I have a dataset of drugs represented as a graph, each of which is described by three non-square matrices:
edge index (A), an 2xe matrix, where e are the bonds of the molecule, the first line indicates the node (atom) from which the edge (bond) starts, and the second one the node where the edge arrives;
node feature matrix (X), an nx9 matrix, where n are the atoms of the molecule and 9 are the features used to describe these (e.g. atomic number, charge, hybridization);
edge feature matrix (E), an 4xe matrix, where e are the bonds of the molecule and 4 are the features used to describe these (e.g. type of bond, geometry).
I would like to plot these data on a Cartesian space to see if clusters are created based on their activity label. I thought, if I can reduce each matrix to a single point in space for each graph I will have three x, y, z coordinates, and then it will be very easy to plot the points. Does this make sense in your opinion? How could I go about turning a matrix into a single point using python? Finally, I leave you with an example of the graph I would like to create
Thank you all!
Assuming:
The nodes in a drug's graph represent features that every drug has to different extents, including zero.
The structure of a drug's graph models the extent to which every feature applies to that that drug
There is an algorithm to calculate from a drug's graph the 'extent' ( a number ) each feature applies the the drug
Then:
Construct a table where each row models a drug and each column is for a feature. Each cell then contains the "extent" to which the column's feature applies to the row's drug.
Apply the K-Means algorithm to the table.
The challenge is, of course: an algorithm to calculate from a drug's graph the 'extent' ( a number ) each feature.
IMHO the first step is to enter your data into a graph theory library. I see you are using Python. Python folks generally use a library called networkx. Are you familiar with this library?
Personally, I much prefer to work with C++ ( it gives the performance required for large problem sets ) Recently, I added a SMILES parser to my C++ graph library.
Convert the SMILES representation of each drug to its graph representation
Calculate the graph edit distance ( GED https://en.wikipedia.org/wiki/Graph_edit_distance ) between every pair of drugs
LOOP GEDMAX from 1 to 10
Add a connection between two drugs if the GED is less than GEDMAX. This forms a new graph we can call "GEDgraph"
Find the components ( clusters of drugs all reachable from each other in the GEDgraph )
SELECT "best" set of components
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?
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
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:
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)
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.
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:
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.
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.
I have a square grid with some points marked off as being the centers of the subparts of the grid. I'd like to be able to assign each location within the grid to the correct subpart. For example, if the subparts of the region were centered on the black dots, I'd like to be able to assign the red dot to the region in the lower right, as it is the closest black dot.
Currently, I do this by iterating over each possible red dot, and comparing its distance to each of the black dots. However, the width, length, and number of black dots in the grid is very high, so I'd like to know if there's a more efficient algorithm.
My particular data is formatted as such, where the numbers are just placeholders to correspond with the given example:
black_dots = [(38, 8), (42, 39), (5, 14), (6, 49)]
grid = [[0 for i in range(0, 50)] for j in range(0, 50)]
For reference, in the sample case, I hope to be able to fill grid up with integers 1, 2, 3, 4, depending on whether they are closest to the 1st, 2nd, 3rd, or 4th entry in black_dots to end up with something that would allow me to create something similar to the following picture where each integer correspond to a color (dots are left on for show).
To summarize, is there / what is the more efficient way to do this?
You can use a breadth-first traversal to solve this problem.
Create a first-in, first-out queue. (A queue makes a traversal breadth-first.)
Create a Visited mask indicating whether a cell in your grid has been added to the queue or not. Set the mask to false.
Create a Parent mask indicating what black dot the cell ultimately belongs to.
Place all the black dots into the queue, flag them in the Visited mask, and assign them unique ids in the Parent mask.
Begin popping cells from the queue one by one. For each cell, iterate of the cell's neighbours. Place each neighbour into the Queue, flag it in Visited, and set its value in Parent to be equal to that of the cell you just popped.
Continue until the queue is empty.
The breadth-first traversal makes a wave which expands outward from each source cell (black dot). Since the waves all travel at the same speed across your grid, each wave gobbles up those cells closest to its source.
This solves the problem in O(N) time.
If I understand correctly what you really need is to construct a Voronoi diagram of your centers:
https://en.m.wikipedia.org/wiki/Voronoi_diagram
Which can be constructed very efficiently with similar computational complexity as calculating its convex hull.
The Voronoi diagram allows you to construct the optimal polygons sorrounding your centers which delimit the regions closest to the centers.
Having the Voronoi diagram the task is reduced to detect in which polygon the red dots lies. Since the Voronoi cells are convex you need an algorithm to decide wether a point is inside a convex polygon. However traversing all polygons has complexity O(n).
There are several algorithms to accelerate the point location so it can be done in O(log n):
https://en.m.wikipedia.org/wiki/Point_location
See also
Nearest Neighbor Searching using Voronoi Diagrams
The "8-way" Voronoi diagram can be constructed efficiently (in linear time wrt the number of pixels) by a two-passes scanline process. (8-way means that distances are evaluated as the length of the shortest 8-connected path between two pixels.)
Assign every center a distinct color and create an array of distances of the same size as the image, initialized with 0 at the centers and "infinity" elsewhere.
In a top-down/left-right pass, update the distances of all pixels as being the minimum of the distances of the four neighbors W, NW, N and NE plus one, and assign the current pixel the color of the neighbor that achieves the minimum.
In a bottom-up/right-left pass, update the distances of all pixels as being the minimum of the current distance and the distances of the four neighbors E, SE, S, SW plus one, and assign the current pixel the color of the neighbor that achieves the minimum (or keep the current color).
It is also possible to compute the Euclidean Voronoi diagram efficiently (in linear time), but this requires a more sophisticated algorithm. It can be based on the wonderful paper "A GENERAL ALGORITHM FOR COMPUTING DISTANCE
TRANSFORMS IN LINEAR TIME" by A. MEIJSTER‚ J.B.T.M. ROERDINK and W.H. HESSELINK, which must be enhanced with some accounting of the neighbor that causes the smallest distance.
What I aim to do is create a graph of the nodes in the first 2 columns, that have edge lengths that are proportional to the values in the 3rd column. My input data looks like:
E06.1644.1 A01.908.1 0.5
E06.1643.1 A01.908.1 0.02
E06.1644.1 A01.2060.1 0.7
I am currently importing it like this:
g=Graph.Read_Ncol("igraph.test.txt",names=True,directed=False,weights=True)
igraph.plot(g, "igraph.pdf", layout="kamada_kawai")
When I print the names or the weights (which I intend them to be the edge lengths), they print out fine with:
print(g.vs["name"])
print(g.es["weight"])
However, the vertices are blank, and the lengths do not seem to be proportional to their values. Also, there are too many nodes (A01.908.1 is duplicated).
What am I doing wrong?
Thanks in advance....
The vertices are blank because igraph does not use the name attribute as vertex labels automatically. If you want to use the names as labels, you have two options:
Copy the name vertex attribute to the label attribute: g.vs["label"] = g.vs["name"]
Tell plot explicitly that you want it to use the names as labels: plot(g, "igraph.pdf", layout="kamada_kawai", vertex_label=g.vs["name"])
I guess the same applies to the weights; igraph does not use the weights automatically to determine the thickness of each edge. If you want to do this, rescale the weight vector to a meaningful thickless range (say, from 0.5 to 3) and then set the rescaled vector as the width edge attribute:
>>> g.es["width"] = rescale(g.es["weight"], out_range=(0.5, 3))
Alternatively, you can also use the edge_width keyword argument in the plot() call:
plot(g, ..., edge_width=rescale(g.es["weight"], out_range=(0.5, 3)))
See help(Graph.__plot__) for more details about the keyword arguments that you can pass to plot().
As for the duplicated node, I strongly suspect that there is a typo in your input file and the two names are not equivalent; one could have a space at the end for instance. Inspect g.vs["name"] carefully to see if this is the case.
Update: if you want the lengths of the edges to be proportional to the prescribed weights, I'm afraid that this cannot be done exactly in the general case - it is easy to come up with a graph where the prescribed lengths cannot be achieved in 2D space. There is a technique called multidimensional scaling (MDS) which could reconstruct the positions of the nodes from a distance matrix - but this requires that a distance is specified for each pair of nodes (i.e. also for disconnected pairs).
The Kamada-Kawai layout algorithm that you have used is able to take edge weights into account to some extent (it is likely to get stuck in local minima so you probably won't get an exact result), but it interprets the weights as similarities, not distances, therefore the larger the weight is, the closer the endpoints will be. However, you still have to tell igraph to use the weights when calculating the layout, like this:
>>> similarities = [some_transformation(weight) for weight in g.es["weight"]]
>>> layout = g.layout_kamada_kawai(weights=similarities)
>>> plot(g, layout=layout, ...)
where some_transformation() is a "reasonable" transformation from distance to similarity. This requires some trial-and-error; I usually use a transformation based on a sigmoid function that transforms the median distance to a similarity of 0.5, the (median + 2 sd) distance to 0.1 and the (median - 2 sd) distance to 0.9 (where sd is the standard deviation of the distance distribution) - but this is not guaranteed to work in all cases.