比较 python 中多维数组的行

Question

我有两个 numpy 数组

a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])
b= np.array([[1,1],[1,3],[5,3]])

我想比较 a 与 b 和 return a-b 这样：

a-b = array([[2,2],
             [3,2],
             [4,2],
             [3,3]])

我试过：

[x for x in a if x not in b]

结果是

[array([2, 2]), array([3, 2]), array([4, 2])] # where clearly [3,3] is missing

我什至尝试在循环中比较 a 和 b 的每一行，但它给了我一个错误

The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

谁能帮我解决这个问题？？？

Answer 1

一种方法是将 numpy 数组转换为元组列表，将 b 转换为元组集，然后执行与它们相同的列表理解。例子-

In [1]: import numpy as np

In [2]: a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])

In [3]: b= np.array([[1,1],[1,3],[5,3]])

In [18]: alist = list(map(tuple, a))

In [19]: bset = set(map(tuple, b))

In [20]: np.array([x for x in alist if x not in bset])
Out[20]:
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

Answer 2

Broadcasting 基于矢量化方法 -

a[~((b[:,None,:] == a).all(2)).any(0)]

使用cdist from scipy.spatial.distance-

from scipy.spatial.distance import cdist

a[~(cdist(a,b)==0).any(1)]

样本运行-

In [89]: a
Out[89]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3],
       [5, 3]])

In [90]: b
Out[90]: 
array([[1, 1],
       [1, 3],
       [5, 3]])

In [91]: a[~((b[:,None,:] == a).all(2)).any(0)]
Out[91]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

In [92]: a[~(cdist(a,b)==0).any(1)]
Out[92]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

比较 python 中多维数组的行

Comparing rows of a multidimesional array in python

python

arrays

numpy

list

scipy