Thank you, Barton, Mario,
indeed, Barton, you have answered a question in advance :)) Surprising use of map and
"[".
I came across that 'switch'-problem matrix <--> list, because the input of a dataset in
plot2d(discrete,..) must be a _list_, whereas it must be a _matrix_ in dataplot(..).
If one wants 2 pictures using both possibilities in the same context, one has to watch the
input format and prepare the dataset XY in 2 formats - could be a small stumble stone for
a beginner like me.
Thanks for your hints.
HTH Wolfgang
"Barton Willis" <willisb at unk.edu> schrieb:
> Mario gave you *the* solution to your Q1; nevertheless, for some other
> problem, you might like to know the trick:
>
> (%i1) x : [8.5,13.7,15,16,16.9,15.1,13.9,12]$
> (%i2) y : [5.2,8.3,9.4,9.5,9.6,9.4,8.8,7.3]$
> (%i3) map("[",x,y);
> (%o3)
> [[8.5,5.2],[13.7,8.300000000000001],[15,9.4],[16,9.5],[16.9,9.6],[15.1,9.4],[13.9,
>
> 8.800000000000001],[12,7.3]]
>
> Barton
>
> -----maxima-bounces at math.utexas.edu wrote: -----
>
> To: "maxima-help" <maxima at math.utexas.edu>
> From: Wolfgang Lindner
> Sent by: maxima-bounces at math.utexas.edu
> Date: 05/28/2007 06:37AM
> Subject: contrib - descriptive - correlation coefficient
>
>
> Q1: what is the most simple method to transform the splitted data
> X:[8.5,13.7,15,16,16.9,15.1,13.9,12];
> Y:[5.2,8.3,9.4,9.5,9.6,9.4,8.8,7.3];
> into the above data_matrix_ XY as input for cov()?