I have these two arrays:
A = [1,2,3,4,5,6,7,8,9,0]
And:
B = [4,5,6,7]
Is there a way to check if B is a sublist in A with the same exact order of items?
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2list order is to be preserved, right?Harry– Harry2013-12-26 18:32:11 +00:00Commented Dec 26, 2013 at 18:32
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The order of the elements of B is the same as the part that includes it in A, if is that your question.iam_agf– iam_agf2013-12-26 18:34:14 +00:00Commented Dec 26, 2013 at 18:34
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I believe he's asking does it matter what order the elements are in? For example... Would B=[4,6,5,7] be just as good, or does order matter?Ryan O'Donnell– Ryan O'Donnell2013-12-26 18:36:22 +00:00Commented Dec 26, 2013 at 18:36
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2The order of the elements in A and B matters. Is like see if B is a cut of the list A.iam_agf– iam_agf2013-12-26 18:42:55 +00:00Commented Dec 26, 2013 at 18:42
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Any ideas if I wanna get the rest array, like C = A - B = [1,2,3,8,9,0]？pandalai– pandalai2021-05-19 06:59:32 +00:00Commented May 19, 2021 at 6:59
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8 Answers
issubset should help you
set(B).issubset(set(A))
e.g.:
>>> A= [1,2,3,4]
>>> B= [2,3]
>>> set(B).issubset(set(A))
True
edit: wrong, this solution does not imply the order of the elements!
How about this:
A = [1,2,3,4,5,6,7,8,9,0]
B = [4,5,6,7]
C = [7,8,9,0]
D = [4,6,7,5]
def is_slice_in_list(s,l):
len_s = len(s) #so we don't recompute length of s on every iteration
return any(s == l[i:len_s+i] for i in xrange(len(l) - len_s+1))
Result:
>>> is_slice_in_list(B,A)
True
>>> is_slice_in_list(C,A)
True
>>> is_slice_in_list(D,A)
False
Comments
Using slicing:
for i in range(len(A) - len(B)):
if A[i:i+len(B)] == B:
return True
return False
Something like that will work if your A is larger than B.
1 Comment
alko
That would work even if A is smaller than B, the only case you should care of is equal lengths. That note shows that you probably should use
range(len(A)-len(B)+1) instead.I prefer to use index to identify the starting point. With this small example, it is faster than the iterative solutions:
def foo(A,B):
n=-1
while True:
try:
n = A.index(B[0],n+1)
except ValueError:
return False
if A[n:n+len(B)]==B:
return True
Times with this are fairly constant regardless of B (long, short, present or not). Times for the iterative solutions vary with where B starts.
To make this more robust I've tested against
A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1]
which is longer, and repeats values.
Comments
You can use scipy.linalg.hankel to create all the sub-arrays in one line and then check if your array is in there. A quick example is as follows:
from scipy import linalg
A = [1,2,3,4,5,6,7,8,9,0]
B = [4,5,6,7]
hankel_mat = linalg.hankel(A, A[::-1][:len(B)])[:-1*len(B)+1] # Creating a matrix with a shift of 1 between rows, with len(B) columns
B in hankel_mat # Should return True if B exists in the same order inside A
This will not work if B is longer than A, but in that case I believe there is no point in checking :)
Comments
import array
def Check_array(c):
count = 0
count2 = 0
a = array.array('i',[4, 11, 20, -4, -3, 11, 3, 0, 50]);
b = array.array('i', [20, -3, 0]);
for i in range(0,len(b)):
for j in range(count2,len(a)):
if a[j]==b[i]:
count = count + 1
count2 = j
break
if count == len(b):
return bool (True);
else:
return bool (False);
res = Check_array(8)
print(res)
1 Comment
Community
As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
By subverting your list to string, you can easily verify if the string "4567" is in the string "1234567890".
stringA = ''.join([str(_) for _ in A])
stringB = ''.join([str(_) for _ in B])
stringB in stringA
>>> True
Dressed as a one line (cause is cooler)
isBinA = ''.join([str(_) for _ in B]) in ''.join([str(_) for _ in A])
isBinA
>>> True
1 Comment
Lenormju
Does not work for some inputs :
A = [45] and B = [4,5] for example.A = [1,2,3,4,5,6,7,8,9,0]
B = [4,5,6,7]
(A and B) == B
True
