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Hi Chris,<br>
<br>
Many thank for contributing!<br>
<br>
Just a small suggestion. If I'm right the longest common substring
should be symmetric. Hence, you may divide computation time by two
if you compute it only once for each comparison (instead of two).
This is done by modifying the second loop in lcsubstr_seqdist<br>
for (i in 1:d_nrow) {<br>
for (j in i:d_nrow){ ## Change start from i instead of
one<br>
d = lcsubstr( charSeqData[i], charSeqData[j] )<br>
dist_matrix[i,j] <- d<br>
dist_matrix[j, i] <- d<br>
}<br>
}<br>
<br>
All the best,<br>
Matthias<br>
<br>
Le 20.03.2012 11:11, Hadrien Commenges a écrit :
<blockquote cite="mid:876c0495-0cd9-4c49-b422-633afc6b8996@clara"
type="cite">
<style> body {height: 100%; color:#000000; font-size:12pt; font-family:Arial, Helvetica, sans-serif;}</style>Yes
it works! <br>
I'll test it with my data (5.000 cases, sequences of max length 8,
alphabet of 6) and tell you how much time it takes.<br>
Thank you again,<br>
Hadrien<br>
<br>
<br>
<br>
----- Mail d'origine -----<br>
De: Chris Cameron <a class="moz-txt-link-rfc2396E" href="mailto:cjc73@cornell.edu"><cjc73@cornell.edu></a><br>
À: Users questions <a class="moz-txt-link-rfc2396E" href="mailto:traminer-users@r-forge.wu-wien.ac.at"><traminer-users@r-forge.wu-wien.ac.at></a><br>
Envoyé: Tue, 20 Mar 2012 03:44:08 +0100 (CET)<br>
Objet: Re: [Traminer-users] longest common substring<br>
<br>
Hi Hadrien -<br>
<br>
These functions will work if your sequence alphabet consists of
single character codes. Because it is in R instead of C, the
computations can take quite a long time. I tested it with
sequences of length 12. Longer sequence lengths could take
substantially longer. <br>
<br>
The number of comparisons necessary to compute a distance matrix
is quite large-- this will only work reasonably for relatively
small data sets. The computational complexity is the square of the
number of cases, so you can estimate the completion time by
calculating the time per comparison and multiplying it by the
square of the number of cases. This table will help you guess how
long it might take.<br>
<br>
Case (Sequence) Count Time to Compute<br>
100 7 seconds<br>
1000 11 mins<br>
2000 50 mins (est)<br>
5000 5 hours (est) <br>
10000 24 hours (est)<br>
<br>
<br>
# Example of how the sequence data can be converted to text
strings and sent to lcsubstr_seqdist( )<br>
data.txt = seqconc(seq_data, sep='')<br>
seq_dist2 = lcsubstr_seqdist( data.txt[1:10] ) #test it with a
subset and see it if looks right!<br>
<br>
#----------------<br>
# code to source into R (or paste)<br>
# Greatest common substring algorithm <br>
lcsubstr = function(s1, s2) {<br>
# Based on pseudocode from
<a class="moz-txt-link-freetext" href="http://en.wikipedia.org/wiki/Longest_common_substring_problem">http://en.wikipedia.org/wiki/Longest_common_substring_problem</a><br>
# Code sections from Gabor Grothendieck
(<a class="moz-txt-link-freetext" href="http://finzi.psych.upenn.edu/R/Rhelp02a/archive/68013.html">http://finzi.psych.upenn.edu/R/Rhelp02a/archive/68013.html</a>)<br>
# This code provided without warranty. It may not work or could
return false values. Use at your own risk.<br>
# Make sure args are strings <br>
a <- as.vector(s1, 'character'); an <- nchar(s1) <br>
b <- as.vector(s2, 'character'); bn <- nchar(s2) <br>
# If one arg is an empty string, returns the length of the other<br>
if (nchar(a)==0) return(nchar(b)) <br>
if (nchar(b)==0) return(nchar(a))<br>
# end Gabor code<br>
z = 0<br>
long_len = ''<br>
m = matrix(0, nrow=an, ncol=bn)<br>
for (i in 1:an) {<br>
for (j in 1:bn) {<br>
if (substr(a,i,i)==substr(b,j,j)) {<br>
if (i==1 | j==1) {<br>
m[i,j]=1<br>
} else {<br>
m[i,j]=m[i-1,j-1]+1 <br>
}<br>
if (m[i,j] > z) {<br>
z = m[i,j]<br>
long_len = ''<br>
}<br>
if (m[i,j] == z) {<br>
long_len = substr(a,i-z+1,i)<br>
}<br>
} else {<br>
m[i,j]=0<br>
}<br>
}<br>
}<br>
return(nchar(long_len))<br>
} <br>
<br>
<br>
lcsubstr_seqdist = function( charSeqData ) {<br>
d_nrow = length(charSeqData)<br>
dist_matrix = matrix(0, nrow=d_nrow, ncol=d_nrow)<br>
for (i in 1:d_nrow) {<br>
for (j in 1:d_nrow){ <br>
#print(c(i,j))<br>
dist_matrix[i,j] = lcsubstr( charSeqData[i], charSeqData[j] )<br>
}<br>
}<br>
return(dist_matrix)<br>
}<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
On Mar 19, 2012, at 11:30 AM, Hadrien Commenges wrote:<br>
<br>
> Thank you Chris. I checked the TraMineR Extras package, but
there is no function to compute the longest common substring. I
don't know if the developers could give me an easy way to modify
the seqdist function in order to compute longest common substring.
If not, I'm very interested in the ideas you mentioned to do that,
although I'm not a good programmer and I don't know anything about
C.<br>
> Thanks,<br>
> Hadrien<br>
> <br>
> <br>
> <br>
> ----- Mail d'origine -----<br>
> De: Chris Cameron <cjc73@cornell.edu><br>
> À: Users questions <traminer-users@r-forge.wu-wien.ac.at><br>
> Envoyé: Sun, 18 Mar 2012 18:48:56 +0100 (CET)<br>
> Objet: Re: [Traminer-users] longest common substring<br>
> <br>
> Hi Hadrien - <br>
> <br>
> I don't see a documented way to compute the greatest
common substrings, so a custom function is probably necessary.
(Maybe a dev can tell you if there is a hidden flag to
indicate substring vs subsequence. The LCS code is implemented
in C, so there is not an easy function to modify in the
TraMineR package. I have some ideas about how you can make a
custom function to do this if you want to talk about it.
Before you go that route, it would be worth checking the
"TraMineR Extras" package Matthias mentioned on the 16th. <br>
> <br>
> Good Luck<br>
> Chris<br>
> <br>
> <br>
> <br>
> <br>
> On Mar 17, 2012, at 3:02 PM, Hadrien Commenges wrote:<br>
> <br>
> Hi,<br>
> <br>
> I've a new question. I'd like to get a dissimilarity
matrix based on the longest common substring. The longest
prefix (computed with seqdist, method=LCS) doesn't work for me
because I'd like to compare substrings anywhere in the
sequence. And the notion of subsequence (computed with
seqdist, method=LCS) doesn't serve my purpose neither. I'd
like to compute something like the LCS method but with the
stricter notion of substring. Is it possible?<br>
> <br>
> Thanks,<br>
> <br>
> Hadrien<br>
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