kdtree | Blazingly fast , native 2d kdtree
kandi X-RAY | kdtree Summary
kandi X-RAY | kdtree Summary
A kd tree is a data structure that recursively partitions the world in order to rapidly answer nearest neighbor queries. A generic kd tree can support any number of dimensions, and can return either the nearest neighbor or a set of N nearest neighbors. This gem is a blazingly fast, native, 2d kdtree. It's specifically built to find the nearest neighbor when searching millions of points. It's used in production at Urbanspoon and several other companies. The first version of this gem was released back in 2009. See the original blog post for the full story. Wikipedia has a great article on kdtrees.
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QUESTION
Let's say you have the grid:
...ANSWER
Answered 2022-Mar-12 at 09:09You don't need to make it too complicate by using Scipy. This problem can easily done by help of mathematics.
Equation of coordinate inside circle is x^2 + y^2 <= Radius^2
, so just check coordinate that inside the circle.
QUESTION
I have a large Dataframe (189090, 8), I need to calculate Euclidean distance and the similarity.
My approach:
...ANSWER
Answered 2022-Mar-09 at 10:54According to the documentation, pdist "returns a condensed distance matrix". That means it would try to calculate and return a matrix of about 189090^2/2 = 17877514050 entries, causing your computer run out of ram.
If you want to calculate distances between some specific data points, filter them out before using pdist.
If you really want to calculate the entire distance matrix, it's better to calculate distances of a small partition of data points at a time (e.g. 1000), and save the result in the disk.
QUESTION
Firstly, i have an image that I pass in arguments, and i retrieve all of his contours with OpenCV (with the cv.findContours
method).
I parse this list with my parseArray
method to have a well parsed list of x,y contours coordinates of the img [(x1, y1), (x2, y2), ...]
(The size of this list equals 24163
for my unicorn image)
So here is my code:
...ANSWER
Answered 2022-Feb-21 at 13:36I think you spend most of your time in your while loop so I will focus on those lines:
QUESTION
Note : almost duplicate of Numpy vectorization: Find intersection between list and list of lists
Differences :
- I am focused on efficiently when the lists are large
- I'm searching for the largest intersections.
ANSWER
Answered 2022-Jan-22 at 02:49Since y
contains disjoint ranges and the union of them is also a range, a very fast solution is to first perform a binary search on y
and then count the resulting indices and only return the ones that appear at least 10 times. The complexity of this algorithm is O(Nx log Ny)
with Nx
and Ny
the number of items in respectively x
and y
. This algorithm is nearly optimal (since x
needs to be read entirely).
First of all, you need to transform your current y
to a Numpy array containing the beginning value of all ranges (in an increasing order) with N
as the last value (assuming N
is excluded for the ranges of y
, or N+1
otherwise). This part can be assumed as free since y
can be computed at compile time in your case. Here is an example:
QUESTION
I used scipy.spatial.KDTree.query_pairs() which returned a python set of tuples. Let's say, this is the output:
...ANSWER
Answered 2021-Dec-16 at 11:13You need to iterate over and check every tuple in set1
, you can do that using a set comprehension, and any()
:
QUESTION
I would like to have a generic KDTree implementation in C++ that can hold any kind of positionable object. Such objects have a 2D position.
Unfortunately. Positionable classes could have different ways of getting the position.
- Getters
getX()
andgetY()
- std::pair
- sf::Vector2f
- ...
What would be the proper way to wrap such classes into my KDTree?
My Tree is composed of nodes such as:
...ANSWER
Answered 2021-Dec-05 at 14:59Check the design rationale of boost geometry for a solution to this problem. The methodology boils down to these steps:
Declare a class template that extracts position information from a type, e.g.
QUESTION
The Clark-Evans index is one of the most basic statistics to measure point aggregation in spatial analysis. However, I can't find any implementation in Python. So I adapted the R code from the hyperlink above. I want to ask if the statistic and p-value are correct with such irregular study areas:
Function ...ANSWER
Answered 2021-Sep-30 at 17:47I have not worked with the Clark-Evans (CE) index before, but having read the information you linked to and studied your code, my interpretation is this:
- The index value for Dataset2 is less than the index value for Dataset1. This correctly reflects the visual difference in clusteredness, that is, the smaller index value is associated with data that is more clustered.
- It is probably not meaningful to say that two CE index values are similar, other than special cases like observing that two CE index values are both smaller than 1 or both greater than 1, or if A < B < C then AB are more similar than AC.
- The p-value and the index value measure different things. The index value measures degree of clusteredness (if less than 1) or regularity (if greater than 1). The p-value (inversely) measures how certain it is that the data are more clustered than would be expected by chance, or more regular than would be expected by chance. The p-value in particular is sensitive to the sample size as well as the distribution of points.
- The use of pi in calculating SE reflects the assumption of Euclidean distances between points (rather than, say, city block distances). That is, the nearest neighbour of a point is the one at the smallest radial distance. The use of pi in calculating SE does not make any assumptions about the shape of the region of interest.
- Particularly for small datasets (like Dataset2) you will want to track down information about the potential impact of boundary effects on the index value or the p-value.
More speculatively, I wonder if it would be useful to use a convex hull to help determine the region of interest rather than do this subjectively.
QUESTION
I have been following this tutorial on how to find nearest neighbors of a point with scikit.
However, when it comes to displaying the data, the tutorial merely mentions that "the indices can be mapped to useful values and the two arrays merged with the rest of the data"
But there's no actual explanation on how to do this. I'm not very well-versed in Pandas and I don't know how to perform this merge, so I just end up with 2 multidimensional arrays and I don't know how to map them to the original data to study the example and experiment with it.
This is the code
...ANSWER
Answered 2021-Sep-21 at 02:18Each integer index in indices
refers to an index value (row number) of locations_a
. You can use locations_a.loc[]
to convert these indices to their corresponding station names as a numpy array:
QUESTION
I'm trying to construct a KDtree for finding 'K nearest neighbour' I've created a class called 'Point' which holds attributes: pointID, lat (latitude) and Lon (longitude). the input for the build_index is an array 'points' contains all instances of points.
in the code below I'm trying to construct the KDtree but I'm having issues with trying to retrieve just the lat and Lon of each Point to sort, but I understand just using 'points' will not work as it is a an array with just the class instances.
thanks for the help in advance!
...ANSWER
Answered 2021-Sep-11 at 09:21point[axis]
is not working because point
does not support such bracket notation.
There are several solutions:
Define
Point
as a named tuple:Instead of defining
Point
as something like:
QUESTION
I am creating an ArrayList of an array and then returning the data in the form of a Stream from the method. But I need a stream of the objects inside the array. I believe I need to use flatMap() but what I have is not working for me, whereas when I return Stream.of(Arguments.of(kt,r,expectedPoints));
It works, and I see the Objects correctly in my unit test.
ANSWER
Answered 2021-Sep-02 at 23:52If I'm understanding correctly what you're trying to do, then you need to make two changes:
Change the type of
l
fromList to
List
:
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