Python: find a list within members of another list(in order)

Question

If I have this:

a=\'abcdefghij\'
b=\'de\'

Then this finds b in a:

b in a => True

Is there a way of doi

Answer 1

I timed the accepted solution, my earlier solution and a new one with an index. The one with the index is clearly best.

EDIT: I timed nosklo's solution, it's even much better than what I came up with. :)

def is_sublist_index(a, b):
    if not a:
        return True

    index = 0
    for elem in b:
        if elem == a[index]:
            index += 1
            if index == len(a):
                return True
        elif elem == a[0]:
            index = 1
        else:
            index = 0

    return False

def is_sublist(a, b):
    return str(a)[1:-1] in str(b)[1:-1]

def is_sublist_copylist(a, b):
    if a == []: return True
    if b == []: return False
    return b[:len(a)] == a or is_sublist_copylist(a, b[1:])

from timeit import Timer
print Timer('is_sublist([99999], range(100000))', setup='from __main__ import is_sublist').timeit(number=100)
print Timer('is_sublist_copylist([99999], range(100000))', setup='from __main__ import is_sublist_copylist').timeit(number=100)
print Timer('is_sublist_index([99999], range(100000))', setup='from __main__ import is_sublist_index').timeit(number=100)
print Timer('sublist_nosklo([99999], range(100000))', setup='from __main__ import sublist_nosklo').timeit(number=100)

Output in seconds:

4.51677298546

4.5824368

1.87861895561

0.357429027557

Answer 2

This should work with whatever couple of lists, preserving the order. Is checking if b is a sub list of a

def is_sublist(b,a): 

    if len(b) > len(a):
        return False    

    if a == b:
        return True    

    i = 0
    while i <= len(a) - len(b):
        if a[i] == b[0]:
            flag = True
            j = 1
            while i+j < len(a) and j < len(b):
                if a[i+j] != b[j]:
                    flag = False
                j += 1
            if flag:
                return True
        i += 1
    return False

Answer 3

Use the lists' string representation and remove the square braces. :)

def is_sublist(a, b):
    return str(a)[1:-1] in str(b)

EDIT: Right, there are false positives ... e.g. is_sublist([1], [11]). Crappy answer. :)