bitsets | Ordered subsets over a finite domain

 by   xflr6 Python Version: 0.8.4 License: MIT

kandi X-RAY | bitsets Summary

kandi X-RAY | bitsets Summary

bitsets is a Python library. bitsets has no bugs, it has no vulnerabilities, it has build file available, it has a Permissive License and it has low support. You can install using 'pip install bitsets' or download it from GitHub, PyPI.

Ordered subsets over a finite domain
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            kandi-support Support

              bitsets has a low active ecosystem.
              It has 24 star(s) with 5 fork(s). There are 2 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 0 open issues and 1 have been closed. On average issues are closed in 2 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of bitsets is 0.8.4

            kandi-Quality Quality

              bitsets has 0 bugs and 0 code smells.

            kandi-Security Security

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

            kandi-License License

              bitsets 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

              bitsets releases are not available. You will need to build from source code and install.
              Deployable package is available in PyPI.
              Build file is available. You can build the component from source.
              bitsets saves you 369 person hours of effort in developing the same functionality from scratch.
              It has 880 lines of code, 170 functions and 18 files.
              It has high code complexity. Code complexity directly impacts maintainability of the code.

            Top functions reviewed by kandi - BETA

            kandi has reviewed bitsets and discovered the below as its top functions. This is intended to give you an instant insight into bitsets implemented functionality, and help decide if they suit your requirements.
            • Pack bools into chunks
            • Flattens a sequence of integers
            • Return the rank of n
            • Generate indexes of integers
            • Create a BitSet from a list of members
            • Construct a list from a list of bitsets
            • Return the number of tokens in the sequence
            • Count the number of bits in the sequence
            • Create a BitSet from a list of boolean values
            • Create an instance from a sequence of bits
            • Create a molecule from bools
            Get all kandi verified functions for this library.

            bitsets Key Features

            No Key Features are available at this moment for bitsets.

            bitsets Examples and Code Snippets

            No Code Snippets are available at this moment for bitsets.

            Community Discussions

            QUESTION

            How to split bitset of any size into list of 64 bits bitsets?
            Asked 2021-May-04 at 20:59

            I have a bitset of any size and I would like to know the fastest way to get a list of 64 bits bitsets from my original bitset ?

            For example, from bitset<10000> b('010001110 ...'), I would like to get a list of 64 bits bitsets containing the 1st 64th bits, then the next 64th bits from my original bitset, and so on.

            ...

            ANSWER

            Answered 2021-May-04 at 20:59

            Sadly, there isn't any functionality to efficiently do this operation straight from the STL. You'd have to go bit by bit or you can shift and mask, as explained in this answer.

            However, giving credit to this

            You can use boost::dynamic_bitset, which can be converted to a range of "blocks" using boost::to_block_range.

            Source https://stackoverflow.com/questions/67391891

            QUESTION

            Why is bitset throwing out_of_range error?
            Asked 2021-Apr-10 at 15:22

            I am implementing a bloom filter with help of bitsets in c++ for finding out malicious URLs. I have a bitset of 100 bits and a simple hash function. But still I get this error.

            ...

            ANSWER

            Answered 2021-Apr-10 at 15:22

            After reading carefully I thought the reason could be signed characters. If signed characters are involved,

            Source https://stackoverflow.com/questions/67021846

            QUESTION

            Declaring an array of std::bitset using std::string
            Asked 2021-Apr-06 at 01:12

            Im currently trying to declare an array of 17 std::bitsets, each 32 bits long. I'm doing it like this:

            ...

            ANSWER

            Answered 2021-Apr-06 at 01:12

            You need to call the constructor of std::bitset like:

            Source https://stackoverflow.com/questions/66959037

            QUESTION

            Python - Fastest way to calculate log2(int) of a number that is an integral power of 2
            Asked 2021-Jan-12 at 18:02

            I'm using bitsets and want to iterate over the set bits quickly, and without needing to know the max set bit ahead of time. Using this trick, it seems like we can do this without scanning the 0s in the middle. But the linked trick produces set bits as bitsets, not as integers (e.g. the 0th set bit is produced as 0b1, not 0).

            Is there a fast bit trick to calculate log2(x) when I know x will be an exact power of 2 (e.g. as in the above case)?

            What have I tried

            Simplest version, using the standard library and with bits(n) being the linked code:

            ...

            ANSWER

            Answered 2021-Jan-12 at 18:02

            QUESTION

            How to convert a Byte to a Bitstring in Kotlin?
            Asked 2020-Sep-06 at 21:11

            I have an ArrayList of Bytes. Firstly, when i print them i see integers? And the second thing is, i want to convert each Byte to a Bitstring and add it to a new list of bitstrings. How do i do that as there is no "i.toBitString"?

            ...

            ANSWER

            Answered 2020-Sep-06 at 21:11

            You can convert to any base with this method, in your case this should work:

            Source https://stackoverflow.com/questions/63767427

            QUESTION

            Casting complex, non-primitive C++ data types into Erlang/Elixir format, to export methods using NIF
            Asked 2020-Jun-30 at 18:23

            I'm writting a library in C++ which I'll use in Elixir/Erlang. There're C++ methods that accept and return, including via out parameters, and involving pointers, data structures or library from std library such as: tuples, vectors, priority queues, bitsets and so forth. And also methods that accept or return generics. Or my own custom data structures.

            How do I export such methods?

            ...

            ANSWER

            Answered 2020-Jun-30 at 18:23

            In Erlang, types are what are defined in here. Internal Erlang and C/C++ std representations do not match, you cannot return a, for example, int64_t from C and use it directly from Erlang.

            Same for complex structures, is PrioryQueue an Erlang's list() or a {list(), pos_integer()}?

            This means that you need to transform types back and forth, using the enif_get_* and enif_make_* erl_nif functions. For really complex structures this may be tedious, so you really need to consider if it wouldn't be enough using resource objects.

            Resource objects are just pointers to memory, and thus, opaque for Erlang. You can have this opaque hold the pointer to the priority queue memory and include methods to put/2 and get/2 Erlang terms to the queue.

            Why are the functions from erl_nif required?

            Erlang has dynamic typing, where each reference held by a variable includes its type (either in the value for the immediate terms, or in the reference for the referenced terms), while C/C++ have static typing, where the variable is the one that states the type only at compile time.

            For C/C++, 0xfabada could be an int, uint, a char*, a void* pointing to your custom structure...

            Other reasons for the representations not to match include:

            1. Erlang's integers have variable size
            2. Erlang terms are tagged (some bits of the reference indicate the type)
            3. The closest thing to an atom in C/C++ is an enum, and they are quite different
            4. Binaries (both short an long binaries) and subbinaries
            5. Memory management ... and so on.

            Source https://stackoverflow.com/questions/62660417

            QUESTION

            How to filter a std::integer_sequence
            Asked 2020-May-25 at 19:09

            If I theoretically have a sequence of integers like

            ...

            ANSWER

            Answered 2017-Jan-20 at 14:45
            Edit 2

            After some back and forth on Barry's answer, I've come up with the following answer that merges the concepts and handles some empty-sequence edge cases (Full code):

            We are allowed to pass a predicate to a function only if it is a constexpr lambda, as only literal types are allowed in constexpr functions, and normal free-floating functions aren't literal types (although I suppose you could wrap one within your lambda).

            Our generic filter function will accept a sequence and a predicate, and return a new sequence. We will use constexpr if to handle empty sequence cases (which also requires the maybe_unused attribute on the predicate, because it's unused) :

            Source https://stackoverflow.com/questions/41723704

            QUESTION

            C++ multiple nested If-else statements work but combined doesn't
            Asked 2020-May-06 at 18:03

            I'm a newbie to c++. I attempted a questions related to bitsets(https://www.codechef.com/problems/CHEFQUE), when I use the following nested statements the code passes all the tests

            ...

            ANSWER

            Answered 2020-May-06 at 17:22

            As you can see from this table the operator ! has a higher precedence that & (Bitwise and, not address of), so your code should looks like this:

            Source https://stackoverflow.com/questions/61641146

            QUESTION

            Turning a char* into an uint8_t
            Asked 2020-Apr-12 at 16:53

            Say I have an array that looks like:{'0b00011000','0b10001000'} How do I convert each element within that array into a uint8_t keeping its output in binary format. (e.g. '0b10001000' == 0b10001000).

            Are there functions that can over turn this, or will I have to make my own function?

            p.s. I get a 'precision lost' error for casting, and when I bypass that obviously it loses its precision.

            Edit: I did end up figuring out using bitsets instead of using uint8_t.

            ...

            ANSWER

            Answered 2020-Apr-12 at 09:04

            QUESTION

            Gecode vs. Z3 for Constrained Randomization
            Asked 2020-Jan-27 at 18:05

            I'm looking for a C++-based alternative to the SystemVerilog language. While I doubt anything out there can match the simplicity and flexibility of the SystemVerilog constraint language, I have settled on using either Z3 or Gecode for what I'm working on, primarily because they're both under the MIT license.

            What I'm looking for is:

            1. Support for variable-sized bit vectors AND bit vector arithmetic logic operations. For example:
            ...

            ANSWER

            Answered 2020-Jan-27 at 18:05

            While z3 (or any SMT solver) can handle all of these, getting a nice sampling of satisfying assignments would be rather difficult to control. SMT solvers are optimized for just giving you a model, and they don't have much in terms of how you want to sample the solution space.

            Incidentally, this is an active research area in SMT solving. Here's a paper that appeared only 6 weeks ago on this very topic: https://ieeexplore.ieee.org/document/8894251

            So, I'd say if support for "good sampling" is your primary motivation, using an SMT solver is probably not the best choice. If your goal is to find satisfying assumptions for bit-vectors expressed conveniently (there are high level APIs in any language you can imagine these days), then z3 would be an extremely fine choice.

            From your description, good sampling sounds like the primary motivation though, and for that SMT solvers are probably not that great. At least not for the time being.

            Source https://stackoverflow.com/questions/59936181

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

            Vulnerabilities

            No vulnerabilities reported

            Install bitsets

            You can install using 'pip install bitsets' or download it from GitHub, PyPI.
            You can use bitsets like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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            Install
          • PyPI

            pip install bitsets

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            https://github.com/xflr6/bitsets.git

          • CLI

            gh repo clone xflr6/bitsets

          • sshUrl

            git@github.com:xflr6/bitsets.git

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