Modifying numpy array to get minimum number of values between elements

Question

I have a numpy array of the form: arr = 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1

I would like to modify it such that there are atleast seven 0s between any two 1s. If there are less than seven 0s, then convert the intervining 1's to 0. I am thinking that numpy.where could work here, but not sure how to do it in a succint, pythonic manner:

The output should look like this:

0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

numpy.where(arr[:] > 1.0, 1.0, 0.0)

Answer 1

The following code is a really ugly hack, but it gets the job done in linear time (assuming 7 is fixed) without resorting to Python loops and without needing anything like Numba or Cython. I don't recommend using it, especially if 7 might be 700 next month.

def rolling_window(a, window):
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)

arr2 = numpy.append(1-arr, [0]*7)
numpy.power.at(rolling_window(arr2[1:], 7), np.arange(len(arr)), arr2[:-7, None])
arr = 1 - arr2[:-7]

It works by setting 1s to 0s and vice versa, then for each element x , setting each element y in the next 7 spots to y**x , then undoing the 0/1 switch. The power operation sets everything within 7 spaces of a 0 to 1, in such a way that the effect is immediately visible to power operations further down the array.

Answer 2

Now this is just a simple implementation using for loops and ifs but I am pretty sure it can be condensed.(a lot!) And yeah there's no need to do Numpy for this, it will only complicate things for you.

question = [0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1]
result = [0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]
indexesOf1s = []
for index,item in enumerate(question):     #Here just calculate all the index of 1s
if item == 1:
    indexesOf1s.append(index)
for i in indexesOf1s:               #Iterate over the indexes and change acc to conditions
    sub = i - indexes[indexes.index(i)-1]
    if sub>0 and sub>=7:
        question[i] = 1
    elif sub>0:
        question[i] = 0
print question 
print result