[Traminer-users] seqfplot issue

[Matthias Studer]

Hi all,

I hope I understand well your question. Usually, the problem with 
seqfplot is that the most frequent one are the stable sequence, 
especially when sequences are long (and 77 monthes is long). This is due 
to the fact that every small difference is taken into account. To 
overcome this problem, you may use "representative sequences". This 
procedure tries to regroup sequences that differ only marginally and to 
identify a set of sequences that might represent a given percentage of 
the sequences in that group (see the help of the seqrep function).

The procedure and the concept of representative sequence is described in 
this reference: Gabadinho A, Ritschard G, Studer M, Müller NS (2011). 
"Extracting and Rendering Representative Sequences", In A Fred, JLG 
Dietz, K Liu, J Filipe (eds.), /Knowledge Discovery, Knowledge 
Engineering and Knowledge Management/, volume 128 of /Communications in 
Computer and Information Science (CCIS)/, pp. 94-106. Springer-Verlag.

I have another suggestion regarding your code. You may use the group 
argument (available in all seq*plot) to automatically draw a distinct 
plot of each group of sequences.

seqdplot(fx.seq, group=pam.best.fx$clustering)


Hope this helps.
Matthias Studer

Le 17.05.2011 19:19, Kleinbaum, Adam M. a écrit :
>
> Hi all,
>
> I'm having an issue with seqfplot I'm hoping you can help me with.  I 
> have a series of career sequences over a 77-month observation period 
> for a group of 15,000 or so individuals.  I use the following code to 
> read in my data (including labels), define a color palette, and drop 
> any observations with missing data:
>
> fx <- read.csv(file="Seq_function.txt", header=FALSE)
>
> fx.lab <- c("AD", "CO", "FI", "GM", "HR", "LE", "MF", "MK", "OT", 
> "RD", "SC", "SL", "SV")
>
> fx.seq <- seqdef(fx, 2:78, labels=fx.lab)
>
> attr(fx.seq,"cpal") <- c(brewer.pal(n=12, name="Set3"), "cyan", "white")
>
> fx <- fx[!is.na(fx[,78]), ]
>
> fx.seq <- fx.seq[seqlength(fx.seq)==77,]
>
> Then I use TraMineR to calculate the substitution cost matrix and do 
> optimal matching.  Next I use the "cluster" package to do a cluster 
> analysis and I find that there are 9 prototypical sequences.  I'd like 
> to do a bit of graphing for each of the 9 clusters, so I run the 
> following code:
>
> for (i in 1:k.best.om.fx)
>
> {
>
>   seqdplot(fx.seq[pam.best.fx$clustering==i,], withlegend=F, 
> title=paste("Cluster ", i))
>
>   seqfplot(fx.seq[pam.best.fx$clustering==i,], withlegend=F, 
> title=paste("Cluster ", i))
>
> }
>
> seqlegend(fx.seq)
>
> where fx.seq contains my sequences, as shown above; pam.best.fx is the 
> pam object that came out of the clustering algorithm and 
> pam.best.fx$clustering contains the index of each actor's cluster 
> assignment.
>
> The seqdplot command produces a series of 9 beautiful distribution 
> plots, one for each cluster.  No problem.
>
> What I want seqfplot to do is, for each cluster, graph out the 
> frequency of sequences within each cluster, from most frequent on down 
> the list.  To take one example, the medoid of one cluster is 77 
> observations of all the same job function -- people who stay in sales 
> and don't move.  People assigned to that cluster should have spent 
> nearly all 77 periods in that function.  The output should look like a 
> long sequence of blocks that are mostly the same color, with another 
> sequence above it that's a little bit different, but also mostly all 
> the same color.  But instead, every single block is a different color 
> from the one next to it.  There's obviously some problem with either 
> how I'm calling the function or how I'm defining my color palette, but 
> I can't figure out what it is.  I'm especially perplexed that seqdplot 
> works properly while an identical call to seqfplot does not.  Any 
> ideas?  Thanks in advance,
>
> All the best,
>
> Adam
>
> --
>
> Adam M. Kleinbaum
>
> Assistant Professor
>
> Tuck School of Business
>
> Dartmouth College
>
> http://bit.ly/kleinbaum
>
> 603.646.6447
>
>
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