L1 A quick Python refresher

License: CC-BY-NC-SA 4.0

Author: Murilo M. Marinho (murilo.marinho@manchester.ac.uk)

Prerequisites¶

The user of this notebook is expected to have prior knowledge in

Basic Python [Tutorial]
Numpy
- [Tutorial: basics for beginners]
- [Tutorial: for MATLAB users]
Jupyter Notebook Basics [Tutorial]

I found an issue¶

Thank you! Please report it at https://github.com/MarinhoLab/OpenExecutableBooksRobotics/issues

Variable assignment¶

Let

a\triangleq 10,b\triangleq 5.

(1)

We can replicate the above in Python with

a = 10
b = 5

Output variables¶

Variables can be output using print. For example, for $a$

print(a)

Output text and variables using f-strings¶

To output $a$ and $b$ within a string, we can use print and f-strings as follows

print(f'The value of a = {a} and b = {b}.')

The value of a = 10 and b = 5.

Basic Arithmetics¶

Basic mathematical operations are trivially performed as follows.

Sum¶

c = a + b.

(2)

c = a + b

print(f'c={c}')

c=15

Subtraction¶

c = a - b.

(3)

c = a - b

print(f'c={c}')

c=5

Multiplication¶

c=ab

(4)

c = a * b

print(f'c={c}')

c=50

Division¶

c = \frac{a}{b}

(5)

c = a / b

print(f'c={c}')

c=2.0

Exponentiation¶

c = a^{b}

(6)

c = a ** b

print(f'c={c}')

c=100000

Math functions¶

For the following functions, we will need Python’s built-in math module.

from math import sqrt, exp, log, pi, sin, cos, tan

Square root¶

c = \sqrt{a}

(7)

c = sqrt(a)

n-th root¶

The nth root,

c= \sqrt[n]{a}, n \in \mathbb{N},

(8)

does not seem to have a shorthand version in Python, but can be computed through simple properties such as

c = \sqrt[n]{a} = a^{\frac{1}{n}} = e^{\frac{ln(a)}{n}}.

(9)

For example, suppose that

n = 3.

(10)

Then,

n=3

and we can calculate the n-th root like so

# n-th root using fractional exponent. Might be easier but most languages do not support a similar syntax
c = a ** (1/n)

print(f'c={c}')

c=2.154434690031884

or like so

c = exp(log(a)/n)

print(f'c={c}')

c=2.154434690031884

and both should output the same value.

Trigonometric functions¶

\phi = \frac{\pi}{4},

(11)

s_{\phi} = \sin \left( \phi \right),

(12)

c_{\phi} = \cos \left( \phi \right),

(13)

t_{\phi} = \tan \left( \phi \right).

(14)

phi = pi/4.0
s_phi = sin(phi)
c_phi = cos(phi)
t_phi = tan(phi)

print(f'phi={phi}')
print(f's_phi={s_phi}')
print(f'c_phi={c_phi}')
print(f't_phi={t_phi}')

phi=0.7853981633974483
s_phi=0.7071067811865475
c_phi=0.7071067811865476
t_phi=0.9999999999999999

Linear Algebra with Numpy¶

Installing the library¶

Just in case numpy is not already installed, we can install it with the following command. Nothing will happen if the library is already installed.

%%capture
%pip install numpy

Importing the library¶

import numpy as np

Instantiating vectors¶

A row vector can be instantiated from a list of lists. For instance, for

\mathbf{\boldsymbol{ v }} = \left[\begin{array}{ccc} 1 & 2 \end{array}\right],

(15)

we have

# Note the double [[]] to instanteate a vector with explicit row shape.
v = np.array([[1, 2]])

print(f'v={v}')

v=[[1 2]]

A column vector can be instantiated from a list of singleton lists. For instance, for

\mathbf{\boldsymbol{ u }} = \left[\begin{array}{ccc} 1 \\ 2 \end{array}\right],

(16)

we have

# Note that each row is defined by a single element within a [], while the whole vector is within an external []
u = np.array([[1],
              [2]])

print(f'u={u}')

u=[[1]
 [2]]

Dot product¶

\mathbf{\boldsymbol{ c }} = <\mathbf{\boldsymbol{ u }},\mathbf{\boldsymbol{ u }}>

(17)

c = np.vdot(u,u)

print(f'c={c}')

c=5

Cross product¶

Cross product is defined for vectors in $\mathbb{R}^3$ .

For example, for

\mathbf{\boldsymbol{ u }}_3 = \left[\begin{array}{ccc} 1 \\ 2 \\ 3 \end{array}\right],

(18)

and

\mathbf{\boldsymbol{ v }}_3 = \left[\begin{array}{ccc} 4 \\ 5 \\ 6 \end{array}\right],

(19)

we can obtain the cross product

\mathbf{\boldsymbol{ c }} = \mathbf{\boldsymbol{ u }}_3 \times \mathbf{\boldsymbol{ v }}_3

(20)

u3 = np.array([[1, 2, 3]])
v3 = np.array([[4, 5, 6]])

c = np.cross(u3,v3)

print(f'c={c}')

c=[[-3  6 -3]]

Euclidean norm¶

\mathbf{\boldsymbol{ c }} = ||\mathbf{\boldsymbol{ u }}||

(21)

Note that the function is np.linalg.norm, as the norm calculation is within the module linalg.

c = np.linalg.norm(u)

print(f'c={c}')

c=2.23606797749979

Instantiating matrices¶

For instance, suppose that we want to instantiate two real square matrices

\mathbf{\boldsymbol{ A }} = \left[\begin{array}{ccc} 1 & 2 \\ 3 & 4 \end{array}\right], \mathbf{\boldsymbol{ B }} = \left[\begin{array}{ccc} 5 & 6 \\ 7 & 8 \end{array}\right]

(22)

A = np.array([[1, 2], 
              [3, 4]])
B = np.array([[5, 6], 
              [7, 8]])

print(f'A={A},\n\nB={B}')

A=[[1 2]
 [3 4]],

B=[[5 6]
 [7 8]]

Transpose¶

\mathbf{\boldsymbol{ C }} = \mathbf{\boldsymbol{ A }}^T

(23)

C = A.T

print(f'C={C}')

C=[[1 3]
 [2 4]]

Sum¶

\mathbf{\boldsymbol{ C }} = \mathbf{\boldsymbol{ A }} + \mathbf{\boldsymbol{ B }}

(24)

C = A + B

print(f'C={C}')

C=[[ 6  8]
 [10 12]]

Subtraction¶

\mathbf{\boldsymbol{ C }} = \mathbf{\boldsymbol{ A }} - \mathbf{\boldsymbol{ B }}

(25)

C = A - B

print(f'C={C}')

C=[[-4 -4]
 [-4 -4]]

Matrix multiplication¶

For instance,

C = AB

(26)

is implemented with

The matrix multiplication operator, @, is very unusual. Pay close attention. Mistaking this can be a major source of bugs and confusion.

C = A @ B # Alternatively C = np.matmul(A,B), but that is too verbose

print(f'C={C}')

C=[[19 22]
 [43 50]]

which will naturally work for the vectors we defined. For example

\mathbf{\boldsymbol{ c }} = \mathbf{\boldsymbol{ u }}\mathbf{\boldsymbol{ v }} = \left[\begin{array}{ccc} 1 & 2 \\ 2 & 4 \end{array}\right],

(27)

\mathbf{\boldsymbol{ d }} = \mathbf{\boldsymbol{ v }}\mathbf{\boldsymbol{ u }} = 5.

(28)

c = u @ v 
d = v @ u 

print(f'c={c},\n\nd={d}')

c=[[1 2]
 [2 4]],

d=[[5]]

and, of course, matrices and vectors

\mathbf{\boldsymbol{ c }} = \mathbf{\boldsymbol{ A }}\mathbf{\boldsymbol{ u }}

(29)

We only use the "." sign to denote matrix multiplication when otherwise it would be difficult to read the equation.

c = A @ u 

print(f'c={c},\n')

c=[[ 5]
 [11]],

Diagonal matrices¶

Diagonal matrices get increasingly sparse with size, so it is important to have shorthand commands for creating them. For instance, suppose that we have the following diagonal matrix

\mathbf{\boldsymbol{ D }} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right] ,

(30)

this can be instantiated in numpy with

D = np.diag([1, 2, 3])

print(f'D={D}.')

D=[[1 0 0]
 [0 2 0]
 [0 0 3]].

Identity matrix¶

Among frequently used diagonal matrices, the identity matrix appears frequently. For instance,

\mathbf{\boldsymbol{ I }}_3 = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right],

(31)

can be instantiated in numpy with

I_3 = np.eye(3)

print(f'I_3={I_3}.')

I_3=[[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]].

Zero matrix¶

Another frequently used matrix is the zero matrix. For instance,

\mathbf{\boldsymbol{ O }}_3 = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right],

(32)

The np.zeros function takes a tuple to generate a properly sized matrix. Do not confuse it with np.eye that accepts a scalar.

O_3 = np.zeros((3,3))

print(f'O_3={O_3}.')

O_3=[[0. 0. 0.]
 [0. 0. 0.]
 [0. 0. 0.]].

Exercises¶

Exercise 1¶

For $\phi = \pi/4$ , let

e_1 = \sin(\phi) + 4\cos(\frac{\phi}{5}).

(33)

Using the math module which is already imported, calculate the value of $e_1$ and store it in the variable e1 shown in the cell below.

e1 = None # Replace None with your solution to this exercise.

Exercise 2¶

Given

\mathbf{\boldsymbol{ A }}_2 = \left[\begin{array}{ccc} 5 & 2 \\ 3 & 5 \end{array}\right]

(34)

and

\mathbf{\boldsymbol{ B }}_2 = \left[\begin{array}{ccc} 5 & 3 \\ 3 & 8 \end{array}\right].

(35)

Let

\mathbf{\boldsymbol{ C }}_2 = \mathbf{\boldsymbol{ A }}_2 + \mathbf{\boldsymbol{ B }}_2 + \mathbf{\boldsymbol{ A }}_2\mathbf{\boldsymbol{ B }}_2 - \mathbf{\boldsymbol{ B }}_2\mathbf{\boldsymbol{ A }}_2.

(36)

Using the numpy module which is already imported, calculate the value of $\mathbf{\boldsymbol{ C }}_2$ and store it in the variable C2 shown in the cell below.

C2 = None # replace None with your solution to this exercise.