Python Numpy Array returns multiple values for each index, yet does not allow indexing of those values

Question

When I print the array it looks like this

print(my_array)
[ 0 0 0 1 0 1...  
1 0 2 0 1 ]  
[ 0 0 0 1 0 1...  
1 0 2 0 1 ]  
none

When I index the array with a single index like

print(my_array[0])

it returns

0  
0  
none

but if I try, for example

print(my_array[0,0])

I get

"IndexError: too many indices for array"

Finally, the shape returns

(750,)  
(750,)  
None

and the type returns

class 'numpy.ndarray'   
class 'numpy.ndarray'  
None

This array was given to me as is for a homework assignment, so I don't know how it was made (Coursera Course). How can I split the three arrays? I can see that each 750 data points are repeats, and I just need one set of the 750 to use for the assignment.

The array is 1d with object dtype. Like a list is contains pointers to other arrays. It's not a 2d array (matrix) but rather more like a list of lists. — hpaulj
– hpaulj, Commented Oct 1, 2018 at 23:24

jpp · Accepted Answer · 2018-10-02 09:20:38Z

1

Your array has dtype object because it's not a numeric array with the same number of columns in each row. It looks something like this:

L = [np.array([0, 0, 1]), np.array([0, 0, 0]), None]

A = np.array(L)

array([array([0, 0, 1]),
       array([0, 0, 0]),
       None], dtype=object)

for i in A:
    print(type(i))    

<class 'numpy.ndarray'>
<class 'numpy.ndarray'>
<class 'NoneType'>

That's not how you should use NumPy. You can try removing the None values to construct a regular NumPy numeric array:

B = np.array([i for i in L if i is not None])

array([[0, 0, 1],
       [0, 0, 0]])

If in doubt, check the dtype of an array to check it has the correct type:

A.dtype  # dtype('O')
B.dtype  # dtype('int32')

answered Oct 2, 2018 at 9:20

jpp

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