On Mon, Nov 14, 2011 at 6:28 AM, Andreas Perstinger <andreas.perstin...@gmx.net> wrote: > On 2011-11-11 14:44, lina wrote: >> >> You are right, I did not think of this parts before. and actually the >> initiative wish was to find possible paths, I mean, possible >> substrings, all possible substrings. not the longest one, but at >> least bigger than 3. > > I had some time today and since you have changed your initial task (from > finding the longest common path to finding all common paths with a minimum > length) I've modified the code and came up with the following solution: > > def AllCommonPaths(list1, list2, minimum=3): > """ finds all common paths with a minimum length (default = 3)""" > > # First we have to initialize the necessary variables: > # M is an empty table where we will store all found matches > # (regardless of their length) > > M = [[0] * (len(list2)) for i in range(len(list1))] > > # length is a dictionary where we store the length of each common > # path. The keys are the starting positions ot the paths in list1. > > length = {} > > # result will be a list of of all found paths > > result =[] > > # Now the hard work begins: > # Each element of list1 is compared to each element in list2 > # (x is the index for list1, y is the index for list2). > # If we find a match, we store the distance to the starting point > # of the matching block. If we are in the left-most column (x == 0) > # or in the upper-most row (y == 0) we have to set the starting > # point ourself because we would get negative indexes if we look > # for the predecessor cell (M[x - 1][y - 1]). Else, we are one > # element farther away as the element before, so we add 1 to its > # value. > > for x in range(len(list1)): > for y in range(len(list2)): > if list1[x] == list2[y]: > if (x == 0) or (y == 0): > M[x][y] = 1 > else: > M[x][y] = M[x - 1][y - 1] + 1 > > # To get everything done in one pass, we update the length of > # the found path in our dictionary if it is longer than the minimum > # length. Thus we don't have to get through the whole table a > # second time to get all found paths with the minimum length (we > # don't know yet if we are already at the end of the matching > # block). > > if M[x][y] >= minimum: > length[x + 1 - M[x][y]] = M[x][y] > > > # We now have for all matching blocks their starting > # position in list1 and their length. Now we cut out this parts > # and create our resulting list
This is a very smart way to store their starting position as a key. My mind was choked about how to save the list as a key before. > > for pos in length: > result.append(list1[pos:pos + length[pos]]) > > return result > > I've tried to explain what I have done, but I'm sure you will still have > questions :-). I am confused myself with this matrix/array, about how to define x-axis, y-axis. I must understand some parts wrong, for the following: > > Is this close to what you want? > > Bye, Andreas > > PS: Here's the function again without comments: > > def AllCommonPaths(list1, list2, minimum=3): > """ finds all common paths with a minimum length (default = 3)""" > > M = [[0] * (len(list2)) for i in range(len(list1))] is it correct that the list2 as the x-axis, the list1 as y-axis:? > length = {} > result =[] > > for x in range(len(list1)): Here for each row , > for y in range(len(list2)): This loop go through each column of certain row then, > if list1[x] == list2[y]: > if (x == 0) or (y == 0): > M[x][y] = 1 Here M[x][y] actually means the x-row? y-column, seems conflicts with the x-axis and y-axis. they took y-axis as x row, x-axis as y column. > else: > M[x][y] = M[x - 1][y - 1] + 1 > if M[x][y] >= minimum: > length[x + 1 - M[x][y]] = M[x][y] > > for pos in length: > result.append(list1[pos:pos + length[pos]]) > > return result I have no problem understanding the other parts, except the array and axis entangled in my mind. > _______________________________________________ > Tutor maillist - Tutor@python.org > To unsubscribe or change subscription options: > http://mail.python.org/mailman/listinfo/tutor > _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor