Can itertools be used for an unspecified number of dimensions in this case?

Question

This is my code, working with dim=3, but I would like it to work for any dimensionality without having to manually edit code.

I would like to be able to vary the dimensionality between 3 and 20 eventually without manually having to ad for-loops. I was looking at itertools, but don't know how to select the correct values from the tuples created by itertools.product() to square and add up for my if statement.

arrayshape = (width * 2 + 1,) * dim
funcspace = np.zeros(shape=arrayshape, dtype='b')

x1 = list(range(-int(width), int(width + 1)))
x2 = x1
x3 = x1

for i in range(len(x1)):
    for j in range(len(x2)):
        for k in range(len(x3)):
            if round(np.sqrt(x1[i] ** 2 + x2[j] ** 2 + x3[k] ** 2)) in ranges:
                funcspace[i][j][k] = 1

Answer 1

You can use product on enumerate of your vectors, which will yield the value and the index:

for ((i,v1),(j,v2),(k,v3)) in itertools.product(enumerate(x1),enumerate(x2),enumerate(x3)):
    if round(np.sqrt(v1**2+v2**2+v3**2)) in ranges:
            funcspace[i][j][k]=1

as a bonus, you get rid of the unpythonic range(len()) construct.

I've cooked a more general case when you have a vector of vectors. It's a little harder to read because unpacking isn't done in the for loop.

The square sum is done using sum on the 1 indexes (the values), and if the condition matches, we loop until we find the "deeper" list to set the value to 1 .

for t in itertools.product(*(enumerate(x) for x in x_list)):
    # compute the squared sum of values
    sqsum = sum(v[1]**2 for v in t)
    if round(sqsum) in ranges:
        # traverse the dimensions except the last one
        deeper_list = funcspace
        for i in range(len(t)-1):
            deeper_list = deeper_list[t[i][0]]
        # set the flag using the last dimension list
        deeper_list[t[-1][0]] = 1

as noted in comments, since x1 seems to be repeated you can replace the first statement by:

for t in itertools.product(enumerate(x1), repeat=dim):

Another comments states that since funcspace is a numpy ndarray , we can simplify the "set to 1" loop by passing the list of indexes:

funcspace[[x[0] for x in t]] = 1