Let me explain the 4 combinatorial iterators from the module `itertools` with ice cream:
1. combinations
2. combinations_with_replacement
3. permutations
4. product
1. `combinations`
This iterator will produce tuples of length `r` with all the unique combinations of values from `iterable`.
(โUniqueโ will make use of the original position in `iterable`, and not the value itself.)
E.g., what ice cream flavour combinations can I get?
2. `combinations_with_replacement`
Same as `combinations`, but values can be repeated.
E.g., what ice cream flavour combinations can I get if I allow myself to repeat flavours?
![from itertools import combinations flavours = ["chocolate", "vanilla", "strawberry"]
for scoops in combinations(flavours, 2): print(scoops)](/proxy/post_attachment/296304/6bc078557d.png "from itertools import combinations flavours = [\"chocolate\", \"vanilla\", \"strawberry\"]
for scoops in combinations(flavours, 2): print(scoops)")
![Diagram with a snippet of code showing how to use `combinations_with_replacement`.
from itertools import combinations with _replacement
flavours = ["chocolate", "vanilla", "strawberry"]
for scoops in combinations with_replacement(flavours, 2):
print(scoops)](/proxy/post_attachment/296305/2462116594.png "Diagram with a snippet of code showing how to use `combinations_with_replacement`.
from itertools import combinations with _replacement
flavours = [\"chocolate\", \"vanilla\", \"strawberry\"]
for scoops in combinations with_replacement(flavours, 2):
print(scoops)")
3. `permutations`
All possible combinations of size `r` in all their possible orderings.
E.g., if I get 2 scoops, how can they be served?
This is a very important question because the flavour at the bottom is eaten last!
4. `product`
Matches up all values of all iterables together.
(Computes the cartesian product of the given iterables.)
E.g., if I can get either 2 or 3 scoops, and if the ice cream can be served on a cup or on a cone, how many different orders are there?
![Diagram showing an example usage of `product`.
Code:
from itertools import product
scoop_nums = [2, 3]
served_on = ["cup", "cone"]
for n, on in product(scoop_nums, served_on):
print(f"{n} scoops served on {on}.")](/proxy/post_attachment/296302/a46613656f.png "Diagram showing an example usage of `product`.
Code:
from itertools import product
scoop_nums = [2, 3]
served_on = [\"cup\", \"cone\"]
for n, on in product(scoop_nums, served_on):
print(f\"{n} scoops served on {on}.\")")
![Diagram showing an example usage of `permutations.
Code:
from itertools import permutations
flavours = ["chocolate", "vanilla", "strawberry"]
for scoops in permutations(flavours, 2):
print(scoops)](/proxy/post_attachment/296303/28b3a43ee5.png "Diagram showing an example usage of `permutations.
Code:
from itertools import permutations
flavours = [\"chocolate\", \"vanilla\", \"strawberry\"]
for scoops in permutations(flavours, 2):
print(scoops)")
These are the 4 combinatorial iterators from the module `itertools`.
The module itertools contains a total of 19 (!) iterators!
If you want to learn more about them, take a look at this short article I wrote yesterday: