Numpy array dimensions

PythonArraysNumpyDimensions

Python Problem Overview


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.

Python Solutions


Solution 1 - Python

It is .shape:

> ndarray.shape
> Tuple of array dimensions.

Thus:

>>> a.shape
(2, 2)

Solution 2 - Python

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

Solution 3 - Python

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)

Solution 4 - Python

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

Solution 5 - Python

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)

Solution 6 - Python

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]])

Solution 7 - Python

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

Solution 8 - Python

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

Solution 9 - Python

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)

>

>2

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

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