Numpy.shape Gives Inconsistent Responses - Why?

Why does the program import numpy as np c = np.array([1,2]) print(c.shape) d = np.array([[1],[2]]).transpose() print(d.shape) give (2,) (1,2) as its output? Shouldn't it be (1,2

Solution 1:

When you invoke the .shape attribute of a ndarray, you get a tuple with as many elements as dimensions of your array. The length, ie, the number of rows, is the first dimension (shape[0])

• You start with an array : c=np.array([1,2]). That's a plain 1D array, so its shape will be a 1-element tuple, and shape[0] is the number of elements, so c.shape = (2,)
• Consider c=np.array([[1,2]]). That's a 2D array, with 1 row. The first and only row is [1,2], that gives us two columns. Therefore, c.shape=(1,2) and len(c)=1
• Consider c=np.array([[1,],[2,]]). Another 2D array, with 2 rows, 1 column: c.shape=(2,1) and len(c)=2.
• Consider d=np.array([[1,],[2,]]).transpose(): this array is the same as np.array([[1,2]]), therefore its shape is (1,2).

Another useful attribute is .size: that's the number of elements across all dimensions, and you have for an array cc.size = np.product(c.shape).

More information on the shape in the documentation.

Solution 2:

len(c.shape) is the "depth" of the array.

For c, the array is just a list (a vector), the depth is 1. For d, the array is a list of lists, the depth is 2.

Note:

c.transpose()
# array([1, 2])

which is not d, so this behaviour is not inconsistent.

dt = d.transpose()
# array([[1],
#        [2]])
dt.shape # (2,1)

Solution 3:

Quick Fix: check the .ndim property - if its 2, then the .shape property will work as you expect.

Reason Why: if the .ndim property is 2, then numpy reports a shape value that agrees with the convention. If the .ndim property is 1, then numpy just reports shape in a different way.

More talking: When you pass np.array a lists of lists, the .shape property will agree with standard notions of the dimensions of a matrix: (rows, columns).

If you pass np.array just a list, then numpy doesn't think it has a matrix on its hands, and reports the shape in a different way.

The question is: does numpy think it has a matrix, or does it think it has something else on its hands.

Solution 4:

transpose does not change the number of dimensions of the array. If c.ndim == 1, c.transpose() == c. Try:

c = np.array([1,2])
print c.shape
print c.T.shape
c = np.atleast_2d(c)
print c.shape
print c.T.shape

Solution 5:

Coming from Matlab, I also find it difficult that a single-dimensional array is not organized as (row_count, colum_count)

My function had to respond consistently on a single-dimensional ndarray like [x1, x2, x3] or a list of arrays [[x1, x2, x3], [x1, x2, x3], [x1, x2, x3]].

This worked for me:

dim = np.shape(subtract_matrix)[-1]

Picking the last dimension.