Numpy array dimensions

I'm currently trying to learn Numpy and Python. Given the following array:

import numpy as np
a = np.array([[1,2],[1,2]])

Is there a function that returns the dimensions of a (e.g.a is a 2 by 2 array)?

size() returns 4 and that doesn't help very much.

1

9 Answers

It is .shape:

ndarray.shape
Tuple of array dimensions.

Thus:

>>> a.shape
(2, 2)

4

First:

By convention, in Python world, the shortcut for numpy is np, so:

In [1]: import numpy as np
In [2]: a = np.array([[1,2],[3,4]])

Second：

In Numpy, dimension, axis/axes, shape are related and sometimes similar concepts:

dimension

In Mathematics/Physics, dimension or dimensionality is informally defined as the minimum number of coordinates needed to specify any point within a space. But in Numpy, according to the numpy doc, it's the same as axis/axes:

In Numpy dimensions are called axes. The number of axes is rank.

In [3]: a.ndim # num of dimensions/axes, *Mathematics definition of dimension*
Out[3]: 2

axis/axes

the nth coordinate to index an array in Numpy. And multidimensional arrays can have one index per axis.

In [4]: a[1,0] # to index `a`, we specific 1 at the first axis and 0 at the second axis.
Out[4]: 3 # which results in 3 (locate at the row 1 and column 0, 0-based index)

shape

describes how many data (or the range) along each available axis.

In [5]: a.shape
Out[5]: (2, 2) # both the first and second axis have 2 (columns/rows/pages/blocks/...) data

import numpy as np
>>> np.shape(a)
(2,2)

Also works if the input is not a numpy array but a list of lists

>>> a = [[1,2],[1,2]]
>>> np.shape(a)
(2,2)

Or a tuple of tuples

>>> a = ((1,2),(1,2))
>>> np.shape(a)
(2,2)

1

You can use .shape

In: a = np.array([[1,2,3],[4,5,6]])
In: a.shape
Out: (2, 3)
In: a.shape[0] # x axis
Out: 2
In: a.shape[1] # y axis
Out: 3

You can use .ndim for dimension and .shape to know the exact dimension:

>>> var = np.array([[1,2,3,4,5,6], [1,2,3,4,5,6]])
>>> var.ndim
2
>>> varshape
(2, 6) 

You can change the dimension using .reshape function:

>>> var_ = var.reshape(3, 4)
>>> var_.ndim
2
>>> var_.shape
(3, 4)

0

The shape method requires that a be a Numpy ndarray. But Numpy can also calculate the shape of iterables of pure python objects:

np.shape([[1,2],[1,2]])

a.shape is just a limited version of np.info(). Check this out:

import numpy as np
a = np.array([[1,2],[1,2]])
np.info(a)

Out

class: ndarray
shape: (2, 2)
strides: (8, 4)
itemsize: 4
aligned: True
contiguous: True
fortran: False
data pointer: 0x27509cf0560
byteorder: little
byteswap: False
type: int32

rows = a.shape[0] # 2
cols = a.shape[1] # 2
a.shape #(2,2)
a.size # rows * cols = 4

Execute below code block in python notebook.

import numpy as np
a = np.array([[1,2],[1,2]])
print(a.shape)
print(type(a.shape))
print(a.shape[0])

output

(2, 2)

<class 'tuple'>

2

then you realized that a.shape is a tuple. so you can get any dimension's size by a.shape[index of dimention]