python - Comparing two lists to see if one occurs in another consecutively

Question

I've been trying to make a function that can take two lists of any size (say, list A and list B) and sees if list B occurs in list A, but consecutively and in the same order. If the above is true, it returns True, else it'll return False

e.g.

A:[9,0,**1,2,3,4,5,6,**7,8] and B:[1,2,3,4,5,6] is successful

A:[1,2,0,3,4,0,5,6,0] and B:[1,2,3,4,5,6] is unsuccessful.

A:[1,2,3,4,5,6] and B [6,5,3,2,1,4] fails because despite having the same 
 numbers, they aren't in the same order

I've tried doing this using nested loops so far and am a bit confused as to where to go

Might want to take a look at https://stackoverflow.com/questions/10106901/elegant-find-sub-list-in-list — Pedro von Hertwig Batista, Dec 27 '17 at 14:07
Is this what you need: https://stackoverflow.com/questions/16579085/python-verifying-if-one-list-is-a-subset-of-the-other — Pritesh Ranjan, Dec 27 '17 at 14:07

Albin Paul · Answer 1 · 2017-12-27T14:29:56.357

i converted the entire list into a string and then found a substring of that string

the list when converted to a string it becomes

str(a)='[9,0,1,2,3,4,5,6,7,8]'

which when when we strip the string becomes

str(a).strip('[]')='9,0,1,2,3,4,5,6,7,8'

Now the problem just converted to

checking if there is a substring in the the string so we can us the in operator to check the substring

The solution

a=[9,0,1,2,3,4,5,6,7,8]
b=[1,2,3,4,5,6]
print(str(b).strip('[]') in str(a).strip(']['))

testcase1

testcase2

score 0 · Answer 2 · answered Dec 27 '17 at 14:16

if your arrays are not huge and if you can find a way to map each element in your array to a string you can use:

list1 = [9,0,1,2,3,4,5,6,7,8]
list2 = [1,2,3,4,5,6]

if ''.join(str(e) for e in list2) in ''.join(str(e) for e in list1):
    print 'true'

it just make two string from the lists and than use 'in' to find any accorence

score 0 · Answer 3 · answered Dec 27 '17 at 14:17

0

Use any function

any(A[i:i+len(B)] == B  for i in range(len(A) - len(B) + 1))

demo

answered Dec 27 '17 at 14:17

splash58

26,043
3
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34

score 0 · Accepted Answer · answered Dec 27 '17 at 14:33

0

Just try this:

L1 = [9,0,1,2,3,4,5,6,7,8]
L2 = [1,2,3,4,5,6]
c = 0
w = 0
for a in range(len(L2)):
   for b in range(w+1, len(L1)):
      if L2[a] == L1[b]:
        c = c+1
        w = b
        break
      else:
        c = 0
    if c == len(L2):
       print('yes')
       break

Here you check if the element of l2 is in l1 and if so breaks the first loops remember where you left and of the next element of l2 is the same as the next element of l1 and so on.

And the last part is to check if this happened as much times as the length of l2. if so then you know that the statement is correct!

answered Dec 27 '17 at 14:33

Seppe Mariën

355
1
13

Thank you. I was trying something alongside with the captions but couldn't wrap my head around what the indexes ought to be. – John Diamond Dec 27 '17 at 19:57
No problem if this works then please accept the answer :) – Seppe Mariën Dec 27 '17 at 21:12

score 0 · Answer 5 · answered Dec 27 '17 at 15:37

0

Try this:

L1 = [9,2,1,2,0,4,5,6,7,8]
L2 = [1,2,3,4,5,6]
def sameorder(L1,L2):
    for i in range(len(L1)-len(L2)+1):
        if L1[i:len(L2)+i]==L2:
            return True
    return False

answered Dec 27 '17 at 15:37

Artier

1,648
2
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22

score 0 · Answer 6 · answered Dec 27 '17 at 21:58

You can create sublists of a that can be analyzed:

def is_consecutive(a, b):
   return any(all(c == d for c, d in zip(b, i)) for i in [a[e:e+len(b)] for e in range(len(a)-len(b))])

cases = [[[9, 0, 1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 6]], [[1, 2, 0, 3, 4, 0, 5, 6, 0], [1, 2, 3, 4, 5, 6]], [[1, 2, 3, 4, 5, 6], [6, 5, 3, 2, 1, 4]]]
final_cases = {"case_{}".format(i):is_consecutive(*a) for i, a in enumerate(cases, start=1)}

Output:

{'case_3': False, 'case_2': False, 'case_1': True}

python - Comparing two lists to see if one occurs in another consecutively

6 Answers6