Generating combinations and permutations is usually a tricky task in programming.
In Python π, using the itertools library this becomes much easier, less memory intensive and faster!
Let me tell you what I learned with them
[4.45 min]
1/13π§΅
In Python π, using the itertools library this becomes much easier, less memory intensive and faster!
Let me tell you what I learned with them
[4.45 min]
1/13π§΅
Let's suppose you want to create all possible cards of a regular deck. What you have to do is to join:
β’ All cards ranks: A, 2 to 10, J, Q and K
β’ The 4 suits: β₯οΈβ£οΈβ¦οΈβ οΈ
How to generate all cards?
2/13π§΅
β’ All cards ranks: A, 2 to 10, J, Q and K
β’ The 4 suits: β₯οΈβ£οΈβ¦οΈβ οΈ
How to generate all cards?
2/13π§΅
The operation that can solve this is a cartesian product:
ranks = ['A'] + list(range(2, 11)) + ['J', 'Q', 'K']
suits = ['β₯οΈ', 'β£οΈ', 'β¦οΈ', 'β οΈ']
all_cards = it.product(suits, ranks)
>>> ('A', 'β₯οΈ'),('A', 'β£οΈ'),('A', 'β¦οΈ'),('A', 'β οΈ'),...
len(all_cards)
>>> 52
3/13π§΅
ranks = ['A'] + list(range(2, 11)) + ['J', 'Q', 'K']
suits = ['β₯οΈ', 'β£οΈ', 'β¦οΈ', 'β οΈ']
all_cards = it.product(suits, ranks)
>>> ('A', 'β₯οΈ'),('A', 'β£οΈ'),('A', 'β¦οΈ'),('A', 'β οΈ'),...
len(all_cards)
>>> 52
3/13π§΅
Before using a deck of cards, usually you shuffle it. One question that we can ask is: How many possible shuffles exist?
This is the same as asking: how many deck permutations exist?
4/13π§΅
This is the same as asking: how many deck permutations exist?
4/13π§΅
To generate permutations, itertools have the permutations function:
all_possible_shuffles = it.permutations(all_cards)
Attention! This is an iterable! If you do list(all_possible_shuffles), do you know what would be the length of this list?
5/13π§΅
all_possible_shuffles = it.permutations(all_cards)
Attention! This is an iterable! If you do list(all_possible_shuffles), do you know what would be the length of this list?
5/13π§΅
If you try running this on public Colab:
num_permutation = len(list(all_possible_shuffles))
It will crash after using all RAM!!! π€―
All possible shuffles of a 52 cards deck is equal to 52! (52 factorial)
This number is close to 8 * 10^67
or 8 followed by 67 zeros!
6/13π§΅
num_permutation = len(list(all_possible_shuffles))
It will crash after using all RAM!!! π€―
All possible shuffles of a 52 cards deck is equal to 52! (52 factorial)
This number is close to 8 * 10^67
or 8 followed by 67 zeros!
6/13π§΅
Here you can clearly see the power of iterables.
You can use a collection that doesn't fit in memory because it's lazily evaluated!
7/13π§΅
You can use a collection that doesn't fit in memory because it's lazily evaluated!
7/13π§΅
Now that we've shuffled the deck, you have to draw K cards from it. My question is, how many possible hands can you draw?
The question, in Math language, is: how many combinations of K cards can you draw from a set of 52 cards? π€
8/13π§΅
The question, in Math language, is: how many combinations of K cards can you draw from a set of 52 cards? π€
8/13π§΅
Itertools also can help you with that.
Let's suppose K = 10:
all_combinations = it.combinations(all_cards, 10)
Again this value is very big!!!
9/13π§΅
Let's suppose K = 10:
all_combinations = it.combinations(all_cards, 10)
Again this value is very big!!!
9/13π§΅
The combinations formula is:
c = n! / (k! * (n-k)!)
c = 52! / (10! * 42!)
c ~ 1.5 * 10^10
Again, huge number! This will pressure the memory!
As you can see, knowing the available tools to help you can make a big difference!
10/13π§΅
c = n! / (k! * (n-k)!)
c = 52! / (10! * 42!)
c ~ 1.5 * 10^10
Again, huge number! This will pressure the memory!
As you can see, knowing the available tools to help you can make a big difference!
10/13π§΅
The last question is, if you change the previous problem to:
how many possible hands can you draw if when you draw a card, you write it down, put it back on the deck, shuffle and draw again. π€
This is done using the combination_with_replacement method.
11/13π§΅
how many possible hands can you draw if when you draw a card, you write it down, put it back on the deck, shuffle and draw again. π€
This is done using the combination_with_replacement method.
11/13π§΅
As you can imagine, this is much bigger than the number of combinations!
Something close to 1.44 * 10Λ17
you can read more about this here: en.wikipedia.org/wiki/Combinatiβ¦
12/13π§΅
Something close to 1.44 * 10Λ17
you can read more about this here: en.wikipedia.org/wiki/Combinatiβ¦
12/13π§΅
These functions are the hardest to understand on the itertools lib but it's worth knowing they exist!
I hope I made it easier to understand them
for the full documentation you can find it here: docs.python.org/3/library/iterβ¦
13/13π§΅
I hope I made it easier to understand them
for the full documentation you can find it here: docs.python.org/3/library/iterβ¦
13/13π§΅
β’ β’ β’
Missing some Tweet in this thread? You can try to
force a refresh
γ
