从列表中重塑numpy数组

Question

I have the following problem with shape of ndarray:

out.shape = (20,)
reference.shape = (20,0)
norm = [out[i] / np.sum(out[i]) for i in range(len(out))]
# norm is a list now so I convert it to ndarray:
norm_array = np.array((norm))
norm_array.shape = (20,30)

# error: operands could not be broadcast together with shapes (20,30) (20,) 
diff = np.fabs(norm_array - reference)

How can I change shape of norm_array from (20,30) into (20,) or reference to (20,30), so I can substract them?

EDIT: Can someone explain me, why they have different shape, if I can access both single elements with norm_array[0][0] and reference[0][0] ?

Answer 1

I am not sure what you are trying to do exactly, but here is some information on numpy arrays.

A 1-d numpy array is a row vector with a shape that is a single-valued tuple:

>>> np.array([1,2,3]).shape
(3,) 

You can create multidimensional arrays by passing in nested lists. Each sub-list is a 1-d row vector of length 1, and there are 3 of them.

>>> np.array([[1],[2],[3]]).shape
(3,1)

Here is the weird part. You can create the same array, but leave the lists empty. You end up with 3 row vectors of length 0.

>>> np.array([[],[],[]]).shape
(3,0)

This is what you have for you reference array, an array with structure but no values. This brings me back to my original point:

You can't subtract an empty array.

Answer 2

If I make 2 arrays with the shapes you describe, I get an error

In [1856]: norm_array=np.ones((20,30))
In [1857]: reference=np.ones((20,0))
In [1858]: norm_array-reference
...
ValueError: operands could not be broadcast together with shapes (20,30) (20,0) 

But it's different from yours. But if I change the shape of reference , the error messages match.

In [1859]: reference=np.ones((20,))
In [1860]: norm_array-reference
 ...
ValueError: operands could not be broadcast together with shapes (20,30) (20,) 

So your (20,0) is wrong. I don't know if you mistyped something or not.

But if I make reference 2d with 1 in the last dimension, broadcasting works, producing a difference that matches (20,30) in shape:

In [1861]: reference=np.ones((20,1))
In [1862]: norm_array-reference

If reference = np.zeros((20,)) , then I could use reference[:,None] to add that singleton last dimension.

If reference is (20,), you can't do reference[0][0] . reference[0][0] only works with 2d arrays with at least 1 in the last dim. reference[0,0] is the preferred way of indexing a single element of a 2d array.

So far this is normal array dimensions and broadcasting; something you'll learn with use.

===============

I'm puzzled about the shape of out . If it is (20,), how does norm_array end up as (20,30). out must consist of 20 arrays or lists, each of which has 30 elements.

If out was 2d array, we could normalize without iteration

In [1869]: out=np.arange(12).reshape(3,4)

with the list comprehension:

In [1872]: [out[i]/np.sum(out[i]) for i in range(out.shape[0])]
Out[1872]: 
[array([ 0.        ,  0.16666667,  0.33333333,  0.5       ]),
 array([ 0.18181818,  0.22727273,  0.27272727,  0.31818182]),
 array([ 0.21052632,  0.23684211,  0.26315789,  0.28947368])]
In [1873]: np.array(_)   # and to array
Out[1873]: 
array([[ 0.        ,  0.16666667,  0.33333333,  0.5       ],
       [ 0.18181818,  0.22727273,  0.27272727,  0.31818182],
       [ 0.21052632,  0.23684211,  0.26315789,  0.28947368]])

Instead take row sums, and tell it to keep it 2d for ease of further use

In [1876]: out.sum(axis=1,keepdims=True)
Out[1876]: 
array([[ 6],
       [22],
       [38]])

now divide

In [1877]: out/out.sum(axis=1,keepdims=True)
Out[1877]: 
array([[ 0.        ,  0.16666667,  0.33333333,  0.5       ],
       [ 0.18181818,  0.22727273,  0.27272727,  0.31818182],
       [ 0.21052632,  0.23684211,  0.26315789,  0.28947368]])