从一个数组中删除另一个数组中的元素 [英] Removing elements from an array that are in another array

问题描述

说我有这些二维数组A和B.

Say I have these 2D arrays A and B.

如何从A中删除B中的元素(集合论中的补语:A-B)

How can I remove elements from A that are in B. (Complement in set theory: A-B)

A=np.asarray([[1,1,1], [1,1,2], [1,1,3], [1,1,4]])
B=np.asarray([[0,0,0], [1,0,2], [1,0,3], [1,0,4], [1,1,0], [1,1,1], [1,1,4]])
#output = [[1,1,2], [1,1,3]]

---

更准确地说，我想做这样的事情.

---

To be more precise, I would like to do something like this.

data = some numpy array
label = some numpy array
A = np.argwhere(label==0) #[[1 1 1], [1 1 2], [1 1 3], [1 1 4]]
B = np.argwhere(data>1.5) #[[0 0 0], [1 0 2], [1 0 3], [1 0 4], [1 1 0], [1 1 1], [1 1 4]]
out = np.argwhere(label==0 and data>1.5) #[[1 1 2], [1 1 3]]

推荐答案

基于 this solution 到 Find the row indexes of several values in a numpy array ，这是一个基于NumPy的解决方案具有较少的内存占用，并且在使用大型阵列时可能会有所帮助-

Based on this solution to Find the row indexes of several values in a numpy array, here's a NumPy based solution with less memory footprint and could be beneficial when working with large arrays -

dims = np.maximum(B.max(0),A.max(0))+1
out = A[~np.in1d(np.ravel_multi_index(A.T,dims),np.ravel_multi_index(B.T,dims))]

样品运行-

In [38]: A
Out[38]: 
array([[1, 1, 1],
       [1, 1, 2],
       [1, 1, 3],
       [1, 1, 4]])

In [39]: B
Out[39]: 
array([[0, 0, 0],
       [1, 0, 2],
       [1, 0, 3],
       [1, 0, 4],
       [1, 1, 0],
       [1, 1, 1],
       [1, 1, 4]])

In [40]: out
Out[40]: 
array([[1, 1, 2],
       [1, 1, 3]])

对大型阵列的运行时测试-

Runtime test on large arrays -

In [107]: def in1d_approach(A,B):
     ...:     dims = np.maximum(B.max(0),A.max(0))+1
     ...:     return A[~np.in1d(np.ravel_multi_index(A.T,dims),\
     ...:                     np.ravel_multi_index(B.T,dims))]
     ...: 

In [108]: # Setup arrays with B as large array and A contains some of B's rows
     ...: B = np.random.randint(0,9,(1000,3))
     ...: A = np.random.randint(0,9,(100,3))
     ...: A_idx = np.random.choice(np.arange(A.shape[0]),size=10,replace=0)
     ...: B_idx = np.random.choice(np.arange(B.shape[0]),size=10,replace=0)
     ...: A[A_idx] = B[B_idx]
     ...: 

基于broadcasting的解决方案的计时-

Timings with broadcasting based solutions -

In [109]: %timeit A[np.all(np.any((A-B[:, None]), axis=2), axis=0)]
100 loops, best of 3: 4.64 ms per loop # @Kasramvd's soln

In [110]: %timeit A[~((A[:,None,:] == B).all(-1)).any(1)]
100 loops, best of 3: 3.66 ms per loop

基于更少内存占用空间的解决方案-

Timing with less memory footprint based solution -

In [111]: %timeit in1d_approach(A,B)
1000 loops, best of 3: 231 µs per loop

进一步的性能提升

in1d_approach通过将每一行视为索引元组来减少每一行.通过使用np.dot引入矩阵乘法，我们可以更有效地完成上述任务，就像-

in1d_approach reduces each row by considering each row as an indexing tuple. We can do the same a bit more efficiently by introducing matrix-multiplication with np.dot, like so -

def in1d_dot_approach(A,B):
    cumdims = (np.maximum(A.max(),B.max())+1)**np.arange(B.shape[1])
    return A[~np.in1d(A.dot(cumdims),B.dot(cumdims))]

让我们在更大的数组上针对以前的版本进行测试-

Let's test it against the previous on much larger arrays -

In [251]: # Setup arrays with B as large array and A contains some of B's rows
     ...: B = np.random.randint(0,9,(10000,3))
     ...: A = np.random.randint(0,9,(1000,3))
     ...: A_idx = np.random.choice(np.arange(A.shape[0]),size=10,replace=0)
     ...: B_idx = np.random.choice(np.arange(B.shape[0]),size=10,replace=0)
     ...: A[A_idx] = B[B_idx]
     ...: 

In [252]: %timeit in1d_approach(A,B)
1000 loops, best of 3: 1.28 ms per loop

In [253]: %timeit in1d_dot_approach(A, B)
1000 loops, best of 3: 1.2 ms per loop

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