Given input
A = np.array([[1,2,3],[4,5,6],[7,8,9]])
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
Need output :
array([[2, 3],
[4, 6],
[7, 8]])
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
Approach #1
One approach with masking -
A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)
Sample run -
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]])
Approach #2
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]
Approach #3 (Strides Strikes!)
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)
Runtime test
Approaches as funcs :
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)
Timings -
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
Solution steps:
range(0, len(x_no_diag), len(x) + 1)The function:
import numpy as np
def remove_diag(x):
x_no_diag = np.ndarray.flatten(x)
x_no_diag = np.delete(x_no_diag, range(0, len(x_no_diag), len(x) + 1), 0)
x_no_diag = x_no_diag.reshape(len(x), len(x) - 1)
return x_no_diag
Example:
>>> x = np.random.randint(5, size=(3,3))
array([[0, 2, 3],
[3, 4, 1],
[2, 4, 0]])
>>> remove_diag(x)
array([[2, 3],
[3, 1],
[2, 4]])
I know I'm late to this party, but I have what I believe is a simper solution. So you want to remove the diagonal? Okay cool:
`
arr = np.array([[1,2,3],[4,5,6],[7,8,9]]).astype(np.float)
np.fill_diagonal(arr, np.nan)
arr[~np.isnan(arr)].reshape(arr.shape[0], arr.shape[1] - 1)
Perhaps the cleanest way, based on Divakar's first solution but using len(array) instead of array.shape[0], is:
array_without_diagonal = array[~np.eye(len(array), dtype=bool)].reshape(len(array), -1)
Just with numpy, assuming a square matrix:
new_A = numpy.delete(A,range(0,A.shape[0]**2,(A.shape[0]+1))).reshape(A.shape[0],(A.shape[1]-1))
If you do not mind creating a new array, then you can use list comprehension.
A = np.array([A[i][A[i] != A[i][i]] for i in range(len(A))])
Rerunning the same methods as @Divakar,
A = np.random.randint(11,99,(5000,5000))
skip_diag_masking
85.7 ms ± 1.55 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_broadcasting
163 ms ± 1.77 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_strided
52.5 ms ± 2.54 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_list_comp
101 ms ± 347 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
I love all the answers here, but would like to add one in case your numpy object has more than 2-dimensions. In that case, you can use the following adjustment of Divakar's approach #1:
def remove_diag(A):
removed = A[~np.eye(A.shape[0], dtype=bool)].reshape(A.shape[0], int(A.shape[0])-1, -1)
return np.squeeze(removed)
The other approach is to use numpy.delete(). assuming square matrix, you can use:
numpy.delete(A,range(0,A.shape[0]**2,A.shape[0])).reshape(A.shape[0],A.shape[1]-1)
