the mean (and is also the mode and median with my solution).
> I wonder whether it is what the initial poster expected...
>
>
> Valeri
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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