Looping 2 1d-arrays to create 2d array in numpy

Question

I have to 1d arrays

x = [1,2,3,4,5]

y = [5,6,7,8,9]

and a zero 2d array

2d_array=np.zeros((5, 5))

I have this equation : 50*x + 20*y

I want to make a loop to find all possible answers from x and y and store them in the 2d_array

[0,0,0,0,0

 0,0,0,0,0

 0,0,0,0,0

 0,0,0,0,0

 0,0,0,0,0]

so this should be

[50*x[0]+20*y[0],50*x[1]+20*y[0],50*x[2]+20*y[0],50*x[3]+20*y[0],50*x[4]+20*y[0]

 50*x[0]+20*y[1],50*x[1]+20*y[1]50*x[2]+20*y[1],50*x[3]+20*y[1],50*x[4]+20*y[1].......

And so on, I'm not sure if the explanation is clear, but if it is not tell me and I'll upload the actual file of the problem.

Thanks

Answer 1

You can perform your calculation in a vectorised fashion:

x = np.array([1,2,3,4,5])
y = np.array([5,6,7,8,9])

res = 50*x + 20*y[:, None]

array([[150, 200, 250, 300, 350],
       [170, 220, 270, 320, 370],
       [190, 240, 290, 340, 390],
       [210, 260, 310, 360, 410],
       [230, 280, 330, 380, 430]])

Here, the indexer [:, None] change the shape of y from (1,) to (5, 1). Broadcasting then allows creation of a 2d array result.

You can expect a performance improvement versus a method based on list comprehension, although you may only notice this for larger input arrays:

%timeit 50*x + 20*y[:, None]                                          # 10.1 µs
%timeit np.array([x_*50+y_*20 for y_ in y for x_ in x]).reshape(5,5)  # 30.2 µs

Answer 2

You don't need to pre-make the 2-d array, and you can do it all in this list comprehension:

np.array([x_*50+y_*20 for y_ in y for x_ in x]).reshape(5,5)

Which returns:

array([[150, 200, 250, 300, 350],
       [170, 220, 270, 320, 370],
       [190, 240, 290, 340, 390],
       [210, 260, 310, 360, 410],
       [230, 280, 330, 380, 430]])