Check if a given list is a sublist of a larger list

Question

def sublist(slist, blist):
    m = 0
    is_a_sublist = False
    for n in range(0, len(blist)):
     while (n < len(blist) and m < len(blist)) and is_a_sublist == False:
            spliced_list = blist[n:m]
            if spliced_list == slist:
                is_a_sublist = True
            else:
                m += 1

    return is_a_sublist

def main():
    test_list = ["a",9,1,3.14,2]
    original_list = ["a",1,"a",9,1,3.14,2]

    is_a_sublist = sublist(test_list,original_list)

    if is_a_sublist == True:
        print("It is a sublist!")
    else:
        print("It ain't a sublist!")


main()

The code I wrote above is to check if a given list is a "sublist" of a larger list. So, if a large list is [3,3,10,4,9], and the given list is [3,10,4], then it is a sublist so it must return True. But what's wrong with my code? When I run it, output is "It ain't a sublist" even though it is (function returns False). I only want to use loops, no built-in functions.

Most answers there are using complicated functions, just wanted to rely on loops. — janissary, Mar 25 '20 at 21:40

kederrac · Answer 1 · 2020-03-26T08:25:26.447

0

you can use:

def sublist(l2, l1):
    for i in range(0, len(l2) - len(l1) + 1):
        if l2[i] == l1[0]:
            for j in range(1, len(l1)):
                if l2[i + j] != l1[j]:
                    break
            else:
                return True
    return False


l1 = [3,3,10,4,9]
l2 = [3,10,4]

sublist(l1, l2)
# True

edited Mar 26 '20 at 08:25

answered Mar 25 '20 at 21:35

kederrac

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I really only want to use loops and boolean logic, not built-in functions. – janissary Mar 25 '20 at 21:38
@MohamedTlili check now my answer, I hope is better – kederrac Mar 25 '20 at 21:40
You check only that all of l1 are in l2, they can be non-continuous and it wouldn't be a sublist – azro Mar 25 '20 at 21:41
@azro forget to mention that the sublist has to be continuous. For example [3,9,8] is not a sublist of [2,3,1,9,7,8] – janissary Mar 25 '20 at 21:42
@MohamedTlili ok, now is for sure better – kederrac Mar 25 '20 at 21:58
@kederrac Well yes, but my question was why my code isn't working. – janissary Mar 25 '20 at 21:59

DarrylG · Accepted Answer · 2020-03-25T22:38:20.293

It's erroroneous for your code to have a for loop followed by a while loop. Only for loop is needed to loop over blist.

So this is code is in error:

for n in range(0, len(blist)):
     while (n < len(blist) and m < len(blist)) and is_a_sublist == False:

Simplified Code

By simplyfying looping, removing superfluous variables and steps, your code can be simplified to the following:

def sublist(slist, blist):
  for n in range(len(blist)-len(slist)+1):
    if slist == blist[n:n+len(slist)]:
      return True
  return False

def main():
    test_list = ["a",9,1,3.14,2]
    original_list = ["a",1,"a",9,1,3.14,2]

    if sublist(test_list,original_list):
        print("It is a sublist!")
    else:
        print("It ain't a sublist!")

main()

Output

It is a sublist!

Explanation

Code:

for n in range(len(blist)-len(slist)+1):

# We're looking for a starting index between
# 0 and len(blist)-len(slist)
# The reason it can be at most 
# len(blist)-len(list) 
# is that's the max index that still could 
# have len(slist) elements
# range(len(blist)-len(slist)+1) provides
# numbers from 0 to en(blist)-len(slist)

   if slist == blist[n:n+len(slist)]:

 # Check if we have a match from n to
 # n + len(slist) - 1 (i.e. since slice
 # a:b means index of elements from a to b-1

Can you explain the logic behind the code in lines 2 - and 3? — janissary, Mar 25 '20 at 22:21
@MohamedTlili--added explanation. Let me know if you need additional information. — DarrylG, Mar 25 '20 at 22:38

Check if a given list is a sublist of a larger list

2 Answers2