what does numpy ndarray shape do?

Question

I have a simple question about the .shape function, which confused me a lot.

a = np.array([1, 2, 3])   # Create a rank 1 array
print(type(a))            # Prints "<class 'numpy.ndarray'>"
print(a.shape)            # Prints "(3,)"

b = np.array([[1,2,3],[4,5,6]])    # Create a rank 2 array
print(b.shape)                     # Prints "(2, 3)"

What did the .shape exactly do? count how many rows, how many columns, then the a.shape suppose to be, (1,3), one row three columns, right?

Answer 1

yourarray.shape or np.shape() or np.ma.shape() returns the shape of your ndarray as a tuple; And you can get the (number of) dimensions of your array using yourarray.ndim or np.ndim(). (i.e. it gives the n of the ndarray since all arrays in NumPy are just n-dimensional arrays (shortly called as ndarrays))

For a 1D array, the shape would be (n,) where n is the number of elements in your array.

For a 2D array, the shape would be (n,m) where n is the number of rows and m is the number of columns in your array.

Please note that in 1D case, the shape would simply be (n, ) instead of what you said as either (1, n) or (n, 1) for row and column vectors respectively.

This is to follow the convention that:

For 1D array, return a shape tuple with only 1 element   (i.e. (n,))
For 2D array, return a shape tuple with only 2 elements (i.e. (n,m))
For 3D array, return a shape tuple with only 3 elements (i.e. (n,m,k))
For 4D array, return a shape tuple with only 4 elements (i.e. (n,m,k,j))

and so on.

Also, please see the example below to see how np.shape() or np.ma.shape() behaves with 1D arrays and scalars:

# sample array
In [10]: u = np.arange(10)

# get its shape
In [11]: np.shape(u)    # u.shape
Out[11]: (10,)

# get array dimension using `np.ndim`
In [12]: np.ndim(u)
Out[12]: 1

In [13]: np.shape(np.mean(u))
Out[13]: ()       # empty tuple (to indicate that a scalar is a 0D array).

# check using `numpy.ndim`
In [14]: np.ndim(np.mean(u))
Out[14]: 0

P.S.: So, the shape tuple is consistent with our understanding of dimensions of space, at least mathematically.