numpy.ndarray.shape changing dimension

Question

The tuple holding the dimensions of a numpy array (numpy.ndarray.shape) changes size. E.g:

len(numpy.array([1,2,3]).shape) -> 1, shape=(1,)
len(numpy.array([[1,2,3],[4,5,6]]).shape) -> 2, shape=(2,3)

Is there any other way to get dimensions invariant to the type of the array?

Here is an example of the problem I encounter quite often:

mat3D = np.arange(27).reshape(3,3,3)
mat2D = np.arange(9)

def processMatrix(mat):
  if M.ndim == 2:
    return foo(mat)
  else:
    return np.array([foo(mat[:,:,c]) for c in range(mat.shape[2])])

Having mat2D.shape = (3,3,1) would simplify the code to:

def processMatrix(mat):
    return np.array([foo(mat[:,:,c]) for c in range(mat.shape[2])])

The shape tuple has to change size, to match the number of dimensions. If you have N dimensions, you need N numbers to specify the size. Could you be more specific about what you find annoying about this? Maybe your real issue is different from what it seems. — DSM
– DSM, Commented Jul 30, 2014 at 19:47
Numpy arrays can have N dimensions. E.g. np.array([[[1,2],[3,4]],[[5,6],[7,8]]]).shape is (2,2,2). The "invariant" is that a.ndim is len(a.shape). — Warren Weckesser
– Warren Weckesser, Commented Jul 30, 2014 at 20:00
@DSM I often find myself to use both 1D and 2D arrays interchangeably and the shape tuple changing size requires unnecessary if-else conditions. I guess I should just wrap everything into matrices. Unfortunately most of the functions returns arrays independently from the type of the input. — memecs
– memecs, Commented Jul 30, 2014 at 20:01
Give an example of one of your unnecessary if-else conditions. Maybe there is another way to handle it. — Warren Weckesser
– Warren Weckesser, Commented Jul 30, 2014 at 20:04

ZJS · Accepted Answer · 2014-07-30 19:45:50Z

1

You can use

numpy.array([[1,2,3],[4,5,6]]).ndim

answered Jul 30, 2014 at 19:45

ZJS

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