[FEA]: Need a libcuspatial count_equal_points(lhs, rhs) function.

[thomcom]

Is this a new feature, an improvement, or a change to existing functionality?

New Feature

How would you describe the priority of this feature request

Critical (currently preventing usage)

Please provide a clear description of problem you would like to solve.

As I've been implementing binary predicates I've discovered that there are plentiful cases wherein I need the ability to count the number of points in one buffer that are equal to the number of points in another buffer. While numerous equality comparisons are possible and fast in python, I can't without iteration compare a series of points, pairwise,with all other points in a buffer for equality.

For example,

mp1 = [0, 1, 3, 2, 4, 5, 7, 6]
mp2 = [0, 1, 2, 3, 4, 5, 6, 7, 4, 5]
result = count_equal_points(mp1, mp2)
print(result)
1
0
2
0

This feature is useful for many binary predicates, and can't be trivially implemented in python.

Describe any alternatives you have considered

No response

Additional context

No response

[isVoid]

Not sure if I understand the input.

x = [0, 1, 3, 2, 4, 5, 7, 6]
y = [0, 1, 2, 3, 4, 5, 6, 7, 4, 5]
result = count_equal_points(x, y)
print(result)
1
0
2
0

• x and y doesn't have the same number of input. Is this x/y coordinate?
• Can you explain what does 1 0 2 0 stand for?

[thomcom]

The output is an allpairs count equals result:

1 = x[0:2] == y[0:2]
0 = x[2:4] != any in y
2 = x[4:6] == y[4:6] and y[8:10]
0 = x[7:8] != any in y

One implementation pseudocode might be:

for each even number in len(x) as i:
    for each even number in len(y) as j:
        if x[i] == y[i] and x[i+1] == y[i+1]:
            result[i % 2]++

[thomcom]

These are interleaved xy buffers.

[isVoid]

So this looks like an all pairs equals count for two point arrays, or two multipoints. Do you just need this for single point array or multipoint array as well?

[thomcom]

I only need it for single point array, I can convert any other geometry type into that as needed.

[harrism]

So the X and y arrays are not x- and y-coordinates? Can you make the original request example less confusing by not naming them X and y?

[harrism]

Do you want the count of matches of each point in the LHS to points of the RHS? Or do you just want true/false if each point in the LHS exists in the RHS? Two different algorithms with different costs...

[thomcom]

No, I can see how that is confusing. mp1 and mp2 are multipoint arrays with interleaved xy coordinates.

I'm not sure yet if I need the count of matches or if I just need to know if any point in mp1 exists in mp2. I think I need both, but I'm working through implementations and haven't covered them all yet. One example is

mp1.touches(mp2), where I only need to know if any point in mp1 is equal to any point in mp2.

Another example is

mp1.equals(mp2), where I need to know if every point in mp1 is equal to every point in mp2.

The second example can be implemented easily in python, but the former example can't be implemented without pythonic iteration.

[isVoid]

mp1.touches(mp2), where I only need to know if any point in mp1 is equal to any point in mp2.

For this I think you only need a True and False test? You don't need to create a counting function.

[harrism]

I would think Thrust set operations would be useful for these.