How to get the K most distant points, given their coordinates?

Question

We have boring CSV with 10000 rows of ages (float), titles (enum/int), scores (float), ....

• We have N columns each with int/float values in a table.
• You can imagine this as points in ND space
• We want to pick K points that would have maximised distance between each other.

So if we have 100 points in a tightly packed cluster and one point in the distance we would get something like this for three points: or this

For 4 points it will become more interesting and pick some point in the middle.

So how to select K most distant rows (points) from N (with any complexity)? It looks like an ND point cloud "triangulation" with a given resolution yet not for 3d points.

I search for a reasonably fast approach (approximate - no precise solution needed) for K=200 and N=100000 and ND=6 (probably multigrid or ANN on KDTree based, SOM or triangulation based..).. Does anyone know one?

Glove

I think this is an NP-Hard problem. I don't think that there is a reasonable way to calculate what you're asking for without looping over all possible pairs. If you want a point that maximizes the distances to all other points, then you can use pdist, sum, and sort. – Clavicembalo

No, that's not the case. But it's a non-trivial problem. – Phyla

your first sketch doesn't look correct to me. It should look more like the second with clusters merged – Simoneaux

would an approximate solution be acceptable? – Simoneaux

@WalterTross: Yes it will, and seems to be the only realistic option (in terms of processing time) – Glove

maybe flothesof.github.io/farthest-neighbors.html will help you – Catechin

I'm curious as to how you solved your problem in the end – Simoneaux

Answer 1

From past experience with a pretty similar problem, a simple solution of computing the mean Euclidean distance of all pairs within each group of K points and then taking the largest mean, works very well. As someone noted above, it's probably hard to avoid a loop on all combinations (not on all pairs). So a possible implementation of all this can be as follows:

import itertools
import numpy as np
from scipy.spatial.distance import pdist

Npoints = 3 # or 4 or 5...
# making up some data:
data = np.matrix([[3,2,4,3,4],[23,25,30,21,27],[6,7,8,7,9],[5,5,6,6,7],[0,1,2,0,2],[3,9,1,6,5],[0,0,12,2,7]])
# finding row indices of all combinations:
c = [list(x) for x in itertools.combinations(range(len(data)), Npoints )]

distances = []
for i in c:    
    distances.append(np.mean(pdist(data[i,:]))) # pdist: a method of computing all pairwise Euclidean distances in a condensed way.

ind = distances.index(max(distances)) # finding the index of the max mean distance
rows = c[ind] # these are the points in question

Boutique

It works yet consumes way too much memory on c = [list(x) for x in itertools.combinations(range(len(data)), Npoints )] line. Is there a way to make it more iterative? – Glove

I know.. When the N of rows is very big + multidimensional + K is larger than 2, a greedy search takes a long time. You can loop on the itertool object directly avoiding the list comprehensions but that would not necessarily reduce run time. What you asked is a non-trivial open question.. It reminds the travelling salesman problem a little bit (though it's not the same!). See a nice direction for a different solution (in R) here: larssonjohan.com/post/2016-10-30-farthest-points . – Boutique

Answer 2

I propose an approximate solution. The idea is to start from a set of K points chosen in a way I'll explain below, and repeatedly loop through these points replacing the current one with the point, among the N-K+1 points not belonging to the set but including the current one, that maximizes the sum of the distances from the points of the set. This procedure leads to a set of K points where the replacement of any single point would cause the sum of the distances among the points of the set to decrease.

To start the process we take the K points that are closest to the mean of all points. This way we have good chances that on the first loop the set of K points will be spread out close to its optimum. Subsequent iterations will make adjustments to the set of K points towards a maximum of the sum of distances, which for the current values of N, K and ND appears to be reachable in just a few seconds. In order to prevent excessive looping in edge cases, we limit the number of loops nonetheless.

We stop iterating when an iteration does not improve the total distance among the K points. Of course, this is a local maximum. Other local maxima will be reached for different initial conditions, or by allowing more than one replacement at a time, but I don't think it would be worthwhile.

The data must be adjusted in order for unit displacements in each dimension to have the same significance, i.e., in order for Euclidean distances to be meaningful. E.g., if your dimensions are salary and number of children, unadjusted, the algorithm will probably yield results concentrated in the extreme salary regions, ignoring that person with 10 kids. To get a more realistic output you could divide salary and number of children by their standard deviation, or by some other estimate that makes differences in salary comparable to differences in number of children.

To be able to plot the output for a random Gaussian distribution, I have set ND = 2 in the code, but setting ND = 6, as per your request, is no problem (except you cannot plot it).

import matplotlib.pyplot as plt
import numpy as np
import scipy.spatial as spatial

N, K, ND = 100000, 200, 2
MAX_LOOPS = 20

SIGMA, SEED = 40, 1234
rng = np.random.default_rng(seed=SEED)
means, variances = [0] * ND, [SIGMA**2] * ND
data = rng.multivariate_normal(means, np.diag(variances), N)

def distances(ndarray_0, ndarray_1):
    if (ndarray_0.ndim, ndarray_1.ndim) not in ((1, 2), (2, 1)):
        raise ValueError("bad ndarray dimensions combination")
    return np.linalg.norm(ndarray_0 - ndarray_1, axis=1)

# start with the K points closest to the mean
# (the copy() is only to avoid a view into an otherwise unused array)
indices = np.argsort(distances(data, data.mean(0)))[:K].copy()
# distsums is, for all N points, the sum of the distances from the K points
distsums = spatial.distance.cdist(data, data[indices]).sum(1)
# but the K points themselves should not be considered
# (the trick is that -np.inf ± a finite quantity always yields -np.inf)
distsums[indices] = -np.inf
prev_sum = 0.0
for loop in range(MAX_LOOPS):
    for i in range(K):
        # remove this point from the K points
        old_index = indices[i]
        # calculate its sum of distances from the K points
        distsums[old_index] = distances(data[indices], data[old_index]).sum()
        # update the sums of distances of all points from the K-1 points
        distsums -= distances(data, data[old_index])
        # choose the point with the greatest sum of distances from the K-1 points
        new_index = np.argmax(distsums)
        # add it to the K points replacing the old_index
        indices[i] = new_index
        # don't consider it any more in distsums
        distsums[new_index] = -np.inf
        # update the sums of distances of all points from the K points
        distsums += distances(data, data[new_index])
    # sum all mutual distances of the K points
    curr_sum = spatial.distance.pdist(data[indices]).sum()
    # break if the sum hasn't changed
    if curr_sum == prev_sum:
        break
    prev_sum = curr_sum

if ND == 2:
    X, Y = data.T
    marker_size = 4
    plt.scatter(X, Y, s=marker_size)
    plt.scatter(X[indices], Y[indices], s=marker_size)
    plt.grid(True)
    plt.gca().set_aspect('equal', adjustable='box')
    plt.show()

Output:

Splitting the data into 3 equidistant Gaussian distributions the output is this:

Simoneaux

Answer 3

Assuming that if you read your csv file with N (10000) rows and D dimension (or features) into a N*D martix X. You can calculate the distance between each point and store it in a distance matrix as follows:

import numpy as np
X = np.asarray(X) ### convert to numpy array
distance_matrix = np.zeros((X.shape[0],X.shape[0]))
for i in range(X.shape[0]):
    for j in range(i+1,X.shape[0]): 
    ## We compute triangle matrix and copy the rest. Distance from point A to point B and distance from point B to point A are the same. 
        distance_matrix[i][j]= np.linalg.norm(X[i]-X[j]) ## Here I am calculating Eucledian distance. Other distance measures can also be used.

        #distance_matrix = distance_matrix + distance_matrix.T - np.diag(np.diag(distance_matrix)) ## This syntax can be used to get the lower triangle of distance matrix, which is not really required in your case.
        K = 5 ## Number of points that you want to pick

        indexes = np.unravel_index(np.argsort(distance_matrix.ravel())[-1*K:], distance_matrix.shape)

        print(indexes)

Yuonneyup

Answer 4

Bottom Line Up Front: Dealing with multiple equally distant points and the Curse of Dimensionality are going to be larger problems than just finding the points. Spoiler alert: There's a surprise ending.

I think this an interesting question but I'm bewildered by some of the answers. I think this is, in part, due to the sketches provided. You've no doubt noticed the answers look similar -- 2d, with clusters -- even though you indicated a wider scope was needed. Because others will eventually see this, I'm going to step through my thinking a bit slowly so bear with me for the early part.

It makes sense to start with a simplified example to see if we can generalize a solution with data that's easy to grasp and a linear 2D model is easiest of the easy.

We don't need to calculate all the distances though. We just need the ones at the extremes. So we can then take the top and bottom few values:

right = lin_2_D.nlargest(8, ['x'])
left = lin_2_D.nsmallest(8, ['x'])

graph = sns.scatterplot(x="x", y="y", data=lin_2_D, color = 'gray', marker = '+', alpha = .4)
sns.scatterplot(x = right['x'], y = right['y'],  color = 'red')
sns.scatterplot(x = left['x'], y = left['y'],  color = 'green')

fig = graph.figure
fig.set_size_inches(8,3)

What we have so far: Of 100 points, we've eliminated the need to calculate the distance between 84 of them. Of what's left we can further drop this by ordering the results on one side and checking the distance against the others.

You can imagine a case where you have a couple of data points way off the trend line that could be captured by taking the greatest or least y values, and all that starts to look like Walter Tross's top diagram. Add in a couple of extra clusters and you get what looks his bottom diagram and it appears that we're sort of making the same point.

The problem with stopping here is the requirement you mentioned is that you need a solution that works for any number of dimensions.

The unfortunate part is that we run into four challenges:

Challenge 1: As you increase the dimensions you can run into a large number of cases where you have multiple solutions when seeking midpoints. So you're looking for k furthest points but have a large number of equally valid possible solutions and no way prioritizing them. Here are two super easy examples illustrate this:

A) Here we have just four points and in only two dimensions. You really can't get any easier than this, right? The distance from red to green is trivial. But try to find the next furthest point and you'll see both of the black points are equidistant from both the red and green points. Imagine you wanted the furthest six points using the first graphs, you might have 20 or more points that are all equidistant.

edit: I just noticed the red and green dots are at the edges of their circles rather than at the center, I'll update later but the point is the same.

B) This is super easy to imagine: Think of a D&D 4 sided die. Four points of data in a three-dimensional space, all equidistant so it's known as a triangle-based pyramid. If you're looking for the closest two points, which two? You have 4 choose 2 (aka, 6) combinations possible. Getting rid of valid solutions can be a bit of a problem because invariably you face questions such as "why did we get rid of these and not this one?"

Challenge 2: The Curse of Dimensionality. Nuff Said.

Challenge 3 Revenge of The Curse of Dimensionality Because you're looking for the most distant points, you have to x,y,z ... n coordinates for each point or you have to impute them. Now, your data set is much larger and slower.

Challenge 4 Because you're looking for the most distant points, dimension reduction techniques such as ridge and lasso are not going to be useful.

So, what to do about this?

Nothing.

Wait. What?!?

Not truly, exactly, and literally nothing. But nothing crazy. Instead, rely on a simple heuristic that is understandable and computationally easy. Paul C. Kainen puts it well:

Intuitively, when a situation is sufficiently complex or uncertain, only the simplest methods are valid. Surprisingly, however, common-sense heuristics based on these robustly applicable techniques can yield results which are almost surely optimal.

In this case, you have not the Curse of Dimensionality but rather the Blessing of Dimensionality. It's true you have a lot of points and they'll scale linearly as you seek other equidistant points (k) but the total dimensional volume of space will increase to power of the dimensions. The k number of furthest points you're is insignificant to the total number of points. Hell, even k^2 becomes insignificant as the number of dimensions increase.

Now, if you had a low dimensionality, I would go with them as a solution (except the ones that are use nested for loops ... in NumPy or Pandas).

If I was in your position, I'd be thinking how I've got code in these other answers that I could use as a basis and maybe wonder why should I should trust this other than it lays out a framework on how to think through the topic. Certainly, there should be some math and maybe somebody important saying the same thing.

Let me reference to chapter 18 of Computer Intensive Methods in Control and Signal Processing and an expanded argument by analogy with some heavy(-ish) math. You can see from the above (the graph with the colored dots at the edges) that the center is removed, particularly if you followed the idea of removing the extreme y values. It's a though you put a balloon in a box. You could do this a sphere in a cube too. Raise that into multiple dimensions and you have a hypersphere in a hypercube. You can read more about that relationship here.

Finally, let's get to a heuristic:

• Select the points that have the most max or min values per dimension. When/if you run out of them pick ones that are close to those values if there isn't one at the min/max. Essentially, you're choosing the corners of a box For a 2D graph you have four points, for a 3D you have the 8 corners of the box (2^3).

More accurately this would be a 4d or 5d (depending on how you might assign the marker shape and color) projected down to 3d. But you can easily see how this data cloud gives you the full range of dimensions.

Here is a quick check on learning; for purposes of ease, ignore the color/shape aspect: It's easy to graphically intuit that you have no problem with up to k points short of deciding what might be slightly closer. And you can see how you might need to randomize your selection if you have a k < 2D. And if you added another point you can see it (k +1) would be in a centroid. So here is the check: If you had more points, where would they be? I guess I have to put this at the bottom -- limitation of markdown.

So for a 6D data cloud, the values of k less than 64 (really 65 as we'll see in just a moment) points are pretty easy. But...

• If you don't have a data cloud but instead have data that has a linear relationship, you'll 2^(D-1) points. So, for that linear 2D space, you have a line, for linear 3D space, you'd have a plane. Then a rhomboid, etc. This is true even if your shape is curved. Rather than do this graph myself, I'm using the one from an excellent post on by Inversion Labs on Best-fit Surfaces for 3D Data

• If the number of points, k, is less than 2^D you need a process to decide what you don't use. Linear discriminant analysis should be on your shortlist. That said, you can probably satisfice the solution by randomly picking one.

• For a single additional point (k = 1 + 2^D), you're looking for one that is as close to the center of the bounding space.

• When k > 2^D, the possible solutions will scale not geometrically but factorially. That may not seem intuitive so let's go back to the two circles. For 2D you have just two points that could be a candidate for being equidistant. But if that were 3D space and rotate the points about the line, any point in what is now a ring would suffice as a solution for k. For a 3D example, they would be a sphere. Hyperspheres (n-spheres) from thereon. Again, 2^D scaling.

One last thing: You should seriously look at xarray if you're not already familiar with it.

Hope all this helps and I also hope you'll read through the links. It'll be worth the time.

*It would be the same shape, centrally located, with the vertices at the 1/3 mark. So like having 27 six-sided dice shaped like a giant cube. Each vertice (or point nearest it) would fix the solution. Your original k+1 would have to be relocated too. So you would select 2 of the 8 vertices. Final question: would it be worth calculating the distances of those points against each other (remember the diagonal is slightly longer than the edge) and then comparing them to the original 2^D points? Bluntly, no. Satifice the solution.

Phyla

Answer 5

If you're interested in getting the most distant points you can take advantage of all of the methods that were developed for nearest neighbors, you just have to give a different "metric".

For example, using scikit-learn's nearest neighbors and distance metrics tools you can do something like this

import numpy as np
from sklearn.neighbors import BallTree
from sklearn.neighbors.dist_metrics import PyFuncDistance
from sklearn.datasets import make_blobs
from matplotlib import pyplot as plt


def inverted_euclidean(x1, x2):
    # You can speed this up using cython like scikit-learn does or numba
    dist = np.sum((x1 - x2) ** 2)
    # We invert the euclidean distance and set nearby points to the biggest possible
    # positive float that isn't inf
    inverted_dist = np.where(dist == 0, np.nextafter(np.inf, 0), 1 / dist)
    return inverted_dist

# Make up some fake data
n_samples = 100000
n_features = 200
X, _ = make_blobs(n_samples=n_samples, centers=3, n_features=n_features, random_state=0)

# We exploit the BallTree algorithm to get the most distant points
ball_tree = BallTree(X, leaf_size=50, metric=PyFuncDistance(inverted_euclidean))

# Some made up query, you can also provide a stack of points to query against
test_point = np.zeros((1, n_features))
distance, distant_points_inds = ball_tree.query(X=test_point, k=10, return_distance=True)
distant_points = X[distant_points_inds[0]]

# We can try to visualize the query results
plt.plot(X[:, 0], X[:, 1], ".b", alpha=0.1)
plt.plot(test_point[:, 0], test_point[:, 1], "*r", markersize=9)
plt.plot(distant_points[:, 0], distant_points[:, 1], "sg", markersize=5, alpha=0.8)
plt.show()

Which will plot something like:

There are many points that you can improve on:

1. I implemented the inverted_euclidean distance function with numpy, but you can try to do what the folks of scikit-learn do with their distance functions and implement them in cython. You could also try to jit compile them with numba.
2. Maybe the euclidean distance isn't the metric you would like to use to find the furthest points, so you're free to implement your own or simply roll with what scikit-learn provides.

The nice thing about using the Ball Tree algorithm (or the KdTree algorithm) is that for each queried point you have to do log(N) comparisons to find the furthest point in the training set. Building the Ball Tree itself, I think also requires log(N) comparison, so in the end if you want to find the k furthest points for every point in the ball tree training set (X), it will have almost O(D N log(N)) complexity (where D is the number of features), which will increase up to O(D N^2) with the increasing k.

Julenejulep