2014-05-21 73 views
1

這是一個K-最近鄰算法,適用於R n中的點,它應計算每個點與其k個最近鄰點的平均距離。問題在於,雖然它是矢量化的,但是我重複着自己的意義,它效率低下。我會很高興,如果有人可以幫助我提高這個代碼:Python中的矢量化平均K-最近鄰距離

import numpy as np 
from scipy.spatial.distance import pdist 
from scipy.spatial.distance import squareform 

def nn_args_R_n_squared(points): 
    """Calculate pairwise distances of points and return the matrix together with matrix of indices of the first matrix sorted""" 
    dist_mat=squareform(pdist(points,'sqeuclidean')) 
    return dist_mat,np.argsort(dist_mat,axis=1) 
def knn_avg_dist(X,k): 
    """Calculates for points in rows of X, the average distance of each, to their k-nearest  neighbours""" 
    X_dist_mat,X_sorted_arg=nn_args_R_n_squared(X) 
    X_matrices=(X[X_sorted_arg[:,1:k+1]]-X[...,None,...]).astype(np.float64) 
    return np.mean(np.linalg.norm(X_matrices,axis=2)**2,axis=1) 
X=np.random.randn(30).reshape((10,3)) 
print X 
print knn_avg_dist(X,3) 

輸出:

[[-1.87979713 0.02832699 0.18654558] 
[ 0.95626677 0.4415187 -0.90220505] 
[ 0.86210012 -0.88348927 0.32462922] 
[ 0.42857316 1.66556448 -0.31829065] 
[ 0.26475478 -1.6807253 -1.37694585] 
[-0.08882175 -0.61925033 -1.77264525] 
[-0.24085553 0.64426394 -0.01973027] 
[-0.86926425 0.93439913 -0.31657442] 
[-0.30987468 0.02925649 -1.38556347] 
[-0.41801804 1.40210993 -1.04450895]] 
[ 3.37983833 2.1257945 3.60884158 1.67051682 2.85013297 1.66756279 
    1.2678029 1.20491026 1.54623574 1.30722388] 

正如你可以看到我計算距離的兩倍,但我無法想出一個辦法從X_dist_mat讀取相同的信息,因爲我必須同時從每行讀取多個元素。

+1

如果添加'import's和生成假數據給你的代碼,那麼可以複製並粘貼到一起來看看。否則,你應該能夠從'sklearn'現有的實現中獲得靈感。 – eickenberg

回答

2

使用scipy.spatial.cKDTree

>>> data = np.random.rand(1000, 3) 
>>> import scipy.spatial 

>>> kdt = scipy.spatial.cKDTree(data) 
>>> k = 5 # number of nearest neighbors 
>>> dists, neighs = kdt.query(data, k+1) 
>>> avg_dists = np.mean(dists[:, 1:], axis=1) 
+0

謝謝!你在Python世界搖滾! :) – Cupitor