Reputation: 5635

Numpy Find Indices of Common Points in 2D Array

System

OS: Windows 10 (x64), Build 1909
Python Version: 3.8.10
Numpy Version: 1.21.2

Question

Given two 2D (N, 3) Numpy arrays of (x, y, z) floating-point data points, what is the Pythonic (vectorized) way to find the indices in one array where points are equal to the points in the other array?

(NOTE: My question differs in that I need this to work with real-world data sets where the two data sets may differ by floating point error. Please read on below for details.)

History

Very similar questions have been asked many times:

Previous Attempts

SO Post 1 provides a working list comprehension solution, but I am looking for a solution that will scale well to large data sets (i.e. millions of points):

Code 1:

import numpy as np

if __name__ == "__main__":
    big_array = np.array(
        [
            [1.0, 2.0, 1.2],
            [5.0, 3.0, 0.12],
            [-1.0, 14.0, 0.0],
            [-9.0, 0.0, 13.0],
        ]
    )

    small_array = np.array(
        [
            [5.0, 3.0, 0.12],
            [-9.0, 0.0, 13.0],
        ]
    )

    inds = [
        ndx
        for ndx, barr in enumerate(big_array)
        for sarr in small_array
        if all(sarr == barr)
    ]

    print(inds)

Output 1:

[1, 2]

Attempting the solution of SO Post 3 (similar to SO Post 2), but using floats does not work (and I suspect something using np.isclose will be needed):

Code 3:

import numpy as np

if __name__ == "__main__":
    big_array = np.array(
        [
            [1.0, 2.0, 1.2],
            [5.0, 3.0, 0.12],
            [-1.0, 14.0, 0.0],
            [-9.0, 0.0, 13.0],
        ],
        dtype=float,
    )

    small_array = np.array(
        [
            [5.0, 3.0, 0.12],
            [-9.0, 0.0, 13.0],
        ],
        dtype=float,
    )

    inds = np.nonzero(
        np.in1d(big_array.view("f,f").reshape(-1), small_array.view("f,f").reshape(-1))
    )[0]

    print(inds)

Output 3:

[ 3  4  5  8  9 10 11]

My Attempt

I have tried numpy.isin with np.all and np.argwhere

inds = np.argwhere(np.all(np.isin(big_array, small_array), axis=1)).reshape(-1)

which works (and, I argue, much more readable and understandable; i.e. pythonic), but will not work for real-world data sets containing floating-point errors:

import numpy as np

if __name__ == "__main__":
    big_array = np.array(
        [
            [1.0, 2.0, 1.2],
            [5.0, 3.0, 0.12],
            [-1.0, 14.0, 0.0],
            [-9.0, 0.0, 13.0],
        ],
        dtype=float,
    )

    small_array = np.array(
        [
            [5.0, 3.0, 0.12],
            [-9.0, 0.0, 13.0],
        ],
        dtype=float,
    )

    small_array_fpe = np.array(
        [
            [5.0 + 1e-9, 3.0 + 1e-9, 0.12 + 1e-9],
            [-9.0 + 1e-9, 0.0 + 1e-9, 13.0 + 1e-9],
        ],
        dtype=float,
    )

    inds_no_fpe = np.argwhere(np.all(np.isin(big_array, small_array), axis=1)).reshape(-1)

    inds_with_fpe = np.argwhere(
        np.all(np.isin(big_array, small_array_fpe), axis=1)
    ).reshape(-1)

    print(f"No Floating Point Error: {inds_no_fpe}")
    print(f"With Floating Point Error: {inds_with_fpe}")

    print(f"Are 5.0 and 5.0+1e-9 close?: {np.isclose(5.0, 5.0 + 1e-9)}")

Output:

No Floating Point Error: [1 3]
With Floating Point Error: []
Are 5.0 and 5.0+1e-9 close?: True

How can I make my above solution work (on data sets with floating point error) by incorporating np.isclose? Alternative solutions are welcome.

NOTE: Since small_array is a subset of big_array, using np.isclose directly doesn't work because the shapes won't broadcast:

np.isclose(big_array, small_array_fpe)

yields

ValueError: operands could not be broadcast together with shapes (4,3) (2,3)

Update

Currently, the only working solution I have is

inds_with_fpe = [
    ndx
    for ndx, barr in enumerate(big_array)
    for sarr in small_array_fpe
    if np.all(np.isclose(sarr, barr))
]

Upvotes: 0

Answers (3)

max9111

Reputation: 6482

As @Michael Anderson already mentioned this can be implemented using a kd-tree. In comparsion to your answer this solution is using an absolute error. If this is acceptable or not depends on the problem.

Example

import numpy as np
from scipy import spatial

def find_nearest(big_array,small_array,tolerance):
    tree_big=spatial.cKDTree(big_array)
    tree_small=spatial.cKDTree(small_array)
    return tree_small.query_ball_tree(tree_big,r=tolerance)

Timings

big_array=np.random.rand(100_000,3)
small_array=np.random.rand(1_000,3)
big_array[1000:2000]=small_array

%timeit find_nearest(big_array,small_array,1e-9) #find all pairs within a distance of 1e-9
#55.7 ms ± 830 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

#A. Hendry
%timeit np.argwhere(np.isclose(small_array, big_array[:, None, :]).all(-1).any(-1)).reshape(-1)
#3.24 s ± 19 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Upvotes: 2

adam.hendry

Reputation: 5635

Credit to @AndrasDeak for this answer

The following code snippet

inds_with_fpe = np.argwhere(
    np.isclose(small_array_fpe, big_array[:, None, :]).all(-1).any(-1)
).reshape(-1)

will make the code work. The corresponding output is now:

No Floating Point Error: [1 3]
With Floating Point Error: [1, 3]
Are 5.0 and 5.0+1e-9 close?: True

None in the above creates a new axis (same as np.newaxis). This changes the shape of the big_array array to (4, 1, 3), which adheres to broadcasting rules and permits np.isclose to run. That is, big_array is now a set of 4 1 x 3 points, and since one of the axes in big_array is 1, small_array_fpe can be broadcast to 2 1 x 3 arrays (i.e. shape (2, 1, 3)) and the elements can be compared element-wise.

The result is a (4, 2, 3) boolean array; every element of big_array is compared element-wise to every element of small_array_fpe and the components where they are close (within a specific tolerance) is returned. Since all is called as an object method rather than a numpy function, the first argument to the function is actually the axis rather than the input array. Hence, -1 in the above functions means "the last axis of the array".

We first return the indeces of the (4, 2, 3) array that are all True (i.e. all (x, y, z) components are equal), which yields a 4 x 2 array. Where any of these are True is the corresponding index in big_array where the points are equal, yielding a 4 x 1 array.

argwhere returns indices grouped by element, so its shape is normally (number nonzero items, num dims of input array), hence we flatten it into a 1d array with reshape(-1).

Unfortunately, this requires a quadratic amount memory w.r.t. the number of points in each array, since we must run through every element of big_array and check it against every element of small_array_fpe. For example, to search for 10,000 points in a set of another 10,000 points, for 32-bit floating point data, requires

Memory = 10000 * 10000 * 4 * 8 = 32 GiB RAM!

If anyone can devise a solution with a faster run time and reasonable amount of memory, that would be fantastic!

FYI:

from timeit import timeit

import numpy as np

big_array = np.array(
    [
        [1.0, 2.0, 1.2],
        [5.0, 3.0, 0.12],
        [-1.0, 14.0, 0.0],
        [-9.0, 0.0, 13.0],
    ],
    dtype=float,
)

small_array = np.array(
    [
        [5.0 + 1e-9, 3.0 + 1e-9, 0.12 + 1e-9],
        [10.0, 2.0, 5.8],
        [-9.0 + 1e-9, 0.0 + 1e-9, 13.0 + 1e-9],
    ],
    dtype=float,
)


def approach01():
    return [
        ndx
        for ndx, barr in enumerate(big_array)
        for sarr in small_array
        if np.all(np.isclose(sarr, barr))
    ]


def approach02():
    return np.argwhere(
        np.isclose(small_array, big_array[:, None, :]).all(-1).any(-1)
    ).reshape(-1)


if __name__ == "__main__":
    time01 = timeit(
        "approach01()",
        number=10000,
        setup="import numpy as np; from __main__ import approach01",
    )
    time02 = timeit(
        "approach02()",
        number=10000,
        setup="import numpy as np; from __main__ import approach02",
    )

    print(f"Approach 1 (List Comprehension): {time01}")
    print(f"Approach 2 (Vectorized): {time02}")

Output:

Approach 1 (List Comprehension): 8.1180582
Approach 2 (Vectorized): 0.9656997

Upvotes: 0

Michael Anderson

Reputation: 73490

I'm not going to give any code, but I've dealt with problems similar to this on a large scale. I suspect that to get decent performance with either of these approaches you'll need to implement the core in C (you might get away with using numba).

If both your arrays are huge there are a few approaches that can work. Primarily these boil down to building a structure that can be used to find the nearest neighbor of a point from one of the arrays, and then querying it for each point in the other data set.

To do this I've previously used a Kd Tree approach, and a grid based approach.

The basis of the grid based approach is

find the 3D extents of your first array.
split this region into LNM bins.
For each input point in the second array, find its bin. Any point that matches it will be in that bin.

The edge cases you need to handle are

if the point falls on the edge of a bin, or close enough to the boundary of a bin that points considered equal to it might fall in the other bin - then you need to search more than one bin for its "equal".
if the point falls outside all the bins, but close to the edge, points "equal" to it might fall in a nearby bin.

The downsides are that this is bad for data that is not uniformly distributed.

The upside is that it is relatively simple. Expected run time for uniform data is n1 * n2 / (L*N*M) (compared to n1*n2). Typically you select L,N,M such that this becomes O(n log(n)). You also get some further uplift from sorting the second array to improve reuse of the bins. It is also relatively easy to parallelize (both the binning and searching)

The K-d Tree approach is similar. IIRC it gives O(n log(n)) behavior, but it is trickier to implement, and the building of the data structure is tricky to parallelize). It tends to not be as cache friendly which can mean that although its asymptotic run-time is better than the grid based approach it can runs slower in practice. However it does give better guarantees for non-uniformly distributed data.