DCL | Construction Learning for Fine-grained Image Recognition | Machine Learning library

 by   JDAI-CV Python Version: Current License: Non-SPDX

kandi X-RAY | DCL Summary

kandi X-RAY | DCL Summary

DCL is a Python library typically used in Artificial Intelligence, Machine Learning, Deep Learning, Pytorch applications. DCL has no bugs, it has no vulnerabilities and it has low support. However DCL build file is not available and it has a Non-SPDX License. You can download it from GitHub.

This project is a DCL pytorch implementation of Destruction and Construction Learning for Fine-grained Image Recognition, CVPR2019.
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              DCL has a low active ecosystem.
              It has 476 star(s) with 126 fork(s). There are 16 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 25 open issues and 44 have been closed. On average issues are closed in 19 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of DCL is current.

            kandi-Quality Quality

              DCL has 0 bugs and 0 code smells.

            kandi-Security Security

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

            kandi-License License

              DCL has a Non-SPDX License.
              Non-SPDX licenses can be open source with a non SPDX compliant license, or non open source licenses, and you need to review them closely before use.

            kandi-Reuse Reuse

              DCL releases are not available. You will need to build from source code and install.
              DCL has no build file. You will be need to create the build yourself to build the component from source.
              Installation instructions are not available. Examples and code snippets are available.

            Top functions reviewed by kandi - BETA

            kandi has reviewed DCL and discovered the below as its top functions. This is intended to give you an instant insight into DCL implemented functionality, and help decide if they suit your requirements.
            • Train the model
            • Evaluate a single turn on a given epoch
            • Update the statistics
            • Calculate the loss
            • Return the number of paths
            • Return datetime
            • Parse command line arguments
            • Load data transforms
            • Adjust the gamma of an image
            • Check if the given image is a PIL Image
            • Perform the forward computation
            • Calculates the phi coefficient
            • Deprecated
            • Resize a PIL Image
            Get all kandi verified functions for this library.

            DCL Key Features

            No Key Features are available at this moment for DCL.

            DCL Examples and Code Snippets

            No Code Snippets are available at this moment for DCL.

            Community Discussions

            QUESTION

            Are thread_local objects initialized to 0 in C?
            Asked 2022-Mar-24 at 21:09

            Are non-local thread storage duration objects initialized to 0 in C?

            ...

            ANSWER

            Answered 2022-Mar-24 at 21:09

            Thread local objects without explicit initialization are initialized to “zero.”

            C 2018 7.26.1 3 says thread_local expands to _Thread_local if is included. That is not shown in the question, but presumably it is.

            C 2018 6.2.4 4 says “An object whose identifier is declared with the storage-class specifier _Thread_local has thread storage duration

            C 2018 6.7.9 10 says:

            … If an object that has static or thread storage duration is not initialized explicitly, then:

            — if it has pointer type, it is initialized to a null pointer;

            — if it has arithmetic type, it is initialized to (positive or unsigned) zero;

            — if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

            — if it is a union, the first named member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

            The paragraph ends there; the trailing “;” instead of a period is a typographical error.

            Source https://stackoverflow.com/questions/71609066

            QUESTION

            Introduced intermediate variable in structured binding definition?
            Asked 2022-Mar-19 at 13:13

            In [dcl.struct.bind] 9.6.4, there is definition of structured binding when initializer is a class type with std​::​tuple_­size​::​value properly defined:

            ... variables are introduced with unique names ri as follows:
            S Ui ri = initializer ;
            Each vi is the name of an lvalue of type Ti that refers to the object bound to ri; the referenced type is Ti.

            My question is why is it necessary to introduce ri, can't we define the identifier vi directly as reference to the result of get(e)?

            ...

            ANSWER

            Answered 2022-Mar-19 at 11:43

            The intent is to disallow redeclaring structured bindings as references. See CWG 2313.

            Source https://stackoverflow.com/questions/71535696

            QUESTION

            Clang and GCC has an inconsistent interpretation to the case where the parameter name appears as an unevaluated operand
            Asked 2022-Feb-10 at 07:02
            #include 
            void fun(int a = sizeof(a)){
                std::cout<< a<
            ...

            ANSWER

            Answered 2022-Feb-10 at 07:02

            GCC is wrong. It assumes that a is not yet in scope in the default argument, although generally the point of declaration is immediately after the declarator before the initializer and none of the exceptions apply here. [basic.scope.pdecl]/1

            An older bug report for this is here.

            Note however, that the bug report has equivalent code to yours, but claims it should be ill-formed. That is because it was written before CWG 2082 which allowed the use of a parameter in an unevaluated context in the default argument.

            Related question regarding this GCC bug here.

            Source https://stackoverflow.com/questions/71059957

            QUESTION

            Implementing lazy initialisation in C
            Asked 2022-Jan-02 at 17:00

            In C++, I can implement lazy initialisation with static:

            ...

            ANSWER

            Answered 2022-Jan-02 at 17:00

            In C, there are no references, and you have to type struct, and you have to take care of initialization.

            Source https://stackoverflow.com/questions/70557610

            QUESTION

            Using regex to split a sting into multiple variables SAS
            Asked 2021-Dec-17 at 16:26

            I have a question regarding the usage of regex in SAS.

            My dataset looks like that:

            ID Code 101 K2K5K8F10F26F2 102 L7P13P4 103 L1

            And I would like it to look like this:

            ID Code 101 K2 101 K5 101 K8 101 F10 101 F26 101 F2 102 L7 102 P13 102 P4 103 L1

            At the beginning I thought that it is easier to do it first by assigning new columns and then by rows.

            My attempt looks as follows:

            ...

            ANSWER

            Answered 2021-Dec-17 at 16:12

            I find this a particularly good use case for call scan, regex isn't nearly as efficient. Here I use call scan to find the "word boundary" of the (always single) letter, then grab it plus whatever's before the next letter (or end-of-word).

            Source https://stackoverflow.com/questions/70394683

            QUESTION

            Why is is_trivially_copyable_v different in GCC and MSVC?
            Asked 2021-Dec-08 at 16:06

            When running this simple program, different behaviour is observed depending on the compiler.

            It prints true when compiled by GCC 11.2, and false when compiled by MSVC 19.29.30137 with the (both are the latest release as of today).

            ...

            ANSWER

            Answered 2021-Dec-08 at 16:06

            GCC and Clang report that S is trivially copyable in C++11 through C++23 standard modes. MSVC reports that S is not trivially copyable in C++14 through C++20 standard modes.

            N3337 (~ C++11) and N4140 (~ C++14) say:

            A trivially copyable class is a class that:

            • has no non-trivial copy constructors,
            • has no non-trivial move constructors,
            • has no non-trivial copy assignment operators,
            • has no non-trivial move assignment operators, and
            • has a trivial destructor.

            By this definition, S is trivially copyable.

            N4659 (~ C++17) says:

            A trivially copyable class is a class:

            • where each copy constructor, move constructor, copy assignment operator, and move assignment operator is either deleted or trivial,
            • that has at least one non-deleted copy constructor, move constructor, copy assignment operator, or move assignment operator, and
            • that has a trivial, non-deleted destructor

            By this definition, S is not trivially copyable.

            N4860 (~ C++20) says:

            A trivially copyable class is a class:

            • that has at least one eligible copy constructor, move constructor, copy assignment operator, or move assignment operator,
            • where each eligible copy constructor, move constructor, copy assignment operator, and move assignment operator is trivial, and
            • that has a trivial, non-deleted destructor.

            By this definition, S is not trivially copyable.

            Thus, as published, S was trivally copyable in C++11 and C++14, but not in C++17 and C++20.

            The change was adopted from DR 1734 in February 2016. Implementors generally treat DRs as though they apply to all prior language standards by convention. Thus, by the published standard for C++11 and C++14, S was trivially copyable, and by convention, newer compiler versions might choose to treat S as not trivially copyable in C++11 and C++14 modes. Thus, all compilers could be said to be correct for C++11 and C++14.

            For C++17 and beyond, S is unambiguously not trivially copyable so GCC and Clang are incorrect. This is GCC bug #96288 and LLVM bug #39050

            Source https://stackoverflow.com/questions/70267387

            QUESTION

            What should happen if one calls `std::exit` in a global object's destructor?
            Asked 2021-Oct-28 at 07:46

            Consider the following code:

            ...

            ANSWER

            Answered 2021-Oct-27 at 12:26

            [basic.start.main]/4:

            If std​::​exit is called to end a program during the destruction of an object with static or thread storage duration, the program has undefined behavior.

            Source https://stackoverflow.com/questions/69737959

            QUESTION

            Identity of unnamed enums with no enumerators
            Asked 2021-Oct-26 at 12:10

            Consider a program with the following two translation units:

            ...

            ANSWER

            Answered 2021-Oct-26 at 07:51

            This program is well-formed and prints 1, as seen. Because S is defined identically in both translation units with external linkage, it is as if there is one definition of S ([basic.def.odr]/14) and thus only one enumeration type is defined. (In practice it is mangled based on the name S or S::x.)

            This is just the same phenomenon as static local variables and lambdas being shared among the definitions of an inline function:

            Source https://stackoverflow.com/questions/69714747

            QUESTION

            Accessing captured variables through explicit this parameter in lambda
            Asked 2021-Oct-23 at 13:38

            From declarations / functions / 9.3.4.6 / 6.2 (i apologize on how to cite the specific sentence from standard):

            An explicit-object-parameter-declaration is a parameter-declaration with a this specifier. An explicit-object-parameter-declaration shall appear only as the first parameter-declaration of a parameter-declaration-list of either: (6.1) a member-declarator that declares a member function ([class.mem]), or (6.2) a lambda-declarator ([expr.prim.lambda]).

            If this as explicit object parameter is permitted from lambda expressions, what will happen when we capture variables at the same time?

            Based on my understandings, if we have lambda under the hood:

            ...

            ANSWER

            Answered 2021-Oct-23 at 13:34

            The standard doesn't allow it:

            For each entity captured by copy, an unnamed non-static data member is declared in the closure type.

            If it's "unnamed", then you can't name it. There's specific language that causes the name of a captured entity to be transformed into a this-based expression, but that's it.

            So you can take an explicit this parameter, and names of captured entities will automatically use that. But you can't access those variables through the explicit parameter.

            The only reason to explicitly take this in a lambda is to use the interfaces the standard provides: calling the lambda. AKA: recursively calling the lambda without naming the lambda.

            Source https://stackoverflow.com/questions/69688581

            QUESTION

            Are uninitialized references zero-initialized and uninitialized scalars default-initialized?
            Asked 2021-Sep-17 at 07:32

            Are the following statements correct?

            • An uninitialized reference is considered zero-initialized.
            • An uninitialized scalar is considered default-initialized.
            • Any other uninitialized entity is not considered zero-initialized nor default-initialized.

            They are based on [dcl.init.general/6] (bold emphasis mine):

            To zero-initialize an object or reference of type T means:

            • if T is a scalar type, the object is initialized to the value obtained by converting the integer literal 0 (zero) to T;
            • if T is a (possibly cv-qualified) non-union class type, its padding bits are initialized to zero bits and each non-static data member, each non-virtual base class subobject, and, if the object is not a base class subobject, each virtual base class subobject is zero-initialized;
            • if T is a (possibly cv-qualified) union type, its padding bits are initialized to zero bits and the object's first non-static named data member is zero-initialized;
            • if T is an array type, each element is zero-initialized;
            • if T is a reference type, no initialization is performed.

            and on [dcl.init.general/7] (bold emphasis mine):

            To default-initialize an object of type T means:

            • If T is a (possibly cv-qualified) class type ([class]), constructors are considered. The applicable constructors are enumerated ([over.match.ctor]), and the best one for the initializer () is chosen through overload resolution ([over.match]). The constructor thus selected is called, with an empty argument list, to initialize the object.
            • If T is an array type, each element is default-initialized.
            • Otherwise, no initialization is performed.
            ...

            ANSWER

            Answered 2021-Sep-16 at 10:21

            (Several questions in the same questions: I will answer one)

            Is the following statement correct?

            • An uninitialized reference is considered zero-initialized.

            No. There is no such thing as an uninitialized reference, as it would need to refer to a variable or object, as opposed to a (reference type) parameter or a (reference type) return type, in which context initialization do not apply; as per [dcl.init.ref]/1 and /3:

            /1 A variable whose declared type is “reference to T” ([dcl.ref]) shall be initialized.

            /3 The initializer can be omitted for a reference only in a parameter declaration ([dcl.fct]), in the declaration of a function return type, in the declaration of a class member within its class definition ([class.mem]), and where the extern specifier is explicitly used.

            Regarding the quoted [dcl.init.general]/6:

            To zero-initialize an object or reference of type T means:

            • [...]
            • [...] if T is a reference type, no initialization is performed.

            it refers to the cases where other parts of the standard governs that zero-initialization applies (and what is the effect of it for different entities); for initialization of a reference, particularly consider static initialization, as governed per [basic.start.static]/2:

            [...] If constant initialization is not performed, a variable with static storage duration ([basic.stc.static]) or thread storage duration ([basic.stc.thread]) is zero-initialized ([dcl.init]). Together, zero-initialization and constant initialization are called static initialization; all other initialization is dynamic initialization. All static initialization strongly happens before ([intro.races]) any dynamic initialization.

            Such that static initialization of a reference type variable (which is not constant-initialized) will pertain solely of dynamic initialization, as the effect of the zero initialization of first step of static initialization is no initialization.

            Source https://stackoverflow.com/questions/69206296

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

            Vulnerabilities

            No vulnerabilities reported

            Install DCL

            You can download it from GitHub.
            You can use DCL 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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