Vectorizing a quadruple for loop in python

[paul18fr]

I was not able to visualize what would be my boolean condition since both inner_point and outer_point are tuples

np.asarray might help you

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import numpy as np   MyTuple = (1, 2, 3, 0, 0, 100) print(f"MyTyple type = {type(MyTuple)}")   MyTupleConvertedIntoArray = np.asarray(MyTuple) print(f"MyTupleConvertedIntoArray type = {type(MyTupleConvertedIntoArray)}") print(f"MyTupleConvertedIntoArray = {MyTupleConvertedIntoArray}")

Output:
MyTyple type = <class 'tuple'>
MyTupleConvertedIntoArray type = <class 'numpy.ndarray'>
MyTupleConvertedIntoArray = [  1   2   3   0   0 100]

[winash12]

Great tips.

Presuming we can continue the discussion two issues come up in trying unroll the for loops

1) How to vectorize the creation of outer and inner points ?

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args = [] for i in range (0,100):     for j in range(0,200):         outer_point = [i,j]         op = np.asarray(outer_point)         args.append(op) opC = np.concatenate(args,axis=0)

2) The shape of collection of outer points and inner points is different .

set1d

Use setdiff1d to calculate the diff of outer and inner points and store the difference as a boolean array.

[paul18fr]

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import numpy as np import time   n = 1_000 m = 2_000   t0 = time.time() args = [] for i in range (0,n):     for j in range(0,m):         outer_point = [i,j]         op = np.asarray(outer_point)         args.append(op) opC = np.concatenate(args,axis=0) t1 = time.time() print(f"current snipper = {t1 - t0}")       t2 = time.time() i = np.ones(m) j = np.arange(n) ind1 = np.kron(j, i).reshape(-1, 1)   k = np.ones(n) l = np.arange(m) ind2 = np.kron(k, l).reshape(-1, 1)   op = np.hstack((ind1, ind2)) op = op.flatten(order='C') t3 = time.time() print(f"new snipper = {t3 - t2}") print(f"with n = {n} and m = {m}, vectorization ratio is X {(t1 - t0) / (t3 - t2):.1f}")   Check = np.array_equal(opC, op) if Check: print("Both arrays are identical!!!")

Output:
current snipper = 2.0446
new snipper = 0.0309
with n = 1000 and m = 2000, vectorization ratio is X 66.2
Both arrays are identical!!!

[Pedroski55]

As I see it, all points (i, j) up to i = 9, j = 19 will be in (k, l), because you start each loop at zero.

If your condition is: inner_point != outer_point, I think you just need this:

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for k in range(10,30):      for l in range(20,40):                  inner_point = [k,l]                  print('the x difference is', 9-k)          print('the y difference is', 19-l)