Minimum of Numpy Array Ignoring Diagonal

问题:

I have to find the maximum value of a numpy array ignoring the diagonal elements.

np.amax() provides ways to find it ignoring specific axes. How can I achieve the same ignoring all the diagonal elements?

回答1:

You could use a mask

mask = np.ones(a.shape, dtype=bool)
np.fill_diagonal(mask, 0)
max_value = a[mask].max()

where a is the matrix you want to find the max of. The mask selects the off-diagonal elements, so a[mask] will be a long vector of all the off-diagonal elements. Then you just take the max.

Or, if you don't mind modifying the original array

np.fill_diagonal(a, -np.inf)
max_value = a.max()

Of course, you can always make a copy and then do the above without modifying the original. Also, this is assuming that a is some floating point format.

回答2:

Another possibility is to use NumPy's as_strided to push the diagonal to the first column and then slice it off:

In [1]: import numpy as np
In [2]: from numpy.lib.stride_tricks import as_strided
In [3]: b = np.arange(0,25,1).reshape((5,5))
In [4]: b
Out[4]: array([[ 0,  1,  2,  3,  4],
               [ 5,  6,  7,  8,  9],
               [10, 11, 12, 13, 14],
               [15, 16, 17, 18, 19],
               [20, 21, 22, 23, 24]])
In [5]: n = b.shape[0]
In [6]: np.max(as_strided(b, (n-1,n+1), (b.itemsize*(n+1), b.itemsize))[:,1:])
Out[6]: 23

Where the argument to np.max is the shifted view on b:

In [7]: as_strided(b, (n-1,n+1), (b.itemsize*(n+1), b.itemsize))
Out[7]: array([[ 0,  1,  2,  3,  4,  5],
               [ 6,  7,  8,  9, 10, 11],
               [12, 13, 14, 15, 16, 17],
               [18, 19, 20, 21, 22, 23]])

so that:

In [8]: as_strided(b, (n-1,n+1), (b.itemsize*(n+1), b.itemsize))[:,1:]
Out[8]: array([[ 1,  2,  3,  4,  5],
               [ 7,  8,  9, 10, 11],
               [13, 14, 15, 16, 17],
               [19, 20, 21, 22, 23]])

回答3:

This should work

>>> import numpy as np
>>> import numpy.random

# create sample matrix
>>> a = numpy.random.randint(10,size=(8,8))
>>> a[0,0] = 100
>>> a
array([[100, 8, 6, 5, 5, 7, 4, 5],
   [4, 6, 1, 7, 4, 5, 8, 5],
   [0, 2, 0, 7, 4, 2, 7, 9],
   [5, 7, 5, 9, 8, 3, 2, 8],
   [2, 1, 3, 4, 0, 7, 8, 1],
   [6, 6, 7, 6, 0, 6, 6, 8],
   [6, 0, 1, 9, 7, 7, 9, 3],
   [0, 5, 5, 5, 1, 5, 4, 4]])

# create mask
>>> mask = np.ones((8,8)) 
>>> mask = (mask - np.diag(np.ones(8))).astype(np.bool)
>>> mask
array([[False,  True,  True,  True,  True,  True,  True,  True],
   [ True, False,  True,  True,  True,  True,  True,  True],
   [ True,  True, False,  True,  True,  True,  True,  True],
   [ True,  True,  True, False,  True,  True,  True,  True],
   [ True,  True,  True,  True, False,  True,  True,  True],
   [ True,  True,  True,  True,  True, False,  True,  True],
   [ True,  True,  True,  True,  True,  True, False,  True],
   [ True,  True,  True,  True,  True,  True,  True, False]], dtype=bool)

# calculate the maximum
>>> np.amax(a[mask])
9