# Lab 11: Iterators and Generators

*Due at 11:59pm on Friday, 04/13/2018.*

## Starter Files

Download lab11.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

## Submission

By the end of this lab, you should have submitted the lab with
`python3 ok --submit`

. You may submit more than once before the
deadline; only the final submission will be graded.
Check that you have successfully submitted your code on
okpy.org.

- To receive credit for this lab, you must complete Questions 1-4 and submit through OK. Started code for Questions 1-2 can be found in lab11.py.
- Questions 5-9 are
*extra practice*. They can be found in the lab11_extra.py file. It is recommended that you complete these problems on your own time.

# Topics

## Iterables and Iterators

Iterables are objects that can be iterated through. One construct that we can use to iterate through an iterable is a for loop:

```
for elem in iterable:
# do something
```

`for`

loops work on any object that is *iterable*. We previously described it as
working with any sequence -- all sequences are iterable, but there are other
objects that are also iterable! As it turns out, for loops are actually
translated by the interpreter something similar to the following code:

```
iterator = iter(iterable)
try:
while True:
elem = next(iterator)
# do something
except StopIteration:
pass
```

Here's a breakdown of what's happening:

- First, the built-in
`iter`

function is called on the iterable to create a corresponding*iterator*, an object used to iterate through the iterable by keeping track of which element is next in the sequence. - To get the next element in the sequence, the built-in
`next`

function is called on this iterator. - When
`next`

is called but there are no elements left in the iterator, a`StopIteration`

error is raised. In the for loop construct, this exception is caught and execution can continue.

Calling `iter`

on an iterable multiple times returns a new iterator each time
with distinct states (otherwise, you'd never be able to iterate through a
iterable more than once). You can also call `iter`

on the iterator itself, which
will just return the same iterator without changing its state. However, note
that you cannot call `next`

directly on an iterable.

Let's see the `iter`

and `next`

functions in action with an iterable we're
already familiar with -- a list.

```
>>> lst = [1, 2, 3, 4]
>>> next(lst)
TypeError: 'list' object is not an iterator
>>> list_iter = iter(lst)
>>> list_iter
<list_iterator object ...>
>>> next(list_iter)
1
>>> next(list_iter)
2
>>> next(iter(list_iter)) # Calling iter on an iterator returns itself
3
>>> list_iter2 = iter(lst)
>>> next(list_iter2) # Second iterator has new state
1
>>> next(list_iter) # First iterator is unaffected by second iterator
4
>>> next(list_iter) # No elements left!
StopIteration
>>> lst # Original iterable is unaffected
[1, 2, 3, 4]
```

Since you can call `iter`

on iterators, this tells us that that they are also
iterables. You can use iterators wherever you can use iterables, but note that
since iterators keep their state, they're only good to iterate through once:

```
>>> list_iter = iter([4, 3, 2, 1])
>>> for e in list_iter:
... print(e)
4
3
2
1
>>> next(list_iter)
StopIteration
```

Analogy: An iterable is like a book (one can flip through the pages) and an iterator for a book would be a bookmark (saves the position and can locate the next page). Calling`iter`

on a book gives you a new bookmark independent of other bookmarks, but calling`iter`

on a bookmark gives you the bookmark itself, without changing its position at all. Calling`next`

on the bookmark moves it to the next page, but does not change the pages in the book. Calling`next`

on the book wouldn't make sense semantically.

### Iterable uses

We know that lists are one type of built-in iterable objects. You may
have also encounter the `range(start, end)`

function, which creates an iterable of
ascending integers from start (inclusive) to end (exclusive).

```
>>> for x in range(2, 6):
... print(x)
...
2
3
4
5
```

Ranges are useful for many things, including performing some operations for a particular number of iterations or iterating through the indices of a list.

There are also some built-in functions that take in iterables and return useful results:

`map(f, iterable)`

- Creates iterator over`f(x)`

for`x`

in iterable`filter(f, iterable)`

- Creates iterator over`x`

for`x`

in iterable if`f(x)`

`zip(iter1, iter2)`

- Creates iterator over co-indexed pairs (x, y) from both input iterables`reversed(iterable)`

- Creates iterator sequence containing elements of iterable in reverse order`list(iterable)`

- Creates a list containing all x in iterable`tuple(iterable)`

- Creates a tuple containing all x in iterable`sorted(iterable)`

- Creates a sorted list containing all x in iterable

## Generators

A generator function returns a special type of iterator called a **generator**.
Generator functions have `yield`

statements within the body of the function
instead of `return`

statements. Calling a generator function will return a
generator object and will *not* execute the body of the function.

For example, let's consider the following generator function:

```
def countdown(n):
print("Beginning countdown!")
while n >= 0:
yield n
n -= 1
print("Blastoff!")
```

Calling `countdown`

will return a generator object that counts down from `n`

to 0.
Since generators are iterators, we can call `iter`

on the resulting object,
which will simply return the same object. Note that the body is not executed
at this point; nothing is printed and no numbers are output.

```
>>> c = countdown(5)
>>> c
<generator object countdown ...>
>>> c is iter(c)
True
```

So how is the counting done? Again, since generators are a type of iterator, we
can also call `next`

on them! The first time `next`

is called, execution
begins at the first line of the function body and continues until the `yield`

statement is reached. The result of evaluating the expression in the `yield`

statement is returned. The following interactive session continues from the
one above.

```
>>> next(c)
Beginning countdown!
5
```

Unlike functions we've seen before in this course, generator functions can
remember their state. On any consecutive calls to `next`

, execution picks up
from the line after the `yield`

statement that was previously executed. Like
the first call to `next`

, execution will continue until the next `yield`

statement is reached.

```
>>> next(c)
4
>>> next(c)
3
```

Can you predict what would happen if we continue to call `next`

on `c`

4 more
times?

`next`

will continue to yield consecutive descending
integers until 0. On the following call, a `StopIteration`

error will be
raised because there are no more values to yield (i.e. the end of the function
body was reached before hitting a `yield`

statement).
```
>>> next(c)
2
>>> next(c)
1
>>> next(c)
0
>>> next(c)
Blastoff!
StopIteration
```

Separate calls to `countdown`

will create distinct generator objects with their
own state. Usually, generators shouldn't restart. If you'd like to reset the
sequence, create another generator object by calling the generator function
again.

```
>>> c1, c2 = countdown(5), countdown(5)
>>> c1 is c2
False
>>> next(c1)
5
>>> next(c2)
5
```

Here is a summary of the above:

- A generator function has a
`yield`

statement and returns a generator object. - Calling the
`iter`

function on a generator object returns the same object without modifying its current state. - The body of a generator function is not evaluated until
`next`

is called on a resulting generator object. Calling the`next`

function on a generator object computes and returns the next object in its sequence. If the sequence is exhausted,`StopIteration`

is raised. A generator "remembers" its state for the next

`next`

call. Therefore,the first

`next`

call works like this:- Enter the function and run until the line with
`yield`

. - Return the value in the
`yield`

statement, but remember the state of the function for future`next`

calls.

- Enter the function and run until the line with
And subsequent

`next`

calls work like this:- Re-enter the function, start at
**the line after**, and run until the next`yield`

`yield`

statement. - Return the value in the
`yield`

statement, but remember the state of the function for future`next`

calls.

- Re-enter the function, start at

- A generator should not restart unless it's defined that way. But
calling the generator function returns a brand new generator object (like
calling
`iter`

on an iterable object).

Another useful tool for generators is the `yield from`

statement (introduced in
Python 3.3). `yield from`

will yield all values from an iterator or iterable.

```
>>> def gen_list(lst):
... yield from lst
...
>>> g = gen_list([1, 2, 3, 4])
>>> next(g)
1
>>> next(g)
2
>>> next(g)
3
>>> next(g)
4
>>> next(g)
StopIteration
```

# Required Questions

## WWPD

### Q1: WWPD: Iterators

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

`python3 ok -q iterators -u`

Enter

`Error`

if you believe an error occurs,`StopIteration`

if a`StopIteration`

exception is raised, and`Iterator`

if the output is a iterator object.

```
>>> s = [1, 2, 3, 4]
>>> t = iter(s)
>>> next(s)
______TypeError
>>> next(t)
______1
>>> next(t)
______2
>>> iter(s)
______Iterator
>>> next(iter(s))
______1
>>> next(iter(t))
______3
>>> next(iter(s))
______1
>>> next(iter(t))
______4
>>> next(t)
______StopIteration
```

```
>>> r = range(6)
>>> r_iter = iter(r)
>>> next(r_iter)
______0
>>> [x + 1 for x in r]
______[1, 2, 3, 4, 5, 6]
>>> [x + 1 for x in r_iter]
______[2, 3, 4, 5, 6]
>>> next(r_iter)
______StopIteration
>>> list(range(-2, 4)) # Converts an iterable into a list
______[-2, -1, 0, 1, 2, 3]
```

```
>>> map_iter = map(lambda x : x + 10, range(5))
>>> next(map_iter)
______10
>>> next(map_iter)
______11
>>> list(map_iter)
______[12, 13, 14]
>>> for e in filter(lambda x : x % 2 == 0, range(1000, 1008)):
... print(e)
...
______1000
1002
1004
1006
>>> [x + y for x, y in zip([1, 2, 3], [4, 5, 6])]
______[5, 7, 9]
>>> for e in zip([10, 9, 8], range(3)):
... print(tuple(map(lambda x: x + 2, e)))
...
______(12, 2)
(11, 3)
(10, 4)
```

### Q2: WWPD: Generators

Use Ok to test your knowledge with the following What would Python Display questions:

`python3 ok -q generators -u`

Enter

`Error`

if you believe an error occurs,`Function`

if the output is a function object, and`Generator`

if the output is a generator object.

```
def gen():
print("Starting here")
i = 0
while i < 6:
print("Before yield")
yield i
print("After yield")
i += 1
```

```
>>> next(gen)
______TypeError
>>> gen
______<function gen ...>
>>> g = gen()
>>> g
______<generator object gen ...>
>>> g == iter(g)
______True
>>> next(g)
______Starting here
Before yield
0
>>> next(g)
______After yield
Before yield
1
>>> next(g)
______After yield
Before yield
2
>>> g2 = gen()
>>> next(g2)
______Starting here
Before yield
0
>>> iter(g2)
______<generator object gen ...>
>>> next(iter(g))
______After yield
Before yield
3
>>> next(gen())
______Starting here
Before yield
0
```

## Coding Practice

### Q3: Scale

Implement the generator function `scale(s, k)`

, which yields elements of the
given iterable `s`

, scaled by `k`

. As an extra challenge, try writing this
function using a `yield from`

statement!

```
def scale(s, k):
"""Yield elements of the iterable s scaled by a number k.
>>> s = scale([1, 5, 2], 5)
>>> type(s)
<class 'generator'>
>>> list(s)
[5, 25, 10]
>>> m = scale(naturals(), 2)
>>> [next(m) for _ in range(5)]
[2, 4, 6, 8, 10]
"""
"*** YOUR CODE HERE ***"
for elem in s:
yield elem * k
# Alternate solution
def scale(s, k):
yield from map(lambda x: x*k, s)
```

Use Ok to test your code:

`python3 ok -q scale`

### Q4: Trap

Write a generator function that yields the first `k`

values in iterable `s`

, but
raises a `ValueError`

exception if any more values are requested. You may assume that
`s`

has at least `k`

values.

```
def trap(s, k):
"""Return a generator that yields the first K values in iterable S,
but raises a ValueError exception if any more values are requested.
>>> t = trap([3, 2, 1], 2)
>>> next(t)
3
>>> next(t)
2
>>> next(t)
ValueError
>>> list(trap(range(5), 5))
ValueError
>>> t2 = trap(map(abs, reversed(range(-6, -4))), 2)
>>> next(t2)
5
>>> next(t2)
6
>>> next(t2)
ValueError
"""
"*** YOUR CODE HERE ***"
t = iter(s)
while k > 0:
yield next(t)
k -= 1
raise ValueError("It's a trap!")
```

Use Ok to test your code:

`python3 ok -q trap`

# Optional Questions

### Q5: Hailstone

Write a generator that outputs the hailstone sequence from homework 1.

Here's a quick reminder of how the hailstone sequence is defined:

- Pick a positive integer
`n`

as the start. - If
`n`

is even, divide it by 2. - If
`n`

is odd, multiply it by 3 and add 1. - Continue this process until
`n`

is 1.

```
def hailstone(n):
"""
>>> for num in hailstone(10):
... print(num)
...
10
5
16
8
4
2
1
"""
"*** YOUR CODE HERE ***"
while n > 1:
yield n
if n % 2 == 0:
n //= 2
else:
n = n * 3 + 1
yield n
```

Use Ok to test your code:

`python3 ok -q hailstone`

### Q6: Repeated

Implement a function (not a generator function) that returns the first value in
iterable `t`

that appears `k`

times in a row.

```
def repeated(t, k):
"""Return the first value in iterable T that appears K times in a row.
>>> s = [3, 2, 1, 2, 1, 4, 4, 5, 5, 5]
>>> repeated(trap(s, 7), 2)
4
>>> repeated(trap(s, 10), 3)
5
>>> print(repeated([4, None, None, None], 3))
None
"""
assert k > 1
"*** YOUR CODE HERE ***"
count = 0
last_item = None
for item in t:
if item == last_item:
count += 1
else:
last_item = item
count = 1
if count == k:
return item
```

Use Ok to test your code:

`python3 ok -q repeated`

### Q7: Merge

Implement `merge(s0, s1)`

, which takes two iterables `s0`

and `s1`

whose
elements are ordered. `merge`

yields elements from `s0`

and `s1`

in sorted
order, eliminating repetition. You may assume `s0`

and `s1`

themselves do not
contain repeats, and that none of the elements of either are `None`

.
You may **not** assume that the iterables are finite; either may produce an infinite
stream of results.

You will probably find it helpful to use the two-argument version of the built-in
`next`

function: `next(s, v)`

is the same as `next(s)`

, except that instead of
raising `StopIteration`

when `s`

runs out of elements, it returns `v`

.

See the doctest for examples of behavior.

```
def merge(s0, s1):
"""Yield the elements of strictly increasing iterables s0 and s1, removing
repeats. Assume that s0 and s1 have no repeats. s0 or s1 may be infinite
sequences.
>>> m = merge([0, 2, 4, 6, 8, 10, 12, 14], [0, 3, 6, 9, 12, 15])
>>> type(m)
<class 'generator'>
>>> list(m)
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
>>> def big(n):
... k = 0
... while True: yield k; k += n
>>> m = merge(big(2), big(3))
>>> [next(m) for _ in range(11)]
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
"""
i0, i1 = iter(s0), iter(s1)
e0, e1 = next(i0, None), next(i1, None)
"*** YOUR CODE HERE ***"
while True:
if e0 is None and e1 is None:
return
elif e0 is None:
yield e1
e1 = next(i1, None)
elif e1 is None:
yield e0
e0 = next(i0, None)
else:
yield min(e0, e1)
if e0 < e1:
e0 = next(i0, None)
elif e1 < e0:
e1 = next(i1, None)
else:
e0, e1 = next(i0, None), next(i1, None)
```

Use Ok to test your code:

`python3 ok -q merge`

### Q8: Remainder Generator

Like functions, generators can also be*higher-order*. For this problem, we will be writing

`remainders_generator`

, which yields a
series of generator objects.
`remainders_generator`

takes in an integer `m`

, and yields `m`

different
generators. The first generator is a generator of multiples of `m`

, i.e. numbers
where the remainder is 0. The second, a generator of natural numbers with
remainder 1 when divided by `m`

. The last generator yield natural numbers with
remainder `m - 1`

when divided by `m`

.

```
def remainders_generator(m):
"""
Takes in an integer m, and yields m different remainder groups
of m.
>>> remainders_mod_four = remainders_generator(4)
>>> for rem_group in remainders_mod_four:
... for _ in range(3):
... print(next(rem_group))
0
4
8
1
5
9
2
6
10
3
7
11
"""
"*** YOUR CODE HERE ***"
def remainder_group(rem):
start = rem
while True:
yield start
start += m
for rem in range(m):
yield remainder_group(rem)
```

Note that if you have implemented this correctly, each of the
generators yielded by `remainder_generator`

will be *infinite* - you
can keep calling `next`

on them forever without running into a
`StopIteration`

exception.

**Hint**: Consider defining an inner generator function. What arguments
should it take in? Where should you call it?

Use Ok to test your code:

`python3 ok -q remainders_generator`

### Q9: Zip Generator

For this problem, we will be writing `zip_generator`

, which yields a
series of lists, each containing the nth items of each iterable.
It should stop when the smallest iterable runs out of elements.

```
def zip_generator(*iterables):
"""
Takes in any number of iterables and zips them together.
Returns a generator that outputs a series of lists, each
containing the nth items of each iterable.
>>> z = zip_generator([1, 2, 3], [4, 5, 6], [7, 8])
>>> for i in z:
... print(i)
...
[1, 4, 7]
[2, 5, 8]
"""
"*** YOUR CODE HERE ***"
iterators = [iter(iterable) for iterable in iterables]
while True:
yield [next(iterator) for iterator in iterators]
```

Use Ok to test your code:

`python3 ok -q zip_generator`