Finding the distance of the 5th closest point to each of the points in a list of points

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 xy[0].sort()
 DT=[]
 for i in range(5982):
       xf=np.abs(xy[0][i]-xy[0][i+1])
       yf=np.abs(xy[1][i]-xy[1][i+1])
       D=((xf**2)+(yf**2))**0.5
       DT.append(D)`

 DT.sort()
 R5=[]
 for i in range(5977):
    R=np.abs(DT[i]-DT[i+4])
    R5.append(R)`

As the title suggests I'm trying to find the 5th closest point to each of the points in the list xy which consists of all the points in the shape (2,5983), however the method Im using returns the 5th closest distance without taking into account points in both the x and y direction. Any help on resolving this would be greatly appreciated!

If I understand correctly what you want, this should give you the index of the 5th closest:

fifth = []
for i in range(len(xy)):
    distances = ((xy.T-xy.T[i])**2).sum(axis=1)
    fifth.append(np.argsort(distances)[5])

This works if xy is in the shape (samples,dimensions) so I am using transport. Otherwise it makes the iterations less readable imo. But you could change this. You calculate all the distances at ones, there is no need to do sqrt because sqrt is monotonous, and you are looking for order only. Then I use the 6th element (index 5) because for simplicity, I am not skipping the self, which will be distance 0. Hope this helps and is clear

Find K Closest Points to the Origin, Given a list of points on the 2-D plane and an integer K. The task is to find K Approach : The idea is to calculate the euclidean distance from the origin for every given point Get hold of all the important DSA concepts with the DSA Self Paced 5th Floor, A-118,; Sector-136, Noida, Uttar Pradesh - 201305� If I understand correctly what you want, this should give you the index of the 5th closest: fifth = [] for i in range(len(xy)): distances = ( (xy.T-xy.T[i])**2).sum(axis=1) fifth.append(np.argsort(distances) [5]) This works if xy is in the shape (samples,dimensions) so I am using transport.

Use:

from scipy.spatial.distance import pdist,squareform
Y = pdist(xy, 'euclidean')

Example:

>>a = [[1,2],[3,4],[5,6]]
>>Y = pdist(a, 'euclidean')
>>Y
array([2.82842712, 5.65685425, 2.82842712])

The first element is the distance between a0 and a1, the second is distance between a0 and a2 and the third is distance between a1 and a2.

Or you can use the square form:

>>square = squareform(pdist(a))
>>square
array([[0.        , 2.82842712, 5.65685425],
       [2.82842712, 0.        , 2.82842712],
       [5.65685425, 2.82842712, 0.        ]])

After that, use:

np.argsort(square)

And you will get what you want.

Check the pdist documentation.

Closest Pair of Points using Divide and Conquer algorithm , The Brute force solution is O(n^2), compute the distance between each pair and return the Duration: 8:44 Posted: Nov 28, 2012 Distance, returned as a column vector containing the Euclidean distances between each query point and the closest input point. The length of dist is equal to the number of query points. Algorithms

Trying not to change your code too much. The first part computes the square of the distance between each pair of points (as pointed by Tacratis the square root is monotonous). Then it finds the fifth closest point to each point and returns an array where the ith element is the fifth closest point the the point i.

DT = []
    for i in range(5983):
        d_i = []
        for j in range(5983):
            xf=np.abs(xy[0][i]-xy[0][j])
            yf=np.abs(xy[1][i]-xy[1][j])
            d_ij =(xf**2)+(yf**2)
            d_i.append(d_ij)
        DT.append(d_i)

R5 = []
for i in range(5983):
    R = DT[i].index(sorted(DT[i])[5]) 
    R5.append(R)

print R5 

How to calculate closest pair of points in 3D data sets?, Basically, I'll have to calculate the distance between the closest points (once they are in the correct order). So far I For (a) you need to calculate all pair-wise distances between all points and find the smallest number. Asked 5th Feb, 2018 List of expected new journals reported in journals citation report (JCR 2019). The closest pair of points problem or closest pair problem is a problem of computational geometry: given n points in metric space, find a pair of points with the smallest distance between them. The closest pair problem for points in the Euclidean plane [1] was among the first geometric problems that were treated at the origins of the systematic

Finding the closest point to a list of points, It will certainly be faster if you vectorize the distance calculations: def closest_node(node, nodes): nodes = np.asarray(nodes) dist_2� Drag the points: Three or More Dimensions. It works perfectly well in 3 (or more!) dimensions. Square the difference for each axis, then sum them up and take the square root: Distance = √[ (x A − x B) 2 + (y A − y B) 2 + (z A − z B) 2] Example: the distance between the two points (8,2,6) and (3,5,7) is:

[PDF] Chapter 5 Distance Analysis I and II, Then, the next point is taken and the distance to all other points (including the first point In calculating the nearest neighbor index, network distance can be used to calculate Fifth, for each point, i, the observed minimum distance is compared to the nearest edge of will be output that lists the intermediate calculations. Hi guys, Long time reader, first time poster, but I'm having a problem which i can't find a solution to on the forum. I have a list of 700 locations (names - Column A) and their latititude (Column B) and longitude (Column C), and I'm looking to return the nearest three locations to the right, along with their distance (i.e. Nearest (Column D), Distance to nearest (Column E), second nearest

How to leverage Geopandas for faster snapping of points to lines , How to leverage Geopandas for faster snapping of points to lines For instance, in order to calculate the travel distance to the nearest electric scooter, you adding a list of values to it that is the same number of columns allows us to quickly add To find the closest line for each point, we can use groupby : Given 2 list of points with x and respective y coordinates, produce a minimal distance between a pair of 2 points. Every battle with a hardcore algorithm should start somewhere.

Comments
  • Thankyou! this is exactly what I needed!
  • Can you check the question as answered ?