Given a list like:
A = [18, 7, 0, 0, 0, 9, 12, 0, 0, 11, 2, 3, 3, 0, 0, 7, 8]
is there a simple way to create subarrays, with those elements that are separated by zeros (or at least by NaNs)? I mean, like:
A1 = [18, 7]
A2 = [9, 12]
A3 = [11, 2, 3, 3]
A4 = [7, 8]
I've written:
q=0
for i in range(0,len(A)):
if A[i]-A[i-1] < 1:
q=q+1
to retrieve the number of zeros-packets present in the list. But I need to populate subarrays, as long as I encounter them through the list... maybe something with the split function? Thank you in advance.
Well, itertools has the solution for you: groupby(list, filter).
If you want to group by zeroes, start with doing:
B = itertools.groupby(A, lambda x:x == 0)
The lambda expression "decides" which of the values should be a separator. You could separate by Nones using lambda x: x == None (for example). That will return you an iterable object. So, using a list comprehension, let's iterate through it (every iteration gives us a 2 values tuple):
C = [(i, list(j)) for i, j in B] # j is cast to a list because it's originally an object, not a list.
Output will be something like:
[(False, [18, 7]), (True, [0]), (True, [0]), (True, [0]), ... ]
Now, every list j that is the separator has a value True for i. So we can filter it:
C = [list(j) for i, j in B if not i]
Now, the result is a 2d list:
[[18, 7], [9, 12], [11, 2, 3, 3], [7, 8]]
So a one liner function:
def splitArr():
return [list(j) for i, j in itertools.groupby(A, lambda x:x == 0) if not i]