Find sub list inside a list in python

Question

I have a list of numbers

l = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
[0, 0, 2, 1, 1, 2, 0, 0, 0, 0]
[0, 0, 2, 1, 1, 2, 2, 0, 0, 1]
[0, 0, 1, 2, 2, 0, 1, 0, 0, 2]
[1, 0, 1, 1, 1, 2, 1, 0, 2, 1]]

For example , i have to search a pattern '2,1,1,2' , as we can see that is present in row 6 and 7 . in order to find that sequence i tried converting each list into str and tried to search the pattern , but for some reason the code isnt working.

import re
for i in l:
 if re.search('2,1,1,2' , str(i).strip('[').strip(']')): print " pattern found"

am i missing something in here ?

Answer 1

Converting your list in string is really not a good idea.

How about something like this:

def getsubidx(x, y):
    l1, l2 = len(x), len(y)
    for i in range(l1):
        if x[i:i+l2] == y:
            return i

Answer 2

I suggest you to use the Knuth-Morris-Pratt algorithm. I suppose you are implicitly assuming that your pattern is present in the list just one time, or you are just interested in knowing if it's in or not.

If you want the list of each first element which starts the sequence, then you can use KMP. Think about it as a sort of string.find() for lists.

I hope this will help.