Is there any fast implementation of the Longest Common Subsequence algorithm for trajectory matching in Python? Ideally it would work with trajectories of different length in 2d spaces.

  • Do you have a quick link to save people from looking it up, specifically the reference you were examining – user681 Jan 25 '15 at 23:04
  • This is what I was working off of: cs.bu.edu/groups/dblab/pub_pdfs/icde02.pdf – Drew Jan 25 '15 at 23:45
  • The vertices of each matching sub sequence, are they equal? – huckfinn Jul 5 '16 at 16:52

The python module Machine Learning Python (mlpy) has an LCS method including an LCS for real series:


Perhaps you could also adapt the LCS algorithm for strings and test your own implementation against mlpy.

A quick google search results in the folloing Wikipedia page:


I copied the content for backup purpose:

Computing the length of the LCS

def LCS(X, Y):
m = len(X)
n = len(Y)
# An (m+1) times (n+1) matrix
C = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m+1):
    for j in range(1, n+1):
        if X[i-1] == Y[j-1]: 
            C[i][j] = C[i-1][j-1] + 1
            C[i][j] = max(C[i][j-1], C[i-1][j])
return C

Reading out an LCS

def backTrack(C, X, Y, i, j):
if i == 0 or j == 0:
    return ""
elif X[i-1] == Y[j-1]:
    return backTrack(C, X, Y, i-1, j-1) + X[i-1]
    if C[i][j-1] > C[i-1][j]:
        return backTrack(C, X, Y, i, j-1)
        return backTrack(C, X, Y, i-1, j)

Reading out all LCSs

def backTrackAll(C, X, Y, i, j):
if i == 0 or j == 0:
    return set([""])
elif X[i-1] == Y[j-1]:
    return set([Z + X[i-1] for Z in backTrackAll(C, X, Y, i-1, j-1)])
    R = set()
    if C[i][j-1] >= C[i-1][j]:
        R.update(backTrackAll(C, X, Y, i, j-1))
    if C[i-1][j] >= C[i][j-1]:
        R.update(backTrackAll(C, X, Y, i-1, j))
    return R

Usage example

m = len(X)
n = len(Y)
C = LCS(X, Y)

print "Some LCS: '%s'" % backTrack(C, X, Y, m, n)
print "All LCSs: %s" % backTrackAll(C, X, Y, m, n)

It prints the following:

Some LCS: 'AAC'
All LCSs: set(['ACC', 'AAC'])

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