Lin | Lin is an Android Lint tool | Code Analyzer library

 by   Serchinastico Kotlin Version: 0.0.6 License: Apache-2.0

kandi X-RAY | Lin Summary

kandi X-RAY | Lin Summary

Lin is a Kotlin library typically used in Code Quality, Code Analyzer applications. Lin has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Add the detectors module dependencies to your project and the dsl module as part of the lint classpath:.
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              Lin has a low active ecosystem.
              It has 235 star(s) with 7 fork(s). There are 6 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 10 open issues and 17 have been closed. On average issues are closed in 11 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of Lin is 0.0.6

            kandi-Quality Quality

              Lin has 0 bugs and 0 code smells.

            kandi-Security Security

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

            kandi-License License

              Lin is licensed under the Apache-2.0 License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              Lin releases are available to install and integrate.
              Installation instructions are not available. Examples and code snippets are available.

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            Lin Key Features

            No Key Features are available at this moment for Lin.

            Lin Examples and Code Snippets

            No Code Snippets are available at this moment for Lin.

            Community Discussions

            QUESTION

            Output of the model depends on the shape of the weights tensor
            Asked 2022-Apr-02 at 18:36

            I want to train the model to sum the three inputs. So it is as simple as possible.

            Firstly the weights are initialized randomly. It produces bad error estimate (approx. 0.5)

            Then I initialize the weights with zeros. There are two options:

            1. the shape of the weights tensor is [1, 3]
            2. the shape of the weights tensor is [3]

            When I choose the 1st option the model still works bad and can't learn this simple formula.

            When I choose the 2nd option it works perfect with the error of 10e-12.

            Why the result depends on the shape of the weights? Why do I need to initialize the model with zeros to solve this simple problem?

            ...

            ANSWER

            Answered 2022-Apr-02 at 18:36

            First option doesn't learning because it fails with broadcasting: while out.shape == (1024, 1) corresponding targets y has shape of (1024, ). MSELoss, as expected, computes mean of tensor (out - y)^2, which in this case has shape (1024, 1024), clearly wrong objective for this task. At the same time, after applying 2-nd option tensor (out - y)^2 has size (1024, ) and mean of it corresponds to actual mse. Default approach, without explicit changing weights shape (through option 1 and 2), would work if set target shape to (1024, 1) for example by y = y.unsqueeze(-1) after definition of y.

            Source https://stackoverflow.com/questions/71718929

            QUESTION

            gnuplot visual studio invalid command error in console
            Asked 2022-Mar-16 at 17:15

            This is the following code I have for plotting a series of random points in gnuplot. I have no errors in gnuplot-iostream header.

            ...

            ANSWER

            Answered 2022-Mar-16 at 17:15

            For debugging I used 5 points instead of 1000, so it was easier to see the first error:

            Source https://stackoverflow.com/questions/71486992

            QUESTION

            How to find common values in groupby groups?
            Asked 2022-Feb-26 at 19:22

            I have a df of this format, my goal is to find users who participate in more than one tournament and ultimately set their 'val' value to the one they first appear with. Initially, I was thinking I need to groupby 'tour' but then it needs some intersection but I'm not sure how to proceed. Alternatively, I can do pd.crosstab(df.user, df.tour) but I'm not sure how to proceed either.

            ...

            ANSWER

            Answered 2022-Feb-26 at 18:37

            You can groupby on 'user' and filter out groups with only 1 element, and then select the first one, like so:

            Source https://stackoverflow.com/questions/71279210

            QUESTION

            Why is the rtcclient `getWorkitem()` for IBM RTC HTTP GET request returning incorrect response?
            Asked 2022-Feb-23 at 07:28

            I am trying to use the rtcclient Python package to authenticate and retrieve work items from my IBM RTC board. Specifically my goal is to run this command without errors:

            ...

            ANSWER

            Answered 2022-Feb-23 at 07:28

            Check first if this was working with 0.6.0 (depending on your RTC version 6.x?; 7.x).

            dixudx/rtcclient issue 136 mentions:

            v0.7.0 has introduced a change on authentication, refer to #133 for details (and issue 112), which may be not compatible with older Rational Team Concert. From your issue, it seems to be true. Sorry for this.

            My suggestion is just sticking with older rtcclient version, such as pip install rtcclient==0.6.0.

            Source https://stackoverflow.com/questions/71230419

            QUESTION

            Pytorch - vanilla Gradient Visualization
            Asked 2022-Feb-15 at 09:52

            I trained a neural network on MNIST using PyTorch:

            ...

            ANSWER

            Answered 2022-Feb-15 at 09:52

            You do not need to change anything about your conv layer. Each layer computes gradients both w.r.t. parameters (for updates) and w.r.t. inputs (for "downstream" gradients by the chain rule). Therefore, all you need is to set your input image's x gradient property to true:

            Source https://stackoverflow.com/questions/71113245

            QUESTION

            Django-CMS : Unexpected keyword argument 'providing_args'
            Asked 2022-Feb-10 at 03:20

            I have installed a vanilla django-cms on a new server. I installed all requirements. It all went fine, up untill the point where I wanted to migrate to the database (Postgres).

            So this is what I did :

            1. Tried reinstalling and installing it all again. Didn't change it.
            2. Used google to try and find people with the same error.
            3. Try editing the signals file on which the error(shown below) fires, but that meant rewriting it all, which still made it unresponsive.

            Traceback:

            ...

            ANSWER

            Answered 2021-Dec-15 at 16:29

            I found that answer to this problem. If you look at the documentation of Django. It has been Django from 2.0 to 3.0.

            Do formerly it was :

            Source https://stackoverflow.com/questions/70364068

            QUESTION

            error when adding lm to datetime data in ggplot2
            Asked 2022-Jan-27 at 20:42

            When Trying to plot the linear regression of the data the output returns a horizontal line with a warning that the fit is rank-deficient. Can someone please explain how to display a proper linear regression?

            Sample Data

            ...

            ANSWER

            Answered 2022-Jan-27 at 20:42

            This isn't a ggplot problem, but an lm problem:

            Source https://stackoverflow.com/questions/70885010

            QUESTION

            Python doesn't save a text file after closing the program
            Asked 2022-Jan-19 at 08:52

            I'm currently stuck on a saving issue. I want to create levels for a game and I'd like to build and modify them in a pygame program, then save them in the form of a string in a .txt file. For now, I managed to open an existing file named "level_n.txt", delete the old level I want to overwrite, and write the new level matrix in the text file. At this point, everything works perfectly. If I modify the text file, close it, re-open it in the same program run, and print the level, the changes were saved.

            Here is my problem : when I close my whole program and re-open it, the level isn't changed, and when I open the text file of the level manually, the changes aren't here either.

            So from my perspective, it looks like python creates a copy of the file in the RAM, overwrites it, uses it, but when it closes, the text file doesn't save.

            I've included the part where I write the text, and a level example just in case. (I've tried to write the text using w, w+, r+, a+, none of these solve my issue)

            ...

            ANSWER

            Answered 2022-Jan-18 at 23:21

            Use the full path to the file instead of just the name of the file and it should work fine For example:

            Source https://stackoverflow.com/questions/70763393

            QUESTION

            Duplicate word Python
            Asked 2022-Jan-14 at 22:44

            I want to make a program that detects duplicate words as in the following example:
            "At least one value must be entered
            entered in order to compute the average"
            We can see that "entered" is repeated and I want to find a way to detect this type of cases. \

            ...

            ANSWER

            Answered 2022-Jan-14 at 21:49
            text = 'i like donkey donkey'
            words = text.split(' ')
            
            for i in range(0, len(words)):  
                count = 1;  
                for j in range(i+1, len(words)):  
                    if(words[i] == (words[j])):  
                        count = count + 1;  
                        words[j] = '0';  
                          
                if(count > 1 and words[i] != '0'):  
                    print(words[i]);  
            
            # output -> donkey
            

            Source https://stackoverflow.com/questions/70716856

            QUESTION

            Compute concrete in PGF
            Asked 2022-Jan-05 at 07:46

            I have these two files that run perfectly fine within GF shell

            My GF code

            Test.gf

            ...

            ANSWER

            Answered 2022-Jan-05 at 07:46

            I will first give you three alternatives how to make the grammar work. Then I will explain the rest of the mysteries: why cc works with your initial approach, but parsing/linearisation doesn't, and also how to actually use cc from Python (just not with the PGF library).

            1. Fixing your grammar (a) Large lexicon, application grammar as a layer on top of RGL

            In your example, you are opening DictEng, so I assume that you would like your application to have a large lexicon.

            If you want to be able to parse with a large lexicon, it needs to be a part of the abstract syntax of your grammar. The first mistake is the fact that you're opening DictEng as a resource, instead of extending. (See tutorial to refresh your memory.)

            So if you want your abstract syntax to contain a lexicon entry called dog_N, which you can give as an argument to the function MySentence, you will need to modify your grammar as follows.

            Abstract:

            Source https://stackoverflow.com/questions/70587676

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

            Vulnerabilities

            No vulnerabilities reported

            Install Lin

            You can download it from GitHub.

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            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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