陣列中點之間的快速加權歐氏距離

Question

我需要有效地計算每個歐幾里得加權距離x,y的給定陣列中的點到每個其它x,y在另一個陣列點。 這是我的代碼按預期工作：

import numpy as np
import random

def rand_data(integ):
    '''
    Function that generates 'integ' random values between [0.,1.)
    '''
    rand_dat = [random.random() for _ in range(integ)]

    return rand_dat

def weighted_dist(indx, x_coo, y_coo):
    '''
    Function that calculates *weighted* euclidean distances.
    '''
    dist_point_list = []
    # Iterate through every point in array_2.
    for indx2, x_coo2 in enumerate(array_2[0]):
        y_coo2 = array_2[1][indx2]
        # Weighted distance in x.
        x_dist_weight = (x_coo-x_coo2)/w_data[0][indx] 
        # Weighted distance in y.
        y_dist_weight = (y_coo-y_coo2)/w_data[1][indx] 
        # Weighted distance between point from array_1 passed and this point
        # from array_2.
        dist = np.sqrt(x_dist_weight**2 + y_dist_weight**2)
        # Append weighted distance value to list.
        dist_point_list.append(round(dist, 8))

    return dist_point_list


# Generate random x,y data points.
array_1 = np.array([rand_data(10), rand_data(10)], dtype=float)

# Generate weights for each x,y coord for points in array_1.
w_data = np.array([rand_data(10), rand_data(10)], dtype=float)

# Generate second larger array.
array_2 = np.array([rand_data(100), rand_data(100)], dtype=float)


# Obtain *weighted* distances for every point in array_1 to every point in array_2.
dist = []
# Iterate through every point in array_1.
for indx, x_coo in enumerate(array_1[0]):
    y_coo = array_1[1][indx]
    # Call function to get weighted distances for this point to every point in
    # array_2.
    dist.append(weighted_dist(indx, x_coo, y_coo))

最終列表dist包含與第一個數組中的點一樣多的子列表，其中每個子元素的數量與第二個中的點數相同（加權距離）。

我想知道是否有辦法使這個代碼更有效率，也許使用cdist函數，因為當數組有很多元素（在我的情況下它們有）時，當我需要檢查時，這個過程變得相當昂貴許多陣列的距離（我也有）

Answer 1

@Evan和@Martinis Group走在正確的軌道上 - 擴展Evan的答案，這是一個使用廣播來快速計算沒有Python循環的n維加權歐氏距離的函數：

import numpy as np

def fast_wdist(A, B, W):
    """
    Compute the weighted euclidean distance between two arrays of points:

    D{i,j} = 
    sqrt( ((A{0,i}-B{0,j})/W{0,i})^2 + ... + ((A{k,i}-B{k,j})/W{k,i})^2 )

    inputs:
        A is an (k, m) array of coordinates
        B is an (k, n) array of coordinates
        W is an (k, m) array of weights

    returns:
        D is an (m, n) array of weighted euclidean distances
    """

    # compute the differences and apply the weights in one go using
    # broadcasting jujitsu. the result is (n, k, m)
    wdiff = (A[np.newaxis,...] - B[np.newaxis,...].T) / W[np.newaxis,...]

    # square and sum over the second axis, take the sqrt and transpose. the
    # result is an (m, n) array of weighted euclidean distances
    D = np.sqrt((wdiff*wdiff).sum(1)).T

    return D

要檢查這是否正常，我們將它與使用嵌套Python循環的較慢版本進行比較：

def slow_wdist(A, B, W):

    k,m = A.shape
    _,n = B.shape
    D = np.zeros((m, n))

    for ii in xrange(m):
        for jj in xrange(n):
            wdiff = (A[:,ii] - B[:,jj]) / W[:,ii]
            D[ii,jj] = np.sqrt((wdiff**2).sum())
    return D

首先，讓我們確保兩個函數給出相同的答案：

# make some random points and weights
def setup(k=2, m=100, n=300):
    return np.random.randn(k,m), np.random.randn(k,n),np.random.randn(k,m)

a, b, w = setup()
d0 = slow_wdist(a, b, w)
d1 = fast_wdist(a, b, w)

print np.allclose(d0, d1)
# True

不用說，使用廣播而不是Python循環的版本要快幾個數量級：

%%timeit a, b, w = setup()
slow_wdist(a, b, w)
# 1 loops, best of 3: 647 ms per loop

%%timeit a, b, w = setup()
fast_wdist(a, b, w)
# 1000 loops, best of 3: 620 us per loop

Answer 2

如果您不需要加權距離，可以使用cdist 。 如果您需要加權距離和性能，請創建適當輸出大小的數組，並使用Numba或Parakeet等自動加速器，或使用Cython手動調整代碼。

Answer 3

您可以使用類似於以下內容的代碼來避免循環：

def compute_distances(A, B, W):
    Ax = A[:,0].reshape(1, A.shape[0])
    Bx = B[:,0].reshape(A.shape[0], 1)
    dx = Bx-Ax

    # Same for dy
    dist = np.sqrt(dx**2 + dy**2) * W
    return dist

這將在Python運行速度快了很多那個什么，只要你有數組足夠的內存循環。

Answer 4

您可以嘗試刪除平方根，因為如果a> b，則遵循平方> b平方...並且計算機通常在平方根處真的很慢。