List in another list

Question

def list(small,big):
    for i in range(len(big)):
        if small[i]!=big[i]:
            break

    else:
        return False

hi, i am new to python. I want to find if the numbers in a list are in another list. so [7,8,4] and [8,5,1] are subsequence of [7,8,5,4,1] and should return True, but [7,8,8] and [7,1,8] are not and should return False.

Your question is not clear. Can you provide an example with desired input and output? — manvi77, Mar 18 '17 at 15:40
Possible duplicate of [Testing if a list contains another list with Python](http://stackoverflow.com/questions/3847386/testing-if-a-list-contains-another-list-with-python) — Ajay Singh, Mar 18 '17 at 15:53
This is not a test of a sub-sequence inside a list. For example, 8,5,1 does not occur in 7,8,5,4,1. One must consider the '4' or any other number or group of numbers. — Brian O'Donnell, Mar 18 '17 at 16:07
You should add more requirements or description to your problem, does the order matter?, — Iron Fist, Mar 18 '17 at 16:26

repzero · Answer 1 · 2017-03-18T16:13:27.310

0

something like this would work Since you are a newbie and building on your function idea

def lister(small,big):
    for i in range(len(small)):
        if small[i] not in big:
            return False
    return True

print(lister([1,2,3],[1,2,3,4,5]))

define your function as lister since list is already a built in function in python

Edit: to Find a subsequence, It believe it can be done as follows:

def lister(small,big):
    for i in range(len(small)):
        if small[i] not in big:
            return False
    if  " ".join([str(x) for x in small]) in str(" ".join([str(y) for y in big])):
        return True
    else:
        return False


print(lister([3,5],[1,2,3,4,5]))

edited Mar 18 '17 at 16:13

answered Mar 18 '17 at 15:54

repzero

8,254
2
18
40

That gives True for print(lister([7,1,8],[7,8,5,4,1])). – Brian O'Donnell Mar 18 '17 at 15:57
thanks Sir. But actually, if i compare [1,2,2] with [1,2,3,4,5] , it returns True , as it is not a subsequence, so it should have been False – hrk23 Mar 18 '17 at 15:58
answer edited..I think comparing two sets will compare subsequences in specific order and out of order..... – repzero Mar 18 '17 at 16:14

score 0 · Answer 2 · answered Mar 18 '17 at 16:16

Here's a recursive approach. It works for all of your examples. Basically list1 is a subsequence of list2 if A) the first element of list1 is in list2 and B) the remainder of list1 is a subsequence of the remainder of list2.

def subsequence(list1, list2):
     if not list1:
             return True
     head, *tail = list1
     try:
             index = list2.index(head)
             return subsequence(tail, list2[index+1:])
     except ValueError:
             return False

score 0 · Answer 3 · answered Mar 18 '17 at 16:44

Here is another option:

def IsListInList(listA, listB):
    res = False
    for index,val in enumerate(listA):
        if (val in listB):
            listB = listB[listB.index(val)+1:]
            res = True
        else:
            res = False
            break
    return res

List in another list

3 Answers3