CODEWITHSUNDEEP
pythonarraysnumpy
memecs
memecs

Reputation: 7564

numpy.ndarray.shape changing dimension

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])])

Upvotes: 1

Views: 576

Answers (1)

ZJS
ZJS

Reputation: 4051

You can use 

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

Upvotes: 1

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