Using Numpy vectorization instead of parallelly looping over two list

Question

I have two lists or arrays of uneven length (let's say arr_a, arr_b). I want to form a third array/list of tuples where the first element of the tuple take the value of arr_a[1st element] - arr_b[1st element] and the second element of the tuple taking arr_a[1st element] - arr_b[2nd element] and then the next tuple taking (arr_a[2nd_element]-arr_b[2nd_element], arr_a[2nd_element] - arr_b[3rd_element]) and so on. The code below achieves this using a for loop. But I want to find out if this can be achieved through some sort NumPy vectorizing, broadcasting approach. please note arr_a and arr_b are of not equal length ( len(arr_b) = len(arr_a)+1)

arr_a = np.random.randint(low = 0,high = 10,size = 10)
arr_b = np.random.randint(low = 0,high = 10,size = 11)

out:
arr_a : [4 8 6 3 3 7 8 6 0 2]
arr_b : [8 8 9 6 1 6 1 5 8 3 1]

list_c = []
for i,j in zip(range(len(arr_a)),range(len(arr_b))):
    list_c.append((arr_a[i]-arr_b[i],arr_a[i]-arr_b[i+1]))

out:
list_c : [(-4, -4), (0, -1), (-3, 0), (-3, 2), (2, -3), (1, 6), (7, 3), (1, -2), (-8, -3), (-1, 1)]

Numpy isnt going to help if you want tuples

roganjosh
– roganjosh

2020-02-08 06:21:42 +00:00
Commented Feb 8, 2020 at 6:21 — roganjosh
– roganjosh, Commented Feb 8, 2020 at 6:21
I don't mind nested lists or arrays

imantha
– imantha

2020-02-08 06:23:52 +00:00
Commented Feb 8, 2020 at 6:23 — imantha
– imantha, Commented Feb 8, 2020 at 6:23

Sayandip Dutta · Accepted Answer · 2020-02-08 06:23:55Z

2

I think you can try zip with numpy:

>>> list(zip(arr_a - arr_b[:-1], arr_a - arr_b[1:]))
[(-4, -4),
 (0, -1),
 (-3, 0),
 (-3, 2),
 (2, -3),
 (1, 6),
 (7, 3),
 (1, -2),
 (-8, -3),
 (-1, 1)]

answered Feb 8, 2020 at 6:23

Sayandip Dutta

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Alexander · Accepted Answer · 2020-02-08 06:54:21Z

2

In order to keep the data in two columns rather than a tuple (which does not play well with numpy given that the data would be an object rather than a numeric dtype):

>>> (arr_a - np.array([arr_b[:-1], arr_b[1:]])).T
array([[-4, -4],
       [ 0, -1],
       [-3,  0],
       [-3,  2],
       [ 2, -3],
       [ 1,  6],
       [ 7,  3],
       [ 1, -2],
       [-8, -3],
       [-1,  1]])

answered Feb 8, 2020 at 6:54

Alexander

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