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In TraMineR, seqdistmc is used to measure multichannel distances between "state" sequences. I am wondering if there is a function to measure multichannel distances between "event" sequences.

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There is no equivalent of the seqdistmc function for computing multichannel distances between event sequences in TraMineR.

Actually, seqdistmc implements the principle defined by Pollock (Pollock, G., 2007, Holistic trajectories: A study of combined employment, housing and family careers by using multiple-sequence analysis, JRSS, series A, 170(1), 167-183). The method applies to optimal matching distances only and aims at reducing the number of required substitution costs. It consists in deriving the substitution costs between combined states of different domains from the substitution costs in each domain. For two domains with each say 10 states, one needs to specify twice $9 * 10/2 = 45$, i.e., 90 substitution costs for the total of $49 * 50/2 = 1225$ pairs that can be formed with the $10*10/2 = 50$ combined states.

We do not have this problem with event sequences where simultaneous events are allowed. The OME measure computed by the seqedist function considers the following two basic operations (see p 240 in Ritschard, G., Bürgin, R. & Studer, M. , 2013, "Exploratory Mining of Life Event Histories", In McArdle, J.J. & Ritschard, G. (eds) Contemporary Issues in Exploratory Data Mining in the Behavioral Sciences. pp. 221-253. New York: Routledge.)

  • insertion/deletion of an event;

  • changing the time stamp by one unit.

A single cost can be specified for each operation, but event dependent costs can also be specified. So the number of costs would be at most twice the event alphabet size.

For a multichannel analysis of event sequences, we can simply merge the multi-sequences associated with each individual, i.e., consider the event sequences that include the events of both dimensions. For two channels, the number of costs to be specified would then simply be the sum of the number of costs for each channel. With two channels having each an alphabet of size 10, the number of costs would be at most $2 * 2 * 10 = 40$.

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