nearest-neighbor

How to solve nearest neighbor through the R-nearest neighbor?

一个人想着一个人 提交于 2019-12-01 12:02:56
问题 Citing the E2LSH manual ( it's not important that's about this specific library, this quote should be true for NN problem in general ): E 2LSH can be also used to solve the nearest neighbor problem, where, given the query q, the data structure is required the report the point in P that is closest to q. This can be done by creating several R-near neighbor data structures, for R = R1, R2, . . . Rt , where Rt should be greater than the maximum distance from any query point to its nearest

How to solve nearest neighbor through the R-nearest neighbor?

瘦欲@ 提交于 2019-12-01 12:01:34
Citing the E2LSH manual ( it's not important that's about this specific library, this quote should be true for NN problem in general ): E 2LSH can be also used to solve the nearest neighbor problem, where, given the query q, the data structure is required the report the point in P that is closest to q. This can be done by creating several R-near neighbor data structures, for R = R1, R2, . . . Rt , where Rt should be greater than the maximum distance from any query point to its nearest neighbor. The nearest neighbor can be then recovered by querying the data structures in the increasing order

find the nearest point pairs between two sets of of matrix

你。 提交于 2019-12-01 11:16:13
Assume I have two sets of matrix ( A and B ), inside each matrix contains few point coordinates, I want to find out point in B nearest to A and output a cell array C listed the nearest point pair coordinates accordingly and one cell array D register the unpaired spot, how should I do it? To be more specific, here is what I want Two sets of matrix contain spot xy coordinates; A=[ 1 2; 3 4]; B=[1 3; 5 6; 2 1]; want to get C{1,1}=[1 2; 1 3]; C{2,1}= [3 4; 5 6]; D{1,1}=[2 1]; Thanks for the help. There is not exactly one solution to this problem, take for example the (one-dimensional, but

Broadcast Annoy object in Spark (for nearest neighbors)?

[亡魂溺海] 提交于 2019-12-01 10:53:46
As Spark's mllib doesn't have nearest-neighbors functionality, I'm trying to use Annoy for approximate Nearest Neighbors. I try to broadcast the Annoy object and pass it to workers; however, it does not operate as expected. Below is code for reproducibility (to be run in PySpark). The problem is highlighted in the difference seen when using Annoy with vs without Spark. from annoy import AnnoyIndex import random random.seed(42) f = 40 t = AnnoyIndex(f) # Length of item vector that will be indexed allvectors = [] for i in xrange(20): v = [random.gauss(0, 1) for z in xrange(f)] t.add_item(i, v)

find the nearest point pairs between two sets of of matrix

旧时模样 提交于 2019-12-01 09:40:51
问题 Assume I have two sets of matrix ( A and B ), inside each matrix contains few point coordinates, I want to find out point in B nearest to A and output a cell array C listed the nearest point pair coordinates accordingly and one cell array D register the unpaired spot, how should I do it? To be more specific, here is what I want Two sets of matrix contain spot xy coordinates; A=[ 1 2; 3 4]; B=[1 3; 5 6; 2 1]; want to get C{1,1}=[1 2; 1 3]; C{2,1}= [3 4; 5 6]; D{1,1}=[2 1]; Thanks for the help.

Heuristics to sort array of 2D/3D points according their mutual distance

元气小坏坏 提交于 2019-12-01 08:57:23
问题 Consider array of points in 2D,3D,(4D...) space ( e.g. nodes of unstructured mesh ). Initially the index of a point in array is not related to its position in space. In simple case, assume I already know some nearest neighbor connectivity graph. I would like some heuristics which increase probability that two points which are close to each other in space would have similar index (would be close in array). I understand that exact solution is very hard (perhaps similar to Travelling salesman

Drawing decision boundaries in R

可紊 提交于 2019-11-30 14:42:45
I've got a series of modelled class labels from the knn function. I've got a data frame with basic numeric training data, and another data frame for test data. How would I go about drawing a decision boundary for the returned values from the knn function? I'll have to replicate my findings on a locked-down machine, so please limit the use of 3rd party libraries if possible. I only have two class labels, "orange" and "blue". They're plotted on a simple 2D plot with the training data. Again, I just want to draw a boundary around the results from the knn function. Code: library(class) n <- 100

What is the meaning of “from distinct vertex chains” in this nearest neighbor algorithm?

自古美人都是妖i 提交于 2019-11-30 13:11:09
问题 The following pseudo-code is from the first chapter of an online preview version of The Algorithm Design Manual (page 7 from this PDF). The example is of a flawed algorithm, but I still really want to understand it: [...] A different idea might be to repeatedly connect the closest pair of endpoints whose connection will not create a problem, such as premature termination of the cycle. Each vertex begins as its own single vertex chain. After merging everything together, we will end up with a

nearest neighbor - k-d tree - wikipedia proof

删除回忆录丶 提交于 2019-11-30 11:38:59
问题 On the wikipedia entry for k-d trees, an algorithm is presented for doing a nearest neighbor search on a k-d tree. What I don't understand is the explanation of step 3.2. How do you know there isn't a closer point just because the difference between the splitting coordinate of the search point and the current node is greater than the difference between the splitting coordinate of the search point and the current best? Nearest neighbor search Animation of NN searching with a KD Tree in 2D The

What is the meaning of “from distinct vertex chains” in this nearest neighbor algorithm?

女生的网名这么多〃 提交于 2019-11-30 06:14:06
The following pseudo-code is from the first chapter of an online preview version of The Algorithm Design Manual (page 7 from this PDF ). The example is of a flawed algorithm, but I still really want to understand it: [...] A different idea might be to repeatedly connect the closest pair of endpoints whose connection will not create a problem, such as premature termination of the cycle. Each vertex begins as its own single vertex chain. After merging everything together, we will end up with a single chain containing all the points in it. Connecting the final two endpoints gives us a cycle. At any