numpy.shape gives inconsistent responses - why?

Question

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)
(1,2)

instead? I got this in both python 2.7.3 and python 3.2.3

Answer 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 c c.size = np.product(c.shape).

More information on the shape in the documentation.