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Publications Calendar Research Teaching Zoom LDS

Publications Calendar Research Teaching Zoom LDS

Introduction to Loops

import numpy as np

Problem 1

Write code to print the numbers 0 to 4 on each line below.

Solution

print(0)
print(1)
print(2)
print(3)

0
1
2
3

Discussion

• Okay, let's keep going to 10000...(!?)

Loops

• Loops in programing are used to simplify repetative work.
• Above, each line is a print statement.
  • The only thing different on each line is the number we are printing.
• We can write a loop statment to do the above problem as follows

for i in range(4):
    print(i)

0
1
2
3

Discussion

• What are the parts of this: list them?
• i can be any variable, not just "i"
• Is there anything weird or unexpected here?

Problem 2

• We can put any code in the for loop
  • Variables, functions, expressions, math, etc.
• We can put variables outside of the loop that are available inside of the loop

For the ideal gas law, pressure is given by $$P = nRT/V$$

• n = 1 kmol
• R = 8314 J/kmol*K
• T = 300 K

For V = 2, 3, 4, 5, ... 20, use a loop to find and print the values of P for each V.

Solution

n = 1     # kmol
R = 8314  # J/kmol*K
T = 300   # K

for V in range(2,21):
    P = n*R*T/V
    print(V, P)

2 1247100.0
3 831400.0
4 623550.0
5 498840.0
6 415700.0
7 356314.28571428574
8 311775.0
9 277133.3333333333
10 249420.0
11 226745.45454545456
12 207850.0
13 191861.53846153847
14 178157.14285714287
15 166280.0
16 155887.5
17 146717.64705882352
18 138566.66666666666
19 131273.68421052632
20 124710.0

Discussion

• What if you wanted V to not step by integers?
  • Step by 2? Step by 5?
  • Step by 0.2?

Problem 3

Add all the numbers from 0 to 100

• Before you start the code, think about what you would do if solving this in your head?
  • How do you start?
  • What things do you keep track of?
  • What variables to use?
  • What is the loop syntax?
  • How do variables interact with the loop?
• Think through the process of the problem to the solution
• Talk to your neighbor
• If you've solved it, help your neighbor.

Solution

s = 0
for i in range(101):
    s += i
print(s)

5050

Nested loops

• We can put any code in a for loop. Even another loop.
• Look at the code below.
• Before running it, think of what you expect to see.
  • Now run it and compare. Did you get what you thought you would?

for i in range(3):
    print("i =", i)
    for j in range(4):
        print("   j =", j)

• Note, each time through the top (outer) loop, everything inside is done.
• This means the whole inner loop is computed for each outer loop.
• Think of an actual application for a nested loop.

Problem 4a

• Given the following function $$f(x) = x^2 - 3,$$ we can use Newton's method to find $x$ for $f(x)=0$.

• Recall, Newton recurrence formula: $$x_{new} = x_{old} - \frac{f(x_{old})}{f^\prime(x_{old})}$$

We solved this previously in Excel

• Open Excel and solve this problem.
• Think about the steps involved, how we set it up.
• Use the simplest method possible.

Solution-Excel

Excel	Python
cell for guess	??
Newton equation in terms of x cell A2 - (A2^2-3)/(2*A2)	??
repeat: fill down	??

Problem 4b

Now solve in Python using loops

• Try to write the simplest solution you can think of with as few lines of code as possible.
• We want 3 decimal places of accuracy.
• Consider:
  • What variables are needed?
  • How to setup and use the loop.
  • What to print and where?

Solution-Python

x = 1
for i in range(5):
    x = x - (x**2-3)/(2*x)
    print(x)

2.0
1.75
1.7321428571428572
1.7320508100147276
1.7320508075688772

Discussion

• Note that i isn't used, it's just convenient for making the loop
• x is reassigned in place (x is replaced with the updated value on each loop)

Explicitly write out the four steps

• We don't have to code it from top to bottom...
• Compare to Excel

Excel	Python
cell for guess	variable for x
Newton equation in terms of x cell A2 - (A2^2-3)/(2*A2)	Newton equation in terms of x x = x - (x**2-3)/(2*x)
repeat: fill down	repeat: wrap in loop

# step 1: initial guess
x = 1

# step 2: add Newton equation
x=1
x = x - (x**2 - 3)/(2*x)

# step 3: wrap in loop
x=1
for i in range(6):
    x = x - (x**2 - 3)/(2*x)

# step 4: add a print
x=1
for i in range(6):
    x = x - (x**2 - 3)/(2*x)
    print(x)

2.0
1.75
1.7321428571428572
1.7320508100147276
1.7320508075688772
1.7320508075688774

Problem 4c

Redo Solution 1 using functions for $f(x)$ and $f^\prime(x)$.

• Recall: we discussed functions two classes back.

def f(x):
    return x**2-3
def fp(x):
    return 2*x

x = 1
for i in range(5):
    x = x - f(x)/fp(x)
    print(x)

2.0
1.75
1.7321428571428572
1.7320508100147276
1.7320508075688772

Problem 4d

• Now, to get to three decimal places we only needed 3 iterations.
• But we didn't know this at the beginning, so we guessed the number of iterations.
  • If 6 wasn't enough, we would do more.
• This guess and check is not very efficient.

Modify the code above to do just the right number of iterations.

• What Python features do you need to use?
• How do you decide when you are done?
• Talk with your neighbor.

Solution

def f(x):
    return x**2-3
def fp(x):
    return 2*x

eps = 0.001
xold = 1
for i in range(5):
    xnew = xold - f(xold)/fp(xold)
    print(xnew)
    rerr = np.abs((xold - xnew)/xnew)
    if rerr < eps: 
        break
    xold = xnew

2.0
1.75
1.7321428571428572
1.7320508100147276

Discussion

• Keep track of old and new x: xold and xnew.
  • Need to reset xold=xnew at the end to get set for next interation.
• New keyword break
• Define the error and compare it to the desired accuracy.
  • Use absolute value: np.abs()