[Traminer-users] Substitution cost matrix

so.bellit so.bellit at laposte.net
Sun Sep 2 20:33:05 CEST 2012


Hi,

 My substitution cost matrix is not symetric. Would you have an explanation for this?

 Best Regards
 Sonia




> Message du 09/08/12 15:06
> De : "Weldon, Mat"
> A : "Users questions"
> Copie à :
> Objet : Re: [Traminer-users] seqplot
>
> Hi Maximilian,
>
> I'm answering based on experience with seqedplot, but it should be the same. If your group argument is just a vector of cluster assignments such as the result of a PAM clustering, for example:
> cluster4 <- pam(dd, diss=TRUE, k=4, cluster.only=TRUE) #where "dd" is the dissimilarity matrix
>
> then to plot only one group you should be able to input something like:
> seqdplot(mvad.seq, group=cluster4[cluster4==4])
>
> Is that helpful?
>
> Kind regards,
>
> Mat
>
> -----Original Message-----
> From: traminer-users-bounces at lists.r-forge.r-project.org [mailto:traminer-users-bounces at lists.r-forge.r-project.org] On Behalf Of Maximilian Rothfuß
> Sent: 09 August 2012 12:12
> To: traminer-users at lists.r-forge.r-project.org
> Subject: [Traminer-users] seqplot
>
> Hello all together
>
> In my analysis I remarked that there is no possibility within the 'seqdplot´ - function for plotting just one level of the factor given by the `group` argument. Further I couldn´t find any solution for that problem, so I wrote a large function containing some loops, very complicated.
> It works but it's takes so much time...
> Of course, I know that it's possible to get the rigt group by seperating them like:
>
> data(mvad)
> mvad.seq <- seqdef(mvad, 17:86)
>
> ## Using the group argument
> seqdplot(mvad.seq, group=mvad$gcse5eq)
>
> ## Plotting groups separately
> seqdplot(mvad.seq[mvad$gcse5eq=="yes",])
> seqdplot(mvad.seq[mvad$gcse5eq=="no",])
>
> But if the grouping is a result from a cluster analysis, you cannot plot just one certin group.
>
> Did I overlook somthing within the code of seqdplot or TraMineR?
>
> Thank you in anticipation!
>
> Best regrads,
>
> Maximilian
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