Checking if list is a sublist

Question

I need to check if list1 is a sublist of list2 (True; if every integer in list2 that is common with list1 is in the same order of indexes as in list1)

def sublist(lst1,lst2):     for i in range(len(lst1)):         if lst1[i] not in lst2:             return False         for j in range(len(lst2)):             if (lst1[j] in lst2) and (lst2.index(lst1[i+1]) > lst2.index(lst1[i])):                 return True 

Can anybody help me... why isn't this working?

Answer 1

i need to check if list1 is a sublist to list2 (True; if every integer in list2 that is common with list1 is in the same order of indexes as in list1)

Your code isn't working because as soon as a list element in ls1 doesn't occur in ls2 it will return False immediately.

This creates two lists that contain only the common elements (but in their original order) and then returns True when they are the same:

def sublist(lst1, lst2):    ls1 = [element for element in lst1 if element in lst2]    ls2 = [element for element in lst2 if element in lst1]    return ls1 == ls2 

edit: A memory-efficient variant:

def sublist(ls1, ls2):     '''     >>> sublist([], [1,2,3])     True     >>> sublist([1,2,3,4], [2,5,3])     True     >>> sublist([1,2,3,4], [0,3,2])     False     >>> sublist([1,2,3,4], [1,2,5,6,7,8,5,76,4,3])     False     '''     def get_all_in(one, another):         for element in one:             if element in another:                 yield element      for x1, x2 in zip(get_all_in(ls1, ls2), get_all_in(ls2, ls1)):         if x1 != x2:             return False      return True 

Answer 2

An easy way to check if all elements of a list are in other one is converting both to sets:

def sublist(lst1, lst2):     return set(lst1) <= set(lst2)