Python: Count changes in an array

Question

I have an Array of zeros and ones which looks like this:

mat = [0,0,0,1,1,0]

I need to write down a new array of positions where a zero/one starts and ends:

mat_help= [[0,3], [4,5], [6,6]]

I thought I could accomplish this Task by simply loop. But then I got stucked: how can I mark the beginning of each zero?

In principle, I've ended up with this Code which Looks ugly and where I still should add a nestedness to the final list:

mat=[0, 0, 0, 1, 1, 0, 0]
matrix_length = len(mat)
mat_help=[0]
for i in range(len(mat)-1):
    if mat[i]==mat[i+1]:
        continue
    elif mat[i]!=mat[i+1]:
        mat_help.append(i)
        mat_help.append(i+1)
mat_help.append(matrix_length-1)

Maybe, there is a fast pythonic way of doing such things than just using several Loops...?

Please share the code you wrote. Also the array has only 6 elements whereas the positions in new_arr are 7 (from 0 to 6). Not sure if you know the problem statement very well — Anurag
– Anurag, Commented Jul 22, 2019 at 8:53
new_arr contains 6 as an index and yet it will raise IndexError given the length of the list is 6. — Chris
– Chris, Commented Jul 22, 2019 at 8:56
The expected output is inconsistent with the input. It should either be [0, 3], [3, 5], [5, 6] if the upper ends of the ranges are exclusive, or [0, 2], [3, 4], [5, 5] if they are inclusive, or [1, 3], [4, 5], [6, 6] if you start counting the indexes from 1, not 0 as is conventional. Please think more about what you want exactly, then the solution will probably follow naturally. — mkrieger1
– mkrieger1, Commented Jul 22, 2019 at 8:56

Alex_6 · Accepted Answer · 2019-07-22 09:31:20Z

2

I'm not sure I understand what you're searching for, but you could try this :

def sep_groups(arr):
   if not arr:
      return []
   res = []
   start = 0
   type = arr[0]
   index = 0
   for current, next in zip(arr[:-1], arr[1:]):
      if next != type:
         res.append([start, index])
         start = index + 1
         type = 1 - type
      index += 1
   res.append([start, len(arr)-1])
   return res

I hope that will help you !

answered Jul 22, 2019 at 9:31

Alex_6

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