logo
Tags down

shadow

how to find farthest neighbors in Euclidean space?


By : shenzhigang
Date : September 14 2020, 05:00 PM
I wish did fix the issue. To the best of my knowledge, the current standard kNN search algorithm for various trees (R-Trees, quadtrees, kd-trees) was developed by:
code :


Share : facebook icon twitter icon

LSH implementation in python 3 with Euclidean distance and seeing all neighbors in LSHForest


By : Hội Trầm Hương faceb
Date : March 29 2020, 07:55 AM
Does that help For software recommendations, please ask here: Software Recommendations.

Find two farthest points in 2D space (Manhattan distance)


By : Harpal Singh
Date : March 29 2020, 07:55 AM
This might help you We have to maximize manhattan distance, where P=(X,Y) is arbitrary fixed point from the set
code :
MD = Abs(X - x[i]) + Abs(Y - y[i])
MD =  X - x[i] + Y - y[i]
MD =  X - x[i] - Y + y[i]

fastest way to get closest 10 euclidean neighbors of large feature vector in python


By : Eder Lugo
Date : March 29 2020, 07:55 AM
around this issue I have a numpy array that has 10,000 vectors with 3,000 elements in each. I want to return the top 10 indices of the closest pairs with the distance between them. So if row 5 and row 7 have the closest euclidean distance of 0.005, and row 8 and row 10 have the second closest euclidean distance of 0.0052 then I want to return [(8,10,.0052),(5,7,.005)]. The traditional for loop method is very slow. Is there an alternative quicker approach for a way to get euclidean neighbors of large features vectors (stored as np array)?
code :
def topTen(M):
    i,j = np.triu_indices(M.shape[0], 1)
    dist_sq = np.einsum('ij,ij->i', M[i]-M[j], M[i]-M[j])
    max_i=np.argpartition(dist_sq, 10)[:10]
    max_o=np.argsort(dist_sq[max_i])
    return np.vstack((i[max_i][max_o], j[max_i][max_o], dist_sq[max_i][max_o]**.5)).T

Unity: How to find all neighbors recursively (neighbors of neighbors)?


By : user3379736
Date : March 29 2020, 07:55 AM
it helps some times Your pattern is correct. I believe stack overflow exception is caused by returning to an already visited node.
You might want to keep a list of already visited nodes and then don't return them as neighbors. You already have what is needed and it only needs to be rigged up. The code bellow might work for you
code :
private List<Bubble> FindAllRecursiveNeighbors(Vector2Int originPosition, List<Bubble> result = null)
{
    List<Bubble> allNeighbors = FindNeighbors(originPosition);

    if (result == null)
        // Mark (see bellow)
        result = new List<Bubble>();

    var newBubbles = new List<Bubble>();
    foreach (Bubble bubble in allNeighbors)
    {
        if (!result.Contains(bubble))
        {
            result.Add(bubble);
            newBubbles.Add(bubble);
        }
    }

    // Recursion starts here.
    foreach (Bubble bubble in newBubbles)
    {
        List<Bubble> neighbors = FindAllRecursiveNeighbors(FindPositionOfBubble(bubble), result);
    }

    return result;
}

I've an undirected NetworkX graph and want to find all neighbors of node_a that are also neighbors of node_b


By : TheBuvajs
Date : March 29 2020, 07:55 AM
Does that help To find common neighbors between two nodes you can use the function common_neighbors():
code :
import networkx as nx

G = nx.turan_graph(6, 2)

%matplotlib inline # jupyter notebook
nx.draw(G, with_labels=True)

list(nx.common_neighbors(G, 3, 5))
# [0, 1, 2]
import networkx as nx
from itertools import chain

G = nx.grid_graph(dim=[5, 5])

list_A = [(3, i) for i in range(5)]
list_B = [(1, i) for i in range(5)]
intersec = nx.node_boundary(G, list_A) & nx.node_boundary(G, list_B)
# {(2, 1), (2, 0), (2, 3), (2, 2), (2, 4)}

color = []
for i in G.nodes():
    if i in intersec:
        col = 'green'
    elif i in chain(list_A, list_B):
        col = 'red'
    else:
        col = 'pink'
    color.append(col)

%matplotlib inline
nx.draw(G, with_labels=True, node_color=color, node_size=1000)
shadow
Privacy Policy - Terms - Contact Us © 35dp-dentalpractice.co.uk