python sum all next n values in array at each index

This should do the trick using slices.

import numpy as np

sum_value = 0

my_array = [1, 13, 6, 100, 12,23,45]
summed_array = [0, 0, 0, 0, 0,0,0]
n = 3;
print(len(my_array))
for i in range(len(summed_array)):
    summed_array[i] = sum(my_array[i+1:i+1+n])
print(summed_array)

With a being your array:

>>> c = a.cumsum()
>>> np.concatenate((c[3:], [a.sum()] * 3)) - c
array([119, 118, 135,  80,  68,  45,   0])

You can also do it in list comprehension:

sum_array = [ sum(my_array[pos+1:pos+4]) for pos in range(len(my_array)) ]

This way there is no need for declaring the sum_array, as it will always be already created with the correct size.

Edit: Fixed the 'next 3 values', since I hadn't realised it in the first place.

If you are trying to minimize the number of additions performed, it might be a good idea to have a pointer so that you add and subtract once for each index instead of adding 3 times. You might find a different solution more optimum than this though. If so, please let me know :) Let a be your array.

# Initialize sum array sa with first index
val = sum(a[1:4])
sa = [val]
i = 1

# Get sum for all except last 3 indices
for i in range(1, len(a) - 3):
  val = val + a[i + 3] - a[i]
  sa.append(val)

# Account for the last 3 indices
while i < len(a):
  val -= a[i]
  sa.append(val)
  i += 1

# sa is the array needed at this point

Please note that empty array would currently return [0]. If you intend to return an empty array instead, that would be an edge case that can be handled by an if statement in the beginning