使用for循环在numpy数组中查找元素的索引

Question

As an input i have 2d numpy array, lets suppose this one:

my_array = np.matrix([[3, 7, 0, 0],
                      [0, 2, 0, 0],
                      [0, 0, 0, 0],
                      [0, 0, 1, 0]])

I have to find the index of every element in that array in which the sum of elements in that row and column == 0. In this case, the answer would be (2, 3), since the sum of elements in second row = 0 and sum of elements in 3rd column = 0. So far i came up with this:

solution = [(i, j) for i in range(my_array.shape[0]) for j in range(my_array.shape[1]) if 1 not in my_array[i] and 1 not in my_array[:, j]]

The problem is, I want to do this without using for loop.

I've tried using np.where and np.sum , ended up with this:

np.where(np.sum(my_array, axis=1) == 0 and np.sum(my_array, axis=0) == 0)

but i end up with this error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Any suggestion about how can i fix this error or just use other method to find the indexes?

Answer 1

The problem with your where expression occurs inside it, in your attempt to combine two conditions:

In [210]: np.sum(arr, axis=1) == 0 and np.sum(arr, axis=0) == 0                 
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-210-46c837435a31> in <module>
----> 1 np.sum(arr, axis=1) == 0 and np.sum(arr, axis=0) == 0

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
In [211]: (np.sum(arr, axis=1) == 0) & (np.sum(arr, axis=0) == 0)               
Out[211]: array([False, False, False, False])

You have to wrap the == test inside () so it occurs first, and you have to use & to perform element-wise and . and is a scalar operation, and does not play well with boolean arrays.

The row and column tests are:

In [212]: arr.sum(0)==0                                                         
Out[212]: array([False, False, False,  True])
In [213]: arr.sum(1)==0                                                         
Out[213]: array([False, False,  True, False])

but you want a kind of outer or cartesian combination, not a simple element-wise combination (that would be more obvious if there were different numbers of rows and columns).

In [218]: (arr.sum(1)==0)[:,None] & (arr.sum(0)==0)                             
Out[218]: 
array([[False, False, False, False],
       [False, False, False, False],
       [False, False, False,  True],
       [False, False, False, False]])
In [219]: np.where(_)                                                           
Out[219]: (array([2]), array([3]))

Or with the keepdims parameter of sum :

In [220]: arr.sum(0, keepdims=True)==0                                          
Out[220]: array([[False, False, False,  True]])
In [221]: arr.sum(1, keepdims=True)==0                                          
Out[221]: 
array([[False],
       [False],
       [ True],
       [False]])
In [222]: np.where(_220 & _221)             # Out[220] etc                                    
Out[222]: (array([2]), array([3]))

Answer 2

Here is a solution using product from itertools. Creating a list of rows and columns with sums == 0 and finding the combinations between them.

from itertools import product

my_array = np.matrix([[3, 7, 0, 0],
                      [0, 2, 0, 0],
                      [0, 0, 0, 0],
                      [0, 0, 1, 0]])


a = np.argwhere(my_array.sum(axis = 1) == 0)[:,0]
b = np.argwhere(my_array.sum(axis = 0) == 0)[:,1]

np.array(list(product(a,b)))