从一个数组中删除另一个数组中的元素

Question

假设我有这些二维数组 A 和 B。

如何从A 中删除B 中的元素。 （集合论中的补码：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。

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]]

Answer 1

list comprehension 有一个简单的解决方案，

A = [i for i in A if i not in B]

结果

[[1, 1, 2], [1, 1, 3]]

列表推导不会从数组中删除元素，它只是重新分配 - 如果要删除元素，请使用此方法：

for i in B:
     if i in A:
     A.remove(i)

Answer 2

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，内存占用更少，在处理大型数组时可能会很有用 -

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))]

示例运行 -

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]])

大型数组的运行时测试 -

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 的解决方案的时间安排 -

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

基于更少内存占用的时间解决方案 -

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

进一步提升性能

in1d_approach 通过将每一行视为一个索引元组来减少每一行。我们可以通过使用np.dot 引入矩阵乘法来更有效地做同样的事情，就像这样 -

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对照之前的方法进行测试 -

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

Answer 3

这是一个带有广播的Numpythonic方法：

In [83]: A[np.all(np.any((A-B[:, None]), axis=2), axis=0)]
Out[83]: 
array([[1, 1, 2],
       [1, 1, 3]])

这是其他答案的时间：

In [90]: def cal_diff(A, B):
   ....:     A_rows = A.view([('', A.dtype)] * A.shape[1])
   ....:     B_rows = B.view([('', B.dtype)] * B.shape[1])
   ....:     return np.setdiff1d(A_rows, B_rows).view(A.dtype).reshape(-1, A.shape[1])
   ....: 

In [93]: %timeit cal_diff(A, B)
10000 loops, best of 3: 54.1 µs per loop

In [94]: %timeit A[np.all(np.any((A-B[:, None]), axis=2), axis=0)]
100000 loops, best of 3: 9.41 µs per loop

# Even better with Divakar's suggestion
In [97]: %timeit A[~((A[:,None,:] == B).all(-1)).any(1)]
100000 loops, best of 3: 7.41 µs per loop

好吧，如果您正在寻找一种更快的方法，您应该寻找减少比较次数的方法。在这种情况下（不考虑顺序），您可以从您的行中生成一个唯一的数字，并比较可以通过将项目的 2 的幂求和来完成的数字。

这是 Divakar 的 in1d 方法的基准：

In [144]: 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 [146]: %timeit in1d_approach(A, B)
10000 loops, best of 3: 23.8 µs per loop

In [145]: %timeit A[~np.in1d(np.power(A, 2).sum(1), np.power(B, 2).sum(1))]
10000 loops, best of 3: 20.2 µs per loop

您可以使用np.diff 获得与订单无关的结果：

In [194]: B=np.array([[0, 0, 0,], [1, 0, 2,], [1, 0, 3,], [1, 0, 4,], [1, 1, 0,], [1, 1, 1,], [1, 1, 4,], [4, 1, 1]])

In [195]: A[~np.in1d(np.diff(np.diff(np.power(A, 2))), np.diff(np.diff(np.power(B, 2))))]
Out[195]: 
array([[1, 1, 2],
       [1, 1, 3]])

In [196]: %timeit A[~np.in1d(np.diff(np.diff(np.power(A, 2))), np.diff(np.diff(np.power(B, 2))))]
10000 loops, best of 3: 30.7 µs per loop

Divakar 设置的基准测试：

In [198]: B = np.random.randint(0,9,(1000,3))

In [199]: A = np.random.randint(0,9,(100,3))

In [200]: A_idx = np.random.choice(np.arange(A.shape[0]),size=10,replace=0)

In [201]: B_idx = np.random.choice(np.arange(B.shape[0]),size=10,replace=0)

In [202]: A[A_idx] = B[B_idx]

In [203]: %timeit A[~np.in1d(np.diff(np.diff(np.power(A, 2))), np.diff(np.diff(np.power(B, 2))))]
10000 loops, best of 3: 137 µs per loop

In [204]: %timeit A[~np.in1d(np.power(A, 2).sum(1), np.power(B, 2).sum(1))]
10000 loops, best of 3: 112 µs per loop

In [205]: %timeit in1d_approach(A, B)
10000 loops, best of 3: 115 µs per loop

用更大的数组计时（Divakar 的解决方案稍微快一点）：

In [231]: %timeit A[~np.in1d(np.diff(np.diff(np.power(A, 2))), np.diff(np.diff(np.power(B, 2))))]
1000 loops, best of 3: 1.01 ms per loop

In [232]: %timeit A[~np.in1d(np.power(A, 2).sum(1), np.power(B, 2).sum(1))]
1000 loops, best of 3: 880 µs per loop

In [233]:  %timeit in1d_approach(A, B)
1000 loops, best of 3: 807 µs per loop

Answer 4

如果你想用 numpy 的方式来做，

import numpy as np

A = np.array([[1, 1, 1,], [1, 1, 2], [1, 1, 3], [1, 1, 4]])
B = np.array([[0, 0, 0], [1, 0, 2], [1, 0, 3], [1, 0, 4], [1, 1, 0], [1, 1, 1], [1, 1, 4]])
A_rows = A.view([('', A.dtype)] * A.shape[1])
B_rows = B.view([('', B.dtype)] * B.shape[1])

diff_array = np.setdiff1d(A_rows, B_rows).view(A.dtype).reshape(-1, A.shape[1])

正如@Rahul 建议的那样，对于一个非 numpy 简单的解决方案，

diff_array = [i for i in A if i not in B]

Answer 5

另一个非 numpy 解决方案：

[i for i in A if i not in B]