create new column based on condition

[paul18fr]

I'm using here a"vectorized" form instead of using loops; vectorization is much faster as you'll notice.

Remarks:

• The Matrix array is just an example, you can useyour own A,B,C instead
  
• D is intialized to avoid (memory) dynamic allocation; it's a good practise
  
• i is the "index" vector (from 1 to (r-1))
  
• D is obvious here - based on your formula
  
• finally the first row corresponds to index i=0
  
• if your working only with integer, it might have been more relevant in my previous post to directly define the dtype for D/D2
  
• since the first row uses a different formula, so it can be calculated before of after the main block
  

Test the following codes to compare the vectorization and (classical) loops => you'll figured out how it works.

Hope it helps

import numpy as np
import time

Nmax = 10_000
r, c = 1_000_000, 3

# Matrix = np.random.randint(1, Nmax, size=(r,c))
Matrix = np.arange(r*c).reshape(r, c)

# for all the matrix except the first row
t0=time.time()
D = np.zeros((r, 1), dtype=int)
i = np.arange(1, r)
D[i, 0] = Matrix[i-1, 2] - Matrix[i, 0] + Matrix[i, 1]

# specifically for the first row
i = 0
D[i, 0] = Matrix[i, 2] + Matrix[i, 0] - Matrix[i, 1]
t1 = time.time()
print(f"Duration1 = {t1-t0}")


# using loops
t2=time.time()
D2 = np.zeros((r, 1), dtype=int)
for i in range(1, r):
    D2[i, 0] = Matrix[i-1, 2] - Matrix[i, 0] + Matrix[i, 1]
D2[0, 0] = Matrix[0, 2] + Matrix[0, 0] - Matrix[0, 1]
t3=time.time()

print(f"D equals D2? => {np.array_equal(D, D2)}")
print(f"Duration2 = {t3-t2}")
print(f"Duration's ratio = {(t3-t2)/(t1-t0)}")