计算矩阵中一个点到所有其他点的距离 [英] Calculate Distances Between One Point in Matrix From All Other Points

问题描述

我是Python的新手，我需要实现一个聚类算法.为此，我将需要计算给定输入数据之间的距离.

I am new to Python and I need to implement a clustering algorithm. For that, I will need to calculate distances between the given input data.

考虑以下输入数据-

    [[1,2,8],
     [7,4,2],
     [9,1,7],
     [0,1,5],
     [6,4,3]]

我想在这里实现的是，我想计算与所有其他点的距离[1,2,8]，并找到距离最小的点.

What I am looking to achieve here is, I want to calculate distance of [1,2,8] from ALL other points, and find a point where the distance is minimum.

对于其他所有方面，我必须重复此操作.

And I have to repeat this for ALL other points.

我正在尝试使用FOR循环来实现这一点，但是我确信SciPy/NumPy必须具有可以帮助我有效实现此结果的功能.

I am trying to implement this with a FOR loop, but I am sure that SciPy/ NumPy must be having a function which can help me achieve this result efficiently.

我看了网上，但是'pdist'命令无法完成我的工作.

I looked online, but the 'pdist' command could not get my work done.

有人可以引导我吗?

TIA

推荐答案

结合使用np.linalg.norm和广播( numpy外减法)，您可以执行以下操作:

Use np.linalg.norm combined with broadcasting (numpy outer subtraction), you can do:

np.linalg.norm(a - a[:,None], axis=-1)

a[:,None]在a中插入一个新轴，由于广播，a - a[:,None]然后将逐行进行减法. np.linalg.norm计算最后一个轴上的np.sqrt(np.sum(np.square(...))):

a[:,None] insert a new axis into a, a - a[:,None] will then do a row by row subtraction due to broadcasting. np.linalg.norm calculates the np.sqrt(np.sum(np.square(...))) over the last axis:

a = np.array([[1,2,8],
     [7,4,2],
     [9,1,7],
     [0,1,5],
     [6,4,3]])

np.linalg.norm(a - a[:,None], axis=-1)
#array([[ 0.        ,  8.71779789,  8.1240384 ,  3.31662479,  7.34846923],
#       [ 8.71779789,  0.        ,  6.164414  ,  8.18535277,  1.41421356],
#       [ 8.1240384 ,  6.164414  ,  0.        ,  9.21954446,  5.83095189],
#       [ 3.31662479,  8.18535277,  9.21954446,  0.        ,  7.        ],
#       [ 7.34846923,  1.41421356,  5.83095189,  7.        ,  0.        ]])

元素[0,1]，[0,2]例如对应于:

np.sqrt(np.sum((a[0] - a[1]) ** 2))
# 8.717797887081348

np.sqrt(np.sum((a[0] - a[2]) ** 2))
# 8.1240384046359608

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