immutable-tuple | Immutable finite list objects with constant-time | Functional Programming library

 by   benjamn JavaScript Version: 0.4.10 License: MIT

kandi X-RAY | immutable-tuple Summary

kandi X-RAY | immutable-tuple Summary

immutable-tuple is a JavaScript library typically used in Travel, Transportation, Logistics, Programming Style, Functional Programming applications. immutable-tuple has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can install using 'npm i immutable-tuple' or download it from GitHub, npm.

Immutable finite list objects with constant-time equality testing (===) and no memory leaks.
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            kandi-support Support

              immutable-tuple has a low active ecosystem.
              It has 30 star(s) with 4 fork(s). There are 1 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 3 open issues and 2 have been closed. There are 1 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of immutable-tuple is 0.4.10

            kandi-Quality Quality

              immutable-tuple has 0 bugs and 0 code smells.

            kandi-Security Security

              immutable-tuple has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              immutable-tuple code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              immutable-tuple is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              immutable-tuple releases are not available. You will need to build from source code and install.
              Deployable package is available in npm.
              Installation instructions, examples and code snippets are available.
              immutable-tuple saves you 458 person hours of effort in developing the same functionality from scratch.
              It has 1082 lines of code, 0 functions and 14 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

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            immutable-tuple Key Features

            No Key Features are available at this moment for immutable-tuple.

            immutable-tuple Examples and Code Snippets

            No Code Snippets are available at this moment for immutable-tuple.

            Community Discussions

            QUESTION

            How do purely functional languages handle index-based algorithms?
            Asked 2022-Apr-05 at 12:51

            I have been trying to learn about functional programming, but I still struggle with thinking like a functional programmer. One such hangup is how one would implement index-heavy operations which rely strongly on loops/order-of-execution.

            For example, consider the following Java code:

            ...

            ANSWER

            Answered 2022-Mar-07 at 21:17

            This is not an index-heavy operation, in fact you can do this with a one-liner with scanl1 :: (a -> a -> a) -> [a] -> [a]:

            Source https://stackoverflow.com/questions/71387267

            QUESTION

            Haskell comparing two lists' lengths but one of them is infinite?
            Asked 2022-Mar-22 at 20:54

            I want to write a function that checks if the first list is longer than the second list and one of them can be infinite. However I can't find a working solution.

            ...

            ANSWER

            Answered 2022-Mar-22 at 20:54

            Plain old natural numbers will not do the trick, because you can't calculate the natural number length of an infinite list in finite time. However, lazy natural numbers can do it.

            Source https://stackoverflow.com/questions/71565429

            QUESTION

            Why is `forever` in Haskell implemented this way?
            Asked 2022-Feb-05 at 20:34

            Haskell provides a convenient function forever that repeats a monadic effect indefinitely. It can be defined as follows:

            ...

            ANSWER

            Answered 2022-Feb-05 at 20:34

            The execution engine starts off with a pointer to your loop, and lazily expands it as it needs to find out what IO action to execute next. With your definition of forever, here's what a few iterations of the loop like like in terms of "objects stored in memory":

            Source https://stackoverflow.com/questions/70990108

            QUESTION

            Memoize multi-dimensional recursive solutions in haskell
            Asked 2022-Jan-13 at 14:28

            I was solving a recursive problem in haskell, although I could get the solution I would like to cache outputs of sub problems since has over lapping sub-problem property.

            The question is, given a grid of dimension n*m, and an integer k, how many ways are there to reach the gird (n, m) from (1, 1) with not more than k change of direction?

            Here is the code without of memoization

            ...

            ANSWER

            Answered 2021-Dec-16 at 16:23

            In Haskell these kinds of things aren't the most trivial ones, indeed. You would really like to have some in-place mutations going on to save up on memory and time, so I don't see any better way than equipping the frightening ST monad.

            This could be done over various data structures, arrays, vectors, repa tensors. I chose HashTable from hashtables because it is the simplest to use and is performant enough to make sense in my example.

            First of all, introduction:

            Source https://stackoverflow.com/questions/70376569

            QUESTION

            Why is my Haskell function argument required to be of type Bool?
            Asked 2021-Nov-30 at 09:42

            I have a function in Haskell that is defined as follows:

            ...

            ANSWER

            Answered 2021-Nov-30 at 09:42

            Haskell values have types. Each value has a type. One type. It can't be two different types at the same time.

            Thus, since x is returned as the result of if's consequent, the type of the whole if ... then ... else ... expression is the same as x's type.

            An if expression has a type. Thus both its consequent and alternative expression must have that same type, since either of them can be returned, depending on the value of the test. Thus both must have the same type.

            Since x is also used in the test, it must be Bool. Then so must be y.

            Source https://stackoverflow.com/questions/70053894

            QUESTION

            Vector of functions in APL
            Asked 2021-Nov-30 at 09:31

            What is the syntax for a vector (array) of functions in APL?

            I have tried the following but these are interpreted as a 3-train and a 2-train, respectively:

            ...

            ANSWER

            Answered 2021-Nov-28 at 23:26

            Dyalog APL does not officially support function arrays, you can awkwardly emulate them by creating an array of namespaces with identically named functions.

            Source https://stackoverflow.com/questions/70148229

            QUESTION

            What's the theoretical loophole that allows F# (or any functional language) to apply a function mulitple times on the same input
            Asked 2021-Nov-17 at 06:29

            In F# if I write

            ...

            ANSWER

            Answered 2021-Nov-17 at 01:24

            To expand on the answer given in the comments, the first p is an immutable value, while the second p is a function. If you refer to an immutable value multiple times, then (obviously) its value doesn't change over time. But if you invoke a function multiple times, it executes each time, even if the arguments are the same each time.

            Note that this is true even for pure functional languages, such as Haskell. If you want to avoid this execution cost, there's a specific technique called memoization that can be used to return cached results when the same inputs occur again. However, memoization has its own costs, and I'm not aware of any mainstream functional language that automatically memoizes all function calls.

            Source https://stackoverflow.com/questions/69997578

            QUESTION

            Is Control.Monad.Reader.withReader actually Data.Functor.Contravariant.contramap?
            Asked 2021-Nov-03 at 06:39

            I'm working trough the book Haskell in depth and I noticed following code example:

            ...

            ANSWER

            Answered 2021-Nov-03 at 06:39

            Reader's type parameters aren't in the right order for that to be contramap for it. A Contravariant functor always needs to be contravariant in its last type parameter, but Reader is contravariant in its first type parameter. But you can do this:

            Source https://stackoverflow.com/questions/69817514

            QUESTION

            Confused about evaluation of lazy sequences
            Asked 2021-Oct-20 at 15:49

            I am experimenting with clojure's lazy sequences. In order to see when the evaluation of an item would occur, I created a function called square that prints the result before returning it. I then apply this function to a vector using map.

            ...

            ANSWER

            Answered 2021-Oct-20 at 15:49

            Laziness isn't all-or-nothing, but some implementations of seq operate on 'chunks' of the input sequence (see here for an explanation). This is the case for vector which you can test for with chunked-seq?:

            Source https://stackoverflow.com/questions/69648554

            QUESTION

            What is the relation between syntax sugar, laziness and list elements accessed by index in Haskell?
            Asked 2021-Aug-30 at 04:46

            Haskell lists are constructed by a sequence of calls to cons, after desugaring syntax:

            ...

            ANSWER

            Answered 2021-Aug-30 at 04:46

            Lists in Haskell are special in syntax, but not fundamentally.

            Fundamentally, Haskell list is defined like this:

            Source https://stackoverflow.com/questions/68978811

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install immutable-tuple

            First install the package from npm:.

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          • npm

            npm i immutable-tuple

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          • CLI

            gh repo clone benjamn/immutable-tuple

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            git@github.com:benjamn/immutable-tuple.git

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