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>>> a_list = ['The', 'sixth', 'sick', "sheik's", 'sixth', "sheep's", 'sick']

>>> set(a_list)

①

{'sixth', 'The', "sheep's", 'sick', "sheik's"}

>>> a_string = 'EAST IS EAST'

>>> set(a_string)

②

{'A', ' ', 'E', 'I', 'S', 'T'}

>>> words = ['SEND', 'MORE', 'MONEY']

>>> ''.join(words)

③

'SENDMOREMONEY'

>>> set(''.join(words))

④

{'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}

1. Given a list of several strings, the

set()

function will return a set of unique strings from the list. This makes

sense if you think of it like a

for

loop. Takethefirst item from the list, put it in the set. Second. Third.

Fourth. Fifth—wait, that’s in theset already, soit only gets listed once, becausePython sets don’t allow

duplicates. Sixth. Seventh—again, a duplicate, so it only gets listed once. Theend result? All theunique

items in the original list, without any duplicates. The original list doesn’t even need to be sorted first.

2. Thesametechnique works with strings, since a string is just asequence of characters.

3. Given a list of strings,

''.join(a_list)

concatenates all the strings together into one.

4. So, given a list of strings, this line of codereturns all the uniquecharacters across all the strings, with no

duplicates.

Thealphametics solver uses this techniqueto build a set of all the unique characters in the puzzle.

unique_characters = set(''.join(words))

This list is later used to assign digits to characters as thesolver iterates through thepossible solutions.

⁂

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8.4.M

AKING ASSERTIONS

Like many programming languages, Python has an

assert

statement. Here’s how it works.

>>> assert 1 + 1 == 2

①

>>> assert 1 + 1 == 3

②

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

AssertionError

>>> assert 2 + 2 == 5, "Only for very large values of 2"

③

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

AssertionError: Only for very large values of 2

1. The

assert

statement is followed by any valid Python expression. In this case, theexpression

1 + 1 == 2

evaluates to

True

,so the

assert

statement does nothing.

2. However, if the Python expression evaluates to

False

,the

assert

statement will raisean

AssertionError

3. You can also include a human-readable message that is printed if the

AssertionError

is raised.

Therefore, this lineof code:

assert len(unique_characters) <= 10, 'Too many letters'

…is equivalent tothis:

if len(unique_characters) > 10:

raise AssertionError('Too many letters')

Thealphametics solver uses this exact

assert

statement to bail out early if the puzzle contains more than

ten uniqueletters. Sinceeach letter is assigned a unique digit, and there are only ten digits, a puzzle with

more than ten unique letters can not possibly have asolution.

⁂

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☞

8.5. G

ENERATOR EXPRESSIONS

Agenerator expression is like a

generator function without the function.

>>> unique_characters = {'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}

>>> gen = (ord(c) for c in unique_characters)

①

>>> gen

②

>>> next(gen)

③

>>> next(gen)

>>> tuple(ord(c) for c in unique_characters)

④

(69, 68, 77, 79, 78, 83, 82, 89)

1. A generator expression is like an anonymous function that yields values. Theexpression itself looks likea

list

comprehension, but it’s wrapped in parentheses instead of squarebrackets.

2. Thegenerator expression returns… an iterator.

3. Calling

next(gen)

returns thenext value from the iterator.

4. If you like, you can iteratethrough all the possiblevalues and return a tuple, list, or set, by passing the

generator expression to

tuple()

list()

,or

set()

.In these cases, you don’t need an extra set of

parentheses—just pass the “bare”expression

ord(c) for c in unique_characters

to the

tuple()

function,

and Python figures out that it’s a generator expression.

Using a generator expression instead of alist comprehension can save both

CPU

and

RAM

.If you’re building an list just to throw it away (e.g. passing it to

tuple()

set()

), usea generator expression instead!

Here’s another way to accomplish the same thing, using a

generator function:

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def ord_map(a_string):

for c in a_string:

yield ord(c)

gen = ord_map(unique_characters)

Thegenerator expression is more compact but functionally equivalent.

⁂

8.6.C

ALCULATING

ERMUTATIONS

…T

AZY

First of all, what the heckare permutations? Permutations are a mathematical concept. (There areactually

several definitions, dependingon what kind of math you’re doing. HereI’m talking about combinatorics, but if

that doesn’t mean anything to you, don’t worry about it. As always,

Wikipedia is your friend.)

Theidea is that you takea list of things (could be numbers, could be letters, could be dancing bears) and

find all the possible ways to split them up into smaller lists. All thesmaller lists have the same size, which

can be as small as 1 and as largeas thetotal number of items. Oh, and nothing can be repeated.

Mathematicians say things like “let’s find the permutations of 3 different items taken 2 at a time,” which

means you havea sequence of 3 items and you want to find all the possible ordered pairs.

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>>> import itertools

①

>>> perms = itertools.permutations([1, 2, 3], 2)

②

>>> next(perms)

③

(1, 2)

>>> next(perms)

(1, 3)

>>> next(perms)

(2, 1)

④

>>> next(perms)

(2, 3)

>>> next(perms)

(3, 1)

>>> next(perms)

(3, 2)

>>> next(perms)

⑤

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

StopIteration

1. The

itertools

module has all kinds of fun stuff in it, including a

permutations()

function that does all the

hard work of finding permutations.

2. The

permutations()

function takes a sequence (herea list of three integers) and a number, which is the

number of items you want in each smaller group. Thefunction returns an iterator, which you can use in a

for

loop or any old place that iterates. Here I’ll step through theiterator manually to show all thevalues.

3. Thefirst permutation of

[1, 2, 3]

taken 2 at atimeis

(1, 2)

4. Notethat permutations are ordered:

(2, 1)

is different than

(1, 2)

5. That’s it! Those are all the permutations of

[1, 2, 3]

taken 2 at a time. Pairs like

(1, 1)

and

(2, 2)

never

show up, becausethey contain repeats so they aren’t valid permutations. When there are no more

permutations, the iterator raises a

StopIteration

exception.

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The

itertools

module has

all kinds of

fun stuff.

The

permutations()

function doesn’t have to take a list.

It can take any sequence—even a string.

>>> import itertools

>>> perms = itertools.permutations('ABC', 3)

①

>>> next(perms)

('A', 'B', 'C')

②

>>> next(perms)

('A', 'C', 'B')

>>> next(perms)

('B', 'A', 'C')

>>> next(perms)

('B', 'C', 'A')

>>> next(perms)

('C', 'A', 'B')

>>> next(perms)

('C', 'B', 'A')

>>> next(perms)

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

StopIteration

>>> list(itertools.permutations('ABC', 3))

③

[('A', 'B', 'C'), ('A', 'C', 'B'),

('B', 'A', 'C'), ('B', 'C', 'A'),

('C', 'A', 'B'), ('C', 'B', 'A')]

1. A string is just a sequence of characters. For thepurposes of finding permutations, thestring

'ABC'

equivalent to the list

['A', 'B', 'C']

2. Thefirst permutation of the 3 items

['A', 'B', 'C']

,taken 3 at a time, is

('A', 'B', 'C')

.There are five

other permutations—thesamethree characters in every conceivableorder.

3. Since the

permutations()

function always returns an iterator, an easy way todebug permutations is topass

that iterator to the built-in

list()

function to see all thepermutations immediately.

196

⁂

8.7. O

THER

TUFFIN THE

itertools M

ODULE

>>> import itertools

>>> list(itertools.product('ABC', '123'))

①

[('A', '1'), ('A', '2'), ('A', '3'),

('B', '1'), ('B', '2'), ('B', '3'),

('C', '1'), ('C', '2'), ('C', '3')]

>>> list(itertools.combinations('ABC', 2))

②

[('A', 'B'), ('A', 'C'), ('B', 'C')]

1. The

itertools.product()

function returns an iterator containing theCartesian product of two sequences.

2. The

itertools.combinations()

function returns an iterator containing all the possible combinations of the

given sequenceof the given length. This is like the

itertools.permutations()

function, except combinations

don’t include items that are duplicates of other items in a different order. So

itertools.permutations('ABC', 2)

will return both

('A', 'B')

and

('B', 'A')

(among others), but

itertools.combinations('ABC', 2)

will not return

('B', 'A')

becauseit is a duplicate of

('A', 'B')

in a

different order.

197

>>> names = list(open('examples/favorite-people.txt', encoding='utf-8'))

①

>>> names

['Dora\n', 'Ethan\n', 'Wesley\n', 'John\n', 'Anne\n',

'Mike\n', 'Chris\n', 'Sarah\n', 'Alex\n', 'Lizzie\n']

>>> names = [name.rstrip() for name in names]

②

>>> names

['Dora', 'Ethan', 'Wesley', 'John', 'Anne',

'Mike', 'Chris', 'Sarah', 'Alex', 'Lizzie']

>>> names = sorted(names)

③

>>> names

['Alex', 'Anne', 'Chris', 'Dora', 'Ethan',

'John', 'Lizzie', 'Mike', 'Sarah', 'Wesley']

>>> names = sorted(names, key=len)

④

>>> names

['Alex', 'Anne', 'Dora', 'John', 'Mike',

'Chris', 'Ethan', 'Sarah', 'Lizzie', 'Wesley']

1. This idiom returns alist of the lines in a text file.

2. Unfortunately (for this example), the

list(open(filename))

idiom also includes the carriage returns at the

end of each line. This list comprehension uses the

rstrip()

string method to strip trailing whitespacefrom

each line. (Strings also havean

lstrip()

method to strip leadingwhitespace, and a

strip()

method which

strips both.)

3. The

sorted()

function takes a list and returns it sorted. By default, it sorts alphabetically.

4. But the

sorted()

function can also take afunction as the

key

parameter, and it sorts by that key. In this

case, thesort function is

len()

,so it sorts by

len(each item)

.Shorter names comefirst, then longer, then

longest.

What does this haveto do with the

itertools

module? I’m glad you asked.

198

…continuing from the previous interactive shell…

>>> import itertools

>>> groups = itertools.groupby(names, len)

①

>>> groups

<itertools.groupby object at 0x00BB20C0>

>>> list(groups)

[(4, <itertools._grouper object at 0x00BA8BF0>),

(5, <itertools._grouper object at 0x00BB4050>),

(6, <itertools._grouper object at 0x00BB4030>)]

>>> groups = itertools.groupby(names, len)

②

>>> for name_length, name_iter in groups:

③

...

print('Names with {0:d} letters:'.format(name_length))

...

for name in name_iter:

...

print(name)

...

Names with 4 letters:

Alex

Anne

Dora

John

Mike

Names with 5 letters:

Chris

Ethan

Sarah

Names with 6 letters:

Lizzie

Wesley

1. The

itertools.groupby()

function takes a sequence and a key function, and returns an iterator that

generates pairs. Each pair contains theresult of

key_function(each item)

and another iterator containing all

the items that shared that key result.

2. Calling the

list()

function “exhausted” the iterator, i.e. you’ve already generated every item in the iterator

to make the list. There’s no “reset” button on an iterator; you can’t just start over once you’ve exhausted

199

☞

it. If you want to loop through it again (say, in the upcoming

for

loop), you need to call

itertools.groupby()

again to createa new iterator.

3. In this example, given a list of names already sorted by length,

itertools.groupby(names, len)

will put all the

4-letter names in one iterator, all the 5-letter names in another iterator, and so on. The

groupby()

function

is completely generic;it could group strings by first letter, numbers by their number of factors, or any other

key function you can thinkof.

The

itertools.groupby()

function only works if the input sequence is already sorted

by the grouping function. In the example above, you grouped a list of names by the

len()

function. That only worked because the input list was already sorted by length.

Are you watching closely?

>>> list(range(0, 3))

[0, 1, 2]

>>> list(range(10, 13))

[10, 11, 12]

>>> list(itertools.chain(range(0, 3), range(10, 13)))

①

[0, 1, 2, 10, 11, 12]

>>> list(zip(range(0, 3), range(10, 13)))

②

[(0, 10), (1, 11), (2, 12)]

>>> list(zip(range(0, 3), range(10, 14)))

③

[(0, 10), (1, 11), (2, 12)]

>>> list(itertools.zip_longest(range(0, 3), range(10, 14)))

④

[(0, 10), (1, 11), (2, 12), (None, 13)]

1. The

itertools.chain()

function takes two iterators and returns an iterator that contains all the items from

the first iterator, followed by all the items from the second iterator. (Actually, it can take any number of

iterators, and it chains them all in the order they were passed to the function.)

2. The

zip()

function does something prosaic that turns out to be extremely useful: it takes any number of

sequences and returns an iterator which returns tuples of the first items of each sequence, then thesecond

items of each, then the third, and so on.

200

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