cartesian-product | memory footprint function to generate all combinations
kandi X-RAY | cartesian-product Summary
kandi X-RAY | cartesian-product Summary
A simple, low-memory footprint function to generate all combinations from a multi-dimensionnal array.
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- Get an iterator over the set .
- Returns the number of elements in the set .
- Validate a subset .
- Creates a subset
- Get the collection as an array .
cartesian-product Key Features
cartesian-product Examples and Code Snippets
Lines of Code : 40License : Permissive (MIT)
Copyprint_r(cartesian_product($data)->asArray());
Array
(
[0] => Array
(
[hair] => blond
[eyes] => blue
)
[1] => Array
(
[hair] => blond
[eyes] => gree require_once __DIR__ . '/vendor/autoload.php';
use function BenTools\CartesianProduct\cartesian_product;
$data = [
'hair' => [
'blond',
'black'
],
'eyes' => [
'blue',
'green',
function (arra require_once __DIR__ . '/vendor/autoload.php';
use function BenTools\CartesianProduct\cartesian_product;
$data = [
'hair' => [
'blond',
'red',
],
'eyes' => [
'blue',
'green',
'brown',
], function cartesianProduct(sets, index, current) {
if (index === sets.length) {
return result.push(current.slice());
}
for (var i = 0; i < sets[index].length; i += 1) {
current[index] = sets[index][i];
cart def _cartesian_product(first, second):
"""Returns all path combinations of first and second."""
return [os.path.join(f, s) for f in first for s in second] Community Discussions
Trending Discussions on cartesian-product
QUESTION
from this dictionary
...ANSWER
Answered 2022-Feb-22 at 16:15Quick solution would be to do only cartesian product of dictionary values while preserving information about keys, and next recreate dictionaries:
QUESTION
Not sure if title is descriptive, but what I would like is this:
input:
- list of templates(in my case containers) taking 1 (required, can be more optional) type arguments
- list of types
output:
"cartesian product" where each template in first set is instantiated with every type in second set.
example:
...ANSWER
Answered 2022-Feb-12 at 23:12You could use std::tuples. Especially std::tuple_cat is nice for this.
Example:
QUESTION
I've been trying to generate all the possible combinations between arrays, let's say a, b, c, x, y, z where the last 3 (x, y, z) can be arrays OR floats. Thanks to useful comments and answers the task is accomplished (BTW, in a more general way, accepting arrays and floats) by
ANSWER
Answered 2021-Dec-05 at 19:12Multiple types are cumbersome, especially if you store them in the same list. You could simply do:
QUESTION
I am looking for a function that will take a nested dictionary, and produce the combinations / product of the values.
My query is similar to the problem specified here, but I can't seem to adapt the answers their to fit my needs: Cartesian product of nested dictionaries of lists
I wish to have an input like this:
...ANSWER
Answered 2021-Dec-01 at 03:40You can use recursion with itertools.product:
QUESTION
I have an array of objects with names and options and I need all possible combinations of products. The important part is that this array has an N number of objects with N number of options in every object.
I tried to create some kind of recursive algorithm, but the problem is that I failed to push recursively to receive the needed data structure in the end. I also tried the approach from Cartesian product of multiple arrays in JavaScript but it seems it is not relevant to the output needed.
Example:
...ANSWER
Answered 2021-Nov-19 at 15:15Update: Fixed the first cartesian implementation based on suggestion from Mulan.
This is definitely a cartesian product question. The only thing is that you will need to format your input before you call the cartesian product. Here is one version, using a simple, recursive cartesian function.
QUESTION
Base provides ZipList, which is just a wrapper for [] where <*> is based on zip instead of cartesian-product. This isn't the default because it's not consistent with the Monad [] instance, but some people find it more intuitive, and both behaviors are useful in different contexts.
Edward Kmett provides Distributive, the dual of Traversable. A Traversable can be mapped/pushed/distributed into any Applicative Functor; a Distributive can be pulled/factored out of any Functor. (For reasons I haven't unpacked, distribute does not require the outer layer to be Applicative.)
Length-indexed lists are Distributive, and work how you'd expect. Specifically, their Applicative instance is based on zip, just like ZipList! This suggests ZipList could also be Distributive, which would be useful.
The docs for Distributive note two things that must be true of any instance:
- "It [must be] isomorphic to
(->) xfor somex."- A list is isomorphic to a partial function
Int ->.
- A list is isomorphic to a partial function
- "[It] will need to have a way to consistently zip a potentially infinite number of copies of itself."
- This is more-or-less the raison d'être of
ZipList.
- This is more-or-less the raison d'être of
Is that good enough? I spent a couple hours this afternoon trying to write instance Distributive ZipList where distributive = ... and couldn't quite get it to work. For most functors f ZipList a there's an obvious meaning to distribute f, although I worry that might just be because I'm not thinking of enough non-Traversable functors.
Maybeis tricky; shoulddistribute Nothingbe[]orrepeat Nothing? Butdistributeis dual tosequenceA, and theTraversable Maybeinstance says it should berepeat Nothing.(->) emight be the dealbreaker.- Intuitively
distribute $ const (repeat x) = repeat (const x). - We can extend that to any function that's guaranteed to return an infinite list; it looks kinda like
(\i -> (! i) <$> f) <$> [0..]. - We can extend that to functions returning finite lists; we end up with an infinite list of partial functions. It's not obvious to me that that's unacceptable; partial functions show up all the time when working with lists.
- But that means
distribute $ const [] ≅ repeat undefined, which is kinda silly. - The instance
Applicative ZipListcontains an important design decision:length (a <*> b) == min (length a) (length b)(as opposed to an error or whatever). We're not leveraging that at all; the way I can see we might would bedistribute = const [].
- Intuitively
Does anyone see a way forward?
If the partial-function interpretation were "acceptable", could we generalize that in any less-dumb way than distribute f = (\i -> (!! i) <$> f) <$> [0..]?
ANSWER
Answered 2021-Jul-20 at 01:34No, it cannot be distributive.
The obvious Distributive instance for Pair looks like this:
QUESTION
I have a nested loop, which will assemble all variations of params. the more params, the bigger the nesting will become.
...ANSWER
Answered 2021-Jun-09 at 14:28You can use still use itertools.product, just zip back the keys and values
QUESTION
I'm working with a dataset of around 700 observations and would like to calculate how many combinations of 4 observations have a mean value between a high and low value.
I've created code that can do this, but applying it to my larger dataset results in billions of combinations and takes several days to run. Surely there is a faster or more efficient way to do this
Here is what I've tried:
...ANSWER
Answered 2021-Apr-13 at 11:58One helpful idea here is that itertools.combinations returns the combinations
in lexicographical order. So, if the input sequence is sorted, the output
sequence will be sorted, too.
This helps with reducing the sequence of combinations to evaluate, since it
is certain that if item[0] > truehi, then item[0] >= item[1] >= item[2] >= item[3], so there's no more combinations that
will meet the conditions.
This allows us to stop the loop earlier. For some tests I ran, it skipped the last 30% or so of combinations for size 100, based on the test data with Gaussian distribution.
To extend this idea even further, I wrote a custom version of the combinations
that used the same approach for level 1 and 2. That shaves off another 40% or so.
Some other ideas that improve runtime performance is to calculate number of
combinations using the logarithmic version of n take 4, instead of iterating
over all combinations (hinted by comment @crissal),
and to use np.sum, instead of np.avg (comment @Sam cd).
For size 100, this reduces the runtime on my machine from 42s to 7.5s.
With size 200, I measure 217s runtime with the algorithm below, not evaluating 69.3% of the combinations. This is about 29 times longer, even though the number of overall combinations is only about 16.5 times bigger.
With size 300, runtime is about 514s, and in one sample, it skips 85% of the combinations.
With size 700, runtime is about 21,858s, and in one sample, it skips 79% of combinations.
QUESTION
I'm looking for a version of Expand a Set[Set[String]] into Cartesian Product in Scala based on map type :
I would like to start with :
...ANSWER
Answered 2021-Mar-29 at 23:23You could do something like this:
(but note that for larger maps this would consume a lot of memory, it may be worth looking into generating a LazyList instead)
QUESTION
I would like to join three dataframes of the following structure:
...ANSWER
Answered 2021-Mar-13 at 23:57You can merge in two steps. For example for March:
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