删除numpy数组的对角元素 [英] Deleting diagonal elements of a numpy array

问题描述

输入

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

需要输出:

array([[2, 3],
       [4, 6],
       [7, 8]])

使用迭代或循环很容易做到这一点，但是应该有一种简洁的方法可以不使用循环来做到这一点.谢谢

It is easy to use iteration or loop to do this, but there should be a neat way to do this without using loops. Thanks

推荐答案

方法1

使用masking-

A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)

样品运行-

In [395]: A
Out[395]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [396]: A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)
Out[396]: 
array([[2, 3],
       [4, 6],
       [7, 8]])

方法2

使用非对角元素的常规模式，可以通过范围数组的广播加法来跟踪-

Using the regular pattern of non-diagonal elements that could be traced with broadcasted additions with range arrays -

m = A.shape[0]
idx = (np.arange(1,m+1) + (m+1)*np.arange(m-1)[:,None]).reshape(m,-1)
out = A.ravel()[idx]

方法3 (大步向前！)

滥用先前方法中的非对角元素的常规模式，我们可以引入

Abusing the regular pattern of non-diagonal elements from previous approach, we can introduce np.lib.stride_tricks.as_strided and some slicing help, like so -

m = A.shape[0]
strided = np.lib.stride_tricks.as_strided
s0,s1 = A.strides
out = strided(A.ravel()[1:], shape=(m-1,m), strides=(s0+s1,s1)).reshape(m,-1)

---

运行时测试

作为功能的方法:

def skip_diag_masking(A):
    return A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)

def skip_diag_broadcasting(A):
    m = A.shape[0]
    idx = (np.arange(1,m+1) + (m+1)*np.arange(m-1)[:,None]).reshape(m,-1)
    return A.ravel()[idx]

def skip_diag_strided(A):
    m = A.shape[0]
    strided = np.lib.stride_tricks.as_strided
    s0,s1 = A.strides
    return strided(A.ravel()[1:], shape=(m-1,m), strides=(s0+s1,s1)).reshape(m,-1)

时间-

In [528]: A = np.random.randint(11,99,(5000,5000))

In [529]: %timeit skip_diag_masking(A)
     ...: %timeit skip_diag_broadcasting(A)
     ...: %timeit skip_diag_strided(A)
     ...: 
10 loops, best of 3: 56.1 ms per loop
10 loops, best of 3: 82.1 ms per loop
10 loops, best of 3: 32.6 ms per loop

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