Download this page as a jupyter notebook at Lesson 9
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Numpy is the core library for scientific computing in Python. It provides a high-performance multidimensional array object, and tools for working with these arrays. The library’s name is short for “Numeric Python” or “Numerical Python”. If you are curious about NumPy, this cheat sheet is recommended: https://s3.amazonaws.com/assets.datacamp.com/blog_assets/Numpy_Python_Cheat_Sheet.pdf
NumPy library has two fundamental conceptualizations:
The CT concepts expressed within NumPy include:
Decomposition
: Break a problem down into smaller pieces; Data interpretation, manipulation, and analysis of NumPy arrays is an act of decomposition.Abstraction
: Pulling out specific differences to make one solution work for multiple problems; The array is a data representation abstraction that allows for placeholder operations, later substituted with specific contents for a problem; enhances reuse and readability. Generally we leverage the principle of algebraic replacement using these abstractions.Algorithms
: A list of steps that you can follow to finish a task; Data interpretation, manipulation, and analysis of NumPy arrays is generally implemented as part of a supervisory algorithm.In principle, NumPy should be available in a default Anaconda install
How to check
import numpy
if the notebook does not protest (i.e. pink block of error), the youis good to go.import numpy
If you do get an error, that means that you will have to install using conda
or pip
; you are on-your-own here! On the content server the process is:
su
then enter the root passwordsudo -H /opt/jupyterhib/bin/python3 -m pip install numpy
compthink
is not in the sudo
groupimport numpy
as above.The process will be similar on a Macintosh, or Windows. Best is to hope you have a sucessful anaconda install.
If you have to do this kind of install, you will have to do some reading, some references I find useful are:
NumPy is a foundational library (module) for scientific computing. Most if not all all data science libraries (modules) rely on NumPy as one of their building blocks to reduce the need to build certain operations in primative python.
NumPy provides a multi-dimensional array object called ‘ndarray’ (n-dimensional array) and methods for:
As compared to the prior lesson, multi-dimensional arrays (lists) are far easier to manipulate using numpy.
A numpy array is a grid of values, all of the same type, and is indexed by a tuple of nonnegative integers. The number of dimensions is the rank of the array; the shape of an array is a tuple of integers giving the size of the array along each dimension.
Figure 1 is a visual depiction of numpy arrays of rank 1 and rank2.
Figure 1: NumPy array visual representation |
---|
In other words, an array contains information about the raw data, how to locate an element and how to interpret an element.To make a numpy array, you can just use the np.array()
function. All you need to do is pass a list to it. Don’t forget that, in order to work with the np.array()
function, you need to make sure that the numpy
library is present in your environment.
If you want to read more about the differences between a Python list and NumPy array, this link is recommended:
https://webcourses.ucf.edu/courses/1249560/pages/python-lists-vs-numpy-arrays-what-is-the-difference
import numpy #First, we need to impoty "numpy"
mylist = [2000,2001,2002,2003,2004,2005,2006,2007,2008,2009] #Create a list of the years
type(mylist)
list
print(mylist) #Check how it looks
[2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009]
mylist = numpy.array(mylist) #Type cast it as a numpy array
type(mylist)
numpy.ndarray
print(mylist)
[2000 2001 2002 2003 2004 2005 2006 2007 2008 2009]
my2dlist = [[1,2,3],[4,5,6],[7,8,9]]
type(my2dlist)
list
print(my2dlist)
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my2dlist= numpy.array(my2dlist) #Type cast it as a numpy array
type(mylist)
numpy.ndarray
print(my2dlist)
[[1 2 3] [4 5 6] [7 8 9]]
len(my2dlist) #how many rows
3
len(my2dlist[0]) #how many columns
3
numpy.shape(my2dlist) # using the shape method
(3, 3)
We already made a 3X3 array, but let's create a 5x2 array from the 2000s (2000-2009):
myotherlist = [[2000,2001],[2002,2003],[2004,2005],[2006,2007],[2008,2009]] #Since I want a 5x2 array, I should group the years two by two
print(myotherlist) #See how it looks as a list
print(numpy.shape(myotherlist)) #check the shape
print(numpy.array(myotherlist) ) #See how it looks as a numpy array
[[2000, 2001], [2002, 2003], [2004, 2005], [2006, 2007], [2008, 2009]] (5, 2) [[2000 2001] [2002 2003] [2004 2005] [2006 2007] [2008 2009]]
Once you have created the arrays, you can do basic Numpy operations. Numpy offers a variety of operations applicable on arrays. From basic operations such as summation, subtraction, multiplication and division to more advanced and essential operations such as matrix multiplication and other elementwise operations. In the examples below, we will go over some of these:
Lets do scalar multiplication, first as just a list!
victorVector = [0,12,24,36,48,60,72,84,96]
MyScalar = input("Enter scalar value \n")
MyScalar = float(MyScalar) # force a float
# now perform element-by-element multiplication
for i in range(0,len(victorVector),1):
victorVector[i] = MyScalar * victorVector[i] # this will change contents of amatrix
print(victorVector)
[0.0, 24.0, 48.0, 72.0, 96.0, 120.0, 144.0, 168.0, 192.0]
Now using numpy
import numpy #import numpy
Array1 = numpy.array([0,12,24,36,48,60,72,84,96]) #Step1: Define Array1
print(Array1)
[ 0 12 24 36 48 60 72 84 96]
MyScalar = input("Enter scalar value \n")
MyScalar = float(MyScalar) # force a float
# now perform element-by-element multiplication
print(Array1*MyScalar) #Step2: Multiple all elements by 2
[ 0. 24. 48. 72. 96. 120. 144. 168. 192.]
print(Array1**2) #Step3: Take all elements to the power of 2
[ 0 144 576 1296 2304 3600 5184 7056 9216]
print(numpy.power(Array1,2)) #Another way to do the same thing, by using a function in numpy
[ 0 144 576 1296 2304 3600 5184 7056 9216]
print(numpy.max(Array1)) #Step4: Find the maximum value of the array
96
print(numpy.argmax(Array1)) ##Step4: Find the postition of the maximum value
8
print(numpy.min(Array1)) #Step5: Find the minimum value of the array
0
print(numpy.argmin(Array1)) ##Step5: Find the postition of the minimum value
0
Array2 = numpy.array([-12,0,12,24,36,48,60,72,84]) #Step6: Define Array2
print(Array2)
[-12 0 12 24 36 48 60 72 84]
print(Array1+Array2) #Step7: Sum of two arrays
print(Array1-Array2) #Step7: Difference of two arrays
[-12 12 36 60 84 108 132 156 180] [12 12 12 12 12 12 12 12 12]
print(Array1*Array2) #Step8: Find the multiplication of these two arrays
[ 0 0 288 864 1728 2880 4320 6048 8064]
import numpy #import numpy
Array1 = numpy.array([[5,10],[15,20]]) #Step1: Define Array1
print(Array1)
[[ 5 10] [15 20]]
Array2 = numpy.array([[3,6],[9,12]]) #Step2: Define Array2
print(Array2)
[[ 3 6] [ 9 12]]
print(Array1+Array2) #Step3: Find the summation
print(Array1-Array2) #Step3: Find the subtraction
[[ 8 16] [24 32]] [[2 4] [6 8]]
MultArray = Array1@Array2 #Step4: To perform a typical matrix multiplication (or matrix product)
MultArray1 = Array1.dot(Array2) #Step4: Another way To perform a matrix multiplication
print(MultArray)
print(MultArray1)
print(numpy.min(MultArray)) #Step4: Find the minimum value of the multiplication
[[105 150] [225 330]] [[105 150] [225 330]] 105
Recall how we have to do the same if using ordinary lists, just for the first line in the block above:
amatrix = [[5,10],[15,20]]
bmatrix = [[3,6],[9,12]]
cmatrix = [[0 for j in range(len(bmatrix[0]))] for i in range(len(amatrix))] # 2D list to receive input; explicit sizing
# now for the multiplication
for i in range(0,len(amatrix)):
for j in range(0,len(bmatrix[0])):
for k in range(0,len(amatrix[0])):
cmatrix[i][j]=cmatrix[i][j]+amatrix[i][k]*bmatrix[k][j]
for i in range(len(cmatrix)): # print by row
print(cmatrix[i][:])
[105, 150] [225, 330]
There is considerable savings in code crafting using numpy; hence its popularity and utility.
print(numpy.argmax(MultArray)) ##Step5: Find the postition of the maximum value
3
print(numpy.mean(MultArray)) ##Step6: Find the mean of the multiplication of these two arrays
print(numpy.mean(MultArray[0,:])) ##Step7: Find the mean of the first row of the multiplication of these two arrays
202.5 127.5
Other functions to create NumPy arrays easily
Functions to do basic operations on NumPy arrays
Functions to do mathematical operations on NumPy arrays
Indexing: An important step in manipulating and analyzing arrays
There are linear algebra functions too, recall our matrices
\begin{gather} \mathbf{A}= \begin{pmatrix} 2 & 3 \\ 4 & -3 \\ \end{pmatrix} \end{gather}\begin{gather} \mathbf{A}^{-1}= \begin{pmatrix} \frac{1}{6} & \frac{1}{6} \\ ~\\ \frac{2}{9} & -\frac{1}{9} \\ \end{pmatrix} \end{gather}We can get the inverse in numpy from:
amatrix = [[2 , 3 ], # a and b are inverses of each other.
[4 , -3 ]]
bmatrix = [[1/6 , 1/6 ],
[2/9 , -1/9]] # if we invert b should get a back.
amatrix = numpy.array(amatrix) # type cast to numpy arrays
bmatrix = numpy.array(bmatrix)
print(amatrix@bmatrix) # proof of inverse - should get identity
print(numpy.linalg.inv(bmatrix)) # invert b, see if we get a - sho 'nuff
[[1. 0.] [0. 1.]] [[ 2. 3.] [ 4. -3.]]
Comparing two NumPy arrays determines whether they are equivalent by checking if every element at each corresponding index are the same.
import numpy #import numpy
Array1 = numpy.array([1.0,2.5,3.4,7,7]) #Step1: Define Array1
print(Array1)
Array2 = numpy.array([5.0/5.0,5.0/2,6.8/2,21/3,14/2]) #Step2: Define Array1
print(Array2)
print(numpy.equal(Array1, Array2)) #Step3: Compare and see if the two arrays are equal
Array3 = numpy.array([6,1.4,2.2,7.5,7]) #Step4: Define Array3
print(Array3)
print(numpy.greater_equal(Array1, Array3)) #Step3: Compare and see if the two arrays are equal
[1. 2.5 3.4 7. 7. ] [1. 2.5 3.4 7. 7. ] [ True True True True True] [6. 1.4 2.2 7.5 7. ] [False True True False True]
import numpy #import numpy
Array1 = numpy.array([[4,40],[68,72]]) #Step1: Define Array1
print(Array1)
Array2 = numpy.array([[9,23],[39,55]]) #Step2: Define Array2
print(Array2)
print(numpy.less_equal(Array1,Array2))
[[ 4 40] [68 72]] [[ 9 23] [39 55]] [[ True False] [False False]]
numpy.copy() allows us to create a copy of an array. This method is particularly useful when we need to manipulate an array while keeping an original copy in memory.
The numpy.delete() function returns a new array with sub-arrays along an axis deleted. Let's have a look at the examples.
Below we will work a few examples.
import numpy #import numpy
x = numpy.array([1,2,3]) #Step1: Define x
print(x)
[1 2 3]
y = x #Step2: Assign y as y=x
print(y)
[1 2 3]
z = numpy.copy(x) #Step3: Define z as a copy of x
print(z)
[1 2 3]
print('x = ',x) # They look similar but ...
print('y = ',y)
print('z = ',z)
x = [1 2 3] y = [1 2 3] z = [1 2 3]
now lets change a value in x, then print the contents again
x[0] = 100
print('x = ',x) # For Step4: They look similar but ...
print('y = ',y)
print('z = ',z)
x = [100 2 3] y = [100 2 3] z = [1 2 3]
Notice how x and y reflect the change, whats going on? The assignment statement y = x
creates equivalence in numpy; that is x and y are different names for the same object. We can verify using the id() function which will return the address in memory of the head of the object.
print(' x object at hex address ',hex(id(x)))
print(' y object at hex address ',hex(id(y)))
print(' z object at hex address ',hex(id(z)))
x object at hex address 0xffff981cde40 y object at hex address 0xffff981cde40 z object at hex address 0xffff981cd9e0
x = numpy.delete(x, 1) #Step5: Delete the second element of x, remember counting starts at zero
print('x = ',x) # For Step4: They look similar but ...
print('y = ',y)
print('z = ',z)
x = [100 3] y = [100 2 3] z = [1 2 3]
print(' x object at hex address ',hex(id(x))) # lets check the objects - now they are all different
print(' y object at hex address ',hex(id(y)))
print(' z object at hex address ',hex(id(z)))
x object at hex address 0xffff9817f0d0 y object at hex address 0xffff981cde40 z object at hex address 0xffff981cd9e0
Sorting means putting elements in an ordered sequence. Ordered sequence is any sequence that has an order corresponding to elements, like numeric or alphabetical, ascending or descending. If you use the sort() method on a 2-D array, both arrays will be sorted.
Define a 1D array as ['FIFA 2020','Red Dead Redemption','Fallout','GTA','NBA 2018','Need For Speed'] and print it out. Then, sort the array alphabetically._
import numpy #import numpy
games = numpy.array(['FIFA 2020','Red Dead Redemption','Fallout','GTA','NBA 2018','Need For Speed'])
print(games)
print(numpy.sort(games))
['FIFA 2020' 'Red Dead Redemption' 'Fallout' 'GTA' 'NBA 2018' 'Need For Speed'] ['FIFA 2020' 'Fallout' 'GTA' 'NBA 2018' 'Need For Speed' 'Red Dead Redemption']
Define a 3x3 array with 17,-6,2,86,-12,0,0,23,12 and print it out. Then, sort the array.
import numpy #import numpy
a = numpy.array([[17,-6,2],[86,-12,0],[0,23,12]])
print(a)
print ("Along columns : \n", numpy.sort(a,axis = 0) ) #This will be sorting in each column
print ("Along rows : \n", numpy.sort(a,axis = 1) ) #This will be sorting in each row
print ("Sorting by default : \n", numpy.sort(a) ) #Same as above
print ("Along None Axis : \n", numpy.sort(a,axis = None) ) #This will be sorted like a 1D array
[[ 17 -6 2] [ 86 -12 0] [ 0 23 12]] Along columns : [[ 0 -12 0] [ 17 -6 2] [ 86 23 12]] Along rows : [[ -6 2 17] [-12 0 86] [ 0 12 23]] Sorting by default : [[ -6 2 17] [-12 0 86] [ 0 12 23]] Along None Axis : [-12 -6 0 0 2 12 17 23 86]
Slicing in python means taking elements from one given index to another given index.
Define a 1D array as [1,3,5,7,9], slice out the [3,5,7] and print it out.
import numpy #import numpy
a = numpy.array([1,3,5,7,9]) #Define the array
print(a)
aslice = a[1:4] #slice the [3,5,7]
print(aslice) #print it out
[1 3 5 7 9] [3 5 7]
Define a 5x5 array with "Superman, Batman, Jim Hammond, Captain America, Green Arrow, Aquaman, Wonder Woman, Martian Manhunter, Barry Allen, Hal Jordan, Hawkman, Ray Palmer, Spider Man, Thor, Hank Pym, Solar, Iron Man, Dr. Strange, Daredevil, Ted Kord, Captian Marvel, Black Panther, Wolverine, Booster Gold, Spawn " and print it out. Then:
import numpy #import numpy
Superheroes = numpy.array([['Superman', 'Batman', 'Jim Hammond', 'Captain America', 'Green Arrow'],
['Aquaman', 'Wonder Woman', 'Martian Manhunter', 'Barry Allen', 'Hal Jordan'],
['Hawkman', 'Ray Palmer', 'Spider Man', 'Thor', 'Hank Pym'],
['Solar', 'Iron Man', 'Dr. Strange', 'Daredevil', 'Ted Kord'],
['Captian Marvel', 'Black Panther', 'Wolverine', 'Booster Gold', 'Spawn']])
print(Superheroes) #Step1
print(Superheroes[:,0])
print(Superheroes[2,:])
print(Superheroes[4,2])
print(Superheroes[1:4,1:4])
[['Superman' 'Batman' 'Jim Hammond' 'Captain America' 'Green Arrow'] ['Aquaman' 'Wonder Woman' 'Martian Manhunter' 'Barry Allen' 'Hal Jordan'] ['Hawkman' 'Ray Palmer' 'Spider Man' 'Thor' 'Hank Pym'] ['Solar' 'Iron Man' 'Dr. Strange' 'Daredevil' 'Ted Kord'] ['Captian Marvel' 'Black Panther' 'Wolverine' 'Booster Gold' 'Spawn']] ['Superman' 'Aquaman' 'Hawkman' 'Solar' 'Captian Marvel'] ['Hawkman' 'Ray Palmer' 'Spider Man' 'Thor' 'Hank Pym'] Wolverine [['Wonder Woman' 'Martian Manhunter' 'Barry Allen'] ['Ray Palmer' 'Spider Man' 'Thor'] ['Iron Man' 'Dr. Strange' 'Daredevil']]
The figure below is a simply supported, statically determinate truss with pin connections (zero moment transfer connections). Find the forces in each member for the loading shown.
This notebook will illustrate how to leverage a linear systems solver(s) to analyze the truss. The approach uses concepts from statics and computational thinking.
From statics
method of joints (for reactions and internal forcez)
direction cosines
From computational thinking
read input file
construct linear system $\textbf{Ax=b}$; solve for $\textbf{x}$
report results
Before even contemplating writing/using a program we need to build a mathematical model of the truss and assemble the system of linear equations that result from the model.
So the first step is to sketch a free-body-diagram as below and build a node naming convention and force names.
Next we will write the force balance for each of the six nodes ($N1$-$N6$), which will produce a total of 12 equations in the 12 unknowns (the 9 member forces, and 3 reactions).
The figure below is the force balance for node $N1$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
$$\begin{gather} \begin{matrix} \sum F_x = 0 = & +F_1cos(45) & + F_2 & & & & & & & + A_x & & & & & \\ \sum F_y = 0 = & +F_1sin(45) & & & & & & & & & & + A_y & & & \\ \end{matrix} \end{gather}$$
The equation above is the force balance equation pair for the node. The $x$ component equation will later be named $N1_x$ to indicate it arises from Node 1, $x$ component equation. A similar notation convention will also be adopted for the $y$ component equation. There will be an equation pair for each node.
Below is a sketch of the force balance for node $N2$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
\begin{gather} \begin{matrix} \sum F_x = 0 = & & -F_2 & & & & +F_6 & & & & & & & & \\ \sum F_y = 0 = & & & +F_3 & & & & & & & & & & & \\ \end{matrix} \end{gather}
Below is a sketch of the force balance for node $N3$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
\begin{gather} \begin{matrix} \sum F_x = 0 = & & & & & -F_5cos(30) & -F_6 & & +F_8 & & & & & & \\ \sum F_y = 0 = & & & & & F_5sin(30) & & +F_7 & & & & & & & -P_3\\ \end{matrix} \end{gather}
Above is the force balance equation pair for node $N3$.
Below is a sketch of the force balance for node $N4$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
\begin{gather} \begin{matrix} \sum F_x = 0 = & & & & & & & & -F_8 & -F_9cos(45) & & & & & \\ \sum F_y = 0 = & & & & & & & & & F_9sin(45) & & & +B_y & & \\ \end{matrix} \end{gather}
Above is the force balance equation pair for node $N4$.
Below is a sketch of the force balance for node $N5$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
\begin{gather} \begin{matrix} \sum F_x = 0 = & -F_1cos(45) & & & +F_4 & +F_5cos(30) & & & & & & & & & \\ \sum F_y = 0 = & -F_1sin(45) & & -F_3 & & -F_5sin(30) & & & & & & & & & -P_1\\ \end{matrix} \end{gather}
Above is the force balance equation pair for node $N5$.
Below is a sketch of the force balance for node $N6$, the two force equations (for the horizontal, $x$, direction and the vertical, $y$, direction) are listed below the figure.
\begin{gather} \begin{matrix} \sum F_x = 0 = & & & & -F_4 & & & & & F_9sin(45) & & & & & \\ \sum F_y = 0 = & & & & & & & -F_7 & & -F_9cos(45) & & & & & P_2\\ \end{matrix} \end{gather}
Above is the force balance equation pair for node $N6$.
The next step is to gather the equation pairs into a system of linear equations.
We will move the known loads to the right hand side and essentially construct the matrix equation $\mathbf{A}\mathbf{x} = \mathbf{b}$.
The system below is a matrix representation of the equation pairs with the forces moved to the right hand side $\mathbf{b} = RHS$.
\begin{gather} \begin{pmatrix} ~ & F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 & F_9 & A_x & A_y & B_y & | & RHS\\ \hline N1_x & 0.707 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & | & 0\\ N1_y & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & | & 0\\ N2_x & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0\\ N2_y & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0\\ N3_x & 0 & 0 & 0 & 0 & -0.866 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & | & 0\\ N3_y & 0 & 0 & 0 & 0 & 0.5 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & | & P_3\\ N4_x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -0.707 & 0 & 0 & 0 & | & 0\\ N4_y & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.707 & 0 & 0 & 0 & | & 0\\ N5_x & -0.707 & 0 & 0 & 1 & 0.866 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0\\ N5_y & -0.707 & 0 & -1 & 0 & -0.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & P_1\\ N6_x & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0.707 & 0 & 0 & 0 & | & 0\\ N6_y & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -0.707 & 0 & 0 & 0 & | & -P_2\\ \end{pmatrix} \end{gather}In the system, the rows are labeled on the left-most column with their node-related equation name.
Thus each row of the matrix corresponds to an equation derived from a node.
The columns are labeled with their respective unknown force (except the last column, which represents the right-hand-side of the system of linear equations).
Thus the coefficient in each column corresponds to a force in each node equation.
The sign of the coefficient refers to the assumed direction the force acts.
In the analysis all the members were assumed to be in tension (except for the reaction forces).
If a coefficient has a value of zero in a particular row, then that force does no act at the node to which the row corresponds.
From this representation the transition to the formal vector-matrix representation is straightforward.
The various matrices above are entere into text files named A.txt and B.txt
# list contents of the A matrix, uses call to OS host
#!(type A.txt) # windoze method
!(cat A.txt) # linux, macOS
0.707106781 1 0 0 0 0 0 0 0 1 0 0 0.707106781 0 0 0 0 0 0 0 0 0 1 0 0 -1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.866 -1 0 1 0 0 0 0 0 0 0 0 0.5 0 1 0 0 0 0 0 0 0 0 0 0 0 0 -1 -0.707106781 0 0 0 0 0 0 0 0 0 0 0 0.707106781 0 0 1 -0.707106781 0 0 1 0.866 0 0 0 0 0 0 0 -0.707106781 0 -1 0 -0.5 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0.707106781 0 0 0 0 0 0 0 0 0 -1 0 -0.707106781 0 0 0
# list contents of RHS or b vector
#!(type B.txt) # winders method
!(cat B.txt) # linux, macOS
0.0 0.0 0.0 0.0 0.0 500.0 0.0 0.0 0.0 1000.0 0.0 -500.0
Now we use our solver tools, here I have not done much on tidy output, thats left for the reader.
# LinearSolverPivot.py
# linearsolver with pivoting adapted from
# https://stackoverflow.com/questions/31957096/gaussian-elimination-with-pivoting-in-python/31959226
def linearsolver(A,b):
n = len(A)
M = A # this is equivalence, A is destroyed by algorithm
i = 0
for x in M:
x.append(b[i])
i += 1
for k in range(n):
for i in range(k,n):
if abs(M[i][k]) > abs(M[k][k]):
M[k], M[i] = M[i],M[k]
else:
pass
for j in range(k+1,n):
q = float(M[j][k]) / M[k][k]
for m in range(k, n+1):
M[j][m] -= q * M[k][m]
x = [0 for i in range(n)]
x[n-1] =float(M[n-1][n])/M[n-1][n-1]
for i in range (n-1,-1,-1):
z = 0
for j in range(i+1,n):
z = z + float(M[i][j])*x[j]
x[i] = float(M[i][n] - z)/M[i][i]
# print (x)
return(x)
amatrix = [] # null list to store matrix reads
bvector = []
rowNumA = 0
colNumA = 0
rowNumB = 0
afile = open("A.txt","r") # connect and read file for MATRIX A
for line in afile:
amatrix.append([float(n) for n in line.strip().split()])
rowNumA += 1
afile.close() # Disconnect the file
colNumA = len(amatrix[0])
afile = open("B.txt","r") # connect and read file for MATRIX B
for line in afile:
bvector.append(float(line)) # vector read different -- just float the line
rowNumB += 1
afile.close() # Disconnect the file
#print (bvector)
if rowNumA != rowNumB: # check the arrays
print ("row ranks not same -- aborting now")
quit()
else:
print ("row ranks same -- continuing operation")
# print all columns each row
cmatrix = [[0 for j in range(colNumA)]for i in range(rowNumA)]
dmatrix = [[0 for j in range(colNumA)]for i in range(rowNumA)]
xvector = [0 for i in range(rowNumA)]
dvector = [0 for i in range(rowNumA)]
for i in range(0,rowNumA,1):
print (amatrix[i][0:colNumA], bvector[i])
print ("-----------------------------")
# copy amatrix into cmatrix
cmatrix = [[amatrix[i][j] for j in range(colNumA)]for i in range(rowNumA)]
dmatrix = [[amatrix[i][j] for j in range(colNumA)]for i in range(rowNumA)]
dvector = [bvector[i] for i in range(rowNumA)]
xvector = linearsolver(cmatrix,bvector) # send cmatrix the copy of amatrix; the solver destroys the inputs, hence the copies
for i in range(0,rowNumA,1):
print (xvector[i])
print ("-----------------------------")
row ranks same -- continuing operation [0.707106781, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0] 0.0 [0.707106781, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0] 0.0 [0.0, -1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 0.0 [0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 0.0 [0.0, 0.0, 0.0, 0.0, -0.866, -1.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0] 0.0 [0.0, 0.0, 0.0, 0.0, 0.5, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0] 500.0 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0, -0.707106781, 0.0, 0.0, 0.0] 0.0 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.707106781, 0.0, 0.0, 1.0] 0.0 [-0.707106781, 0.0, 0.0, 1.0, 0.866, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 0.0 [-0.707106781, 0.0, -1.0, 0.0, -0.5, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 1000.0 [0.0, 0.0, 0.0, -1.0, 0.0, 0.0, 0.0, 0.0, 0.707106781, 0.0, 0.0, 0.0] 0.0 [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0, -0.707106781, 0.0, 0.0, 0.0] -500.0 ----------------------------- -1035.2710218174145 732.0471596998927 0.0 -267.9528403001072 -535.9056806002143 732.0471596998927 767.9528403001071 267.95284030010714 -378.94254092877543 0.0 732.0471596998927 267.9528403001072 -----------------------------
Now lets do the same tasks using numpy
import numpy
# already read in the arrays
A = numpy.array(amatrix)
Ainv = numpy.linalg.inv(A)
b = numpy.array(bvector)
x = Ainv@b
for i in range(0,len(x)):
print(x[i])
-1035.2710218174143 732.0471596998927 0.0 -267.95284030010714 -535.9056806002143 732.0471596998927 767.9528403001071 267.9528403001071 -378.9425409287754 -3.9202487786118276e-14 732.0471596998927 267.95284030010714
import numpy
# need to read in the arrays
A = numpy.loadtxt("A.txt")
b = numpy.loadtxt("B.txt")
Ainv = numpy.linalg.inv(A)
x = Ainv@b
for i in range(0,len(x)):
print(x[i])
-1035.2710218174143 732.0471596998927 0.0 -267.95284030010714 -535.9056806002143 732.0471596998927 767.9528403001071 267.9528403001071 -378.9425409287754 -3.9202487786118276e-14 732.0471596998927 267.95284030010714
import numpy
# need to read in the arrays
A = numpy.loadtxt("A.txt")
b = numpy.loadtxt("B.txt")
x = numpy.linalg.solve(A,b) # avoid inversion and just solve the linear system
for i in range(0,len(x)):
print(x[i])
-1035.2710218174145 732.0471596998927 0.0 -267.9528403001072 -535.9056806002143 732.0471596998927 767.9528403001071 267.95284030010714 -378.94254092877543 -3.9202487786118276e-14 732.0471596998927 267.95284030010714
Johnson, J. (2020). Python Numpy Tutorial (with Jupyter and Colab). Retrieved September 15, 2020, from https://cs231n.github.io/python-numpy-tutorial/
Willems, K. (2019). (Tutorial) Python NUMPY Array TUTORIAL. Retrieved September 15, 2020, from https://www.datacamp.com/community/tutorials/python-numpy-tutorial?utm_source=adwords_ppc
Willems, K. (2017). NumPy Cheat Sheet: Data Analysis in Python. Retrieved September 15, 2020, from https://www.datacamp.com/community/blog/python-numpy-cheat-sheet
W3resource. (2020). NumPy: Compare two given arrays. Retrieved September 15, 2020, from https://www.w3resource.com/python-exercises/numpy/python-numpy-exercise-28.php
Sorting https://www.programiz.com/python-programming/methods/list/sort
Overland, B. (2018). Python Without Fear. Addison-Wesley ISBN 978-0-13-468747-6.
Grus, Joel (2015). Data Science from Scratch: First Principles with Python O’Reilly Media. Kindle Edition.
Precord, C. (2010) wxPython 2.8 Application Development Cookbook Packt Publishing Ltd. Birmingham , B27 6PA, UK ISBN 978-1-849511-78-0.