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# Programming in PythonIteration

Okuma zamanı: ~25 min

We have already seen one way of doing something to each element in a collection: the list comprehension.

smallest_factor = {2: 2, 3: 3, 4: 2, 5: 5,
6: 2, 7: 7, 8: 2, 9: 3}
[v for (k,v) in smallest_factor.items()]

In this list comprehension, we iterate over the pairs of the dictionary to produce a new list. Although list comprehensions are very useful, they are not flexible enough to cover all our iteration needs. A much more flexible tool is the for loop.

## For statements

The code above could also be rewritten as follows:

smallest_factor = {2: 2, 3: 3, 4: 2, 5: 5,
6: 2, 7: 7, 8: 2, 9: 3}
vals = []
for (k,v) in smallest_factor.items():
vals.append(v)
vals

The statement for item in collection: works as follows: the first element of collection is assigned to item, and the block indented below the for statement is executed. Then, the second element of collection is assigned to item, the indented block is executed again, etc., until the end of the collection is reached.

We can nest for statements. For example, suppose we have a matrix represented as a list of lists, and we want to sum all of the matrix entries. We can do that by iterating over the rows and then iterating over each row:

def sum_matrix_entries(M):
"""
Return the sum of the entries of M
"""
s = 0
for row in M:
for entry in row:
s = s + entry
return s

def test_sum():
M = [[1,2,3],[4,5,6],[7,8,9]]
assert sum_matrix_entries(M) == 45
return "Test passed!"

test_sum()

Exercise
Suppose you have imported a function file_bug_report with two parameters: id and description. Suppose also that you have a dict called bugs whose keys are ids and whose values are strings representing descriptions. Write a loop which performs the action of filing each bug report in the dictionary.

def file_bug_report(id, description):
"A dummy function which represents filing a bug report"
print(f"bug {id} ({description}) successfully filed")

bugs = {"07cc242a":
"trackShipment hangs if trackingNumber is missing",
"100b359a":
"customers not receiving text alerts"}

Solution. We loop over the items:

for id, desc in bugs.items():
file_bug_report(id, desc)

Exercise
Write a function called factorial which takes a positive integer n as an argument and returns its factorial.

def factorial(n):
"Return n!"

def test_factorial():
assert factorial(3) == 6
assert factorial(0) == 1
assert factorial(20) == 2432902008176640000
return "Tests passed!"

test_factorial()

Solution. We loop through range(1, n+1) and multiply as we go.

def factorial(n):
"Return n!"
product = 1
for k in range(1, n+1):
product = k * product
return product

def test_factorial():
assert factorial(3) == 6
assert factorial(0) == 1
assert factorial(20) == 2432902008176640000
return "Tests passed!"

test_factorial()

## While statements

The Collatz conjecture is one of the easiest-to-state unsolved problems in mathematics. Starting from any given positive integer, we halve it if it's even and triple it and add one if it's odd. The Collatz conjecture states that repeatedly applying this rule always gets us to the number 1 eventually. For example, the Collatz sequence starting from 17 is

17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

If we want to write a Python function which returns the Collatz sequence for any given starting number, we face a problem: we don't know from the start how many steps it will take to reach 1, so it isn't clear how we could use a for loop. What we want to do is execute a block of code until a given condition is met. Python provides the while loop for this purpose.

def collatz_sequence(n):
"Return the Collatz sequence starting from n"
sequence = [n]
while n > 1:
if n % 2 == 0:
n = n // 2
else:
n = 3*n + 1
sequence.append(n)
return sequence

def test_collatz():
assert collatz_sequence(17) == [17, 52, 26, 13,
40, 20, 10, 5,
16, 8, 4, 2, 1]
return "Test passed!"

test_collatz()

The expression which appears immediately following the while keyword is called the condition, and the block indented below the while statement is the body of the loop. The rules of the language stipulate the following execution sequence for a while statement: the condition is evaluated, and if it's true, then the body is executed, then condition is evaluated again, and so on. When the condition returns False, the loop is exited. An exit can also be forced from within the body of the while loop with the keyword break.

Exercise
Newton's algorithm for finding the square root of a number n starts from 1 and repeatedly applies the function . For example, applying this algorithm to approximate , we get

1, 3/2, 17/12, 577/408, ...

This algorithm converges very fast: 577/408 approximates with a relative error of about 0.00015%.

Write a function newtonsqrt which takes as an argument the value n to square root and applies Newton's algorithm until the relative difference between consecutive iterates drops below .

Note that can be represented in Python using scientific notation 1e-8.

def newtonsqrt(n):
"""Use Newton's algorithm to approximate √n"""

def test_newton():
assert abs(newtonsqrt(2) - 1.4142135623730951) < 1e-6
assert abs(newtonsqrt(9) - 3) < 1e-6
return "Tests passed!"

test_newton()

Solution. We keep up with two separate variables, which we call x and old_x, to compare the most recent two iterates:

def newtonsqrt(n):
"""Use Newton's algorithm to approximate √n"""
x = 1
while True:
old_x = x
x = 1/2 * (x + n/x)
if abs(x - old_x)/old_x < 1e-8:
return x

## Exercises

Exercise
Write a function which prints an checkerboard pattern of x's and o's.

Note: \n in a string literal represents the "newline" character. You'll need to print this character after each row you've printed.

def checkerboard(n):
"""
Prints an n × n checkerboard, like:

xoxo
oxox
xoxo
oxox
"""

Solution. We loop through the rows and use an if statement to print a different output depending on whether the row is even-numbered or odd-numbered.

def checkerboard(n):
"Prints an n × n checkerboard"
for i in range(n):
if i % 2 == 0:
print("xo" * (n//2))
else:
print("ox" * (n//2))
print("\n")

Exercise
Write a function which prints Pascal's triangle up to the $n$th row, where the top row counts as row zero. You might want to use a helper function print_row(n,row) to manage the responsibility of printing each row, as well as a helper function next_row(row) to calculate each row from the previous one.

Example output, for n = 4:

        1
1   1
1   2   1
1   3   3   1
1   4   6   4   1

Note: there's no solution to this one, but you can do it on your own!

def print_row(n,row):
"""
Prints the nth row (row) of Pascal's triangle
with appropriate spacing.
"""

def next_row(row):
"""
Returns the next row in Pascal's triangle.
Example: next_row([1,3,3,1]) == [1,4,6,4,1]
"""

def pascals_triangle(n):
"""
Print the first n rows of Pascal's triangle
"""
Bruno