rpub | an ePub generation library in Ruby | Media library

 by   avdgaag Ruby Version: Current License: MIT

kandi X-RAY | rpub Summary

kandi X-RAY | rpub Summary

rpub is a Ruby library typically used in Media applications. rpub has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Rpub is a command-line tool that generates a collection of plain text input files into an eBook in ePub format. It provides several related functions to make working with ePub files a little easier:.
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            kandi-support Support

              rpub has a low active ecosystem.
              It has 110 star(s) with 22 fork(s). There are 4 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 4 open issues and 5 have been closed. On average issues are closed in 4 days. There are 2 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of rpub is current.

            kandi-Quality Quality

              rpub has 0 bugs and 0 code smells.

            kandi-Security Security

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

            kandi-License License

              rpub 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

              rpub releases are not available. You will need to build from source code and install.
              Installation instructions, examples and code snippets are available.

            Top functions reviewed by kandi - BETA

            kandi has reviewed rpub and discovered the below as its top functions. This is intended to give you an instant insight into rpub implemented functionality, and help decide if they suit your requirements.
            • Create a new manifest with the given manifest
            • Write content to file
            • Removes a filename from the file .
            • Create an outline of the document
            • Creates a new Context instance
            • Get the title of the title
            • Creates a chapter .
            • Create a new zip file .
            • Store the filename in the archive .
            • Get the content from ERB file
            Get all kandi verified functions for this library.

            rpub Key Features

            No Key Features are available at this moment for rpub.

            rpub Examples and Code Snippets

            No Code Snippets are available at this moment for rpub.

            Community Discussions

            QUESTION

            Titles in plotly subplots using grid
            Asked 2022-Mar-18 at 10:02

            I'm trying to apply the solution for having titles in a plotly subplot to a plotly grid using this Subplots Using Grid example:

            ...

            ANSWER

            Answered 2022-Mar-18 at 10:02

            You can use this code:

            Source https://stackoverflow.com/questions/71517783

            QUESTION

            Why does the output of of a linear mixed model using lme4 show one level of a factor but not another?
            Asked 2022-Feb-10 at 19:43

            I am using the lme4 package and running a linear mixed model but I am confused but the output and expect that I am encountering an error even though I do not get an error message. The basic issue is when I fit a model like lmer(Values ~ stimuli + timeperiod + scale(poly(distance.code,3,raw=FALSE))*habitat + wind.speed + (1|location.code), data=df, REML=FALSE) and then look at the results using something like summary I see all the model fixed (and random) effects as I would expect however the habitat effect is always displayed as habitatForest. Like this:

            ...

            ANSWER

            Answered 2022-Feb-10 at 19:43

            note: although your question is about the lmer() function, this answer also applies to lm() and other R functions that fit linear models.

            The way that coefficient estimates from linear models in R are presented can be confusing. To understand what's going on, you need to understand how R fits linear models when the predictor is a factor variable.

            Coefficients on factor variables in R linear models

            Before we look at factor variables, let's look at the more straightforward situation where the predictor is continuous. In your example dataset, one of the predictors is wind speed (continuous variable). The estimated coefficient is about -0.35. It's easy to interpret this: averaged across the other predictors, for every increase of 1 km/h in wind speed, your response value is predicted to decrease by 0.35.

            But what about if the predictor is a factor? A categorical variable cannot increase or decrease by 1. Instead it can take several discrete values. So what the lmer() or lm() function does by default is automatically code your factor variable as a set of so-called "dummy variables." Dummy variables are binary (they can take values of 0 or 1). If the factor variable has n levels, you need n-1 dummy variables to encode it. The reference level or control group acts like an intercept.

            In the case of your habitat variable, there are only 2 levels so you have only 1 dummy variable which will be 0 if habitat is not Forest and 1 if it is Forest. Now we can interpret the coefficient estimate of -68.8: the average value of your response is expected to be 68.8 less in forest habitat relative to the reference level of grassland habitat. You don't need a second dummy variable for grassland because you only need to estimate the one coefficient to compare the two habitats.

            If you had a third habitat, let's say wetland, there would be a second dummy variable that would be 0 if not wetland and 1 if wetland. The coefficient estimate there would be the expected difference between the value of the response variable in wetland habitat compared to grassland habitat. Grassland will be the reference level for all the coefficients.

            Default setting of reference level

            Now to directly address your question of why habitatForest is the coefficient name.

            Because by default no reference level or control group is specified, the first one in the factor level ordering becomes the reference level to which all other levels are compared. Then the coefficients are named by appending the variable's name to the name of the level being compared to the reference level. Your factor is ordered with grassland first and forest second. So the coefficient is the effect of the habitat being forest habitat, compared to the reference level, which is grassland in this case. If you switched the habitat factor level ordering, Forest would be the reference level and you would get habitatGrassland as the coefficient instead. (Note that default factor level ordering is alphabetical, so without specifically ordering the factor levels as you seem to have done, Forest would be the reference level by default).

            Incidentally, the two links you give in your question (guides to mixed models from Phillip Alday and Tufts) do in fact have the same kind of output as you are getting. For example in Alday's tutorial, the factor recipe has 3 levels: A, B, and C. There are two coefficients in the fixed effects summary, recipeB and recipeC, just as you would expect from dummy coding using A as reference level. You may be confusing the fixed effects summary with the ANOVA table presented elsewhere in his post. The ANOVA table does only have a single line for recipe which gives you the ratio of variance due to recipe (across all its levels) and the total variance. So that would only be one ratio regardless of how many levels recipe has.

            Further reading

            This is not the place for a full discussion of contrast coding in linear models in R. The dummy coding (which you may also see called one-hot encoding) I described here is just one way to do it. These resources may be helpful:

            Source https://stackoverflow.com/questions/71055840

            QUESTION

            How can I extract bigrams from text without removing the hash symbol?
            Asked 2022-Jan-09 at 06:43

            I am using the following function (based on https://rpubs.com/sprishi/twitterIBM) to extract bigrams from text. However, I want to keep the hash symbol for analysis purposes. The function to clean text works fine, but the unnest tokens function removes special characters. Is there any way to run unnest tokens without removing special characters?

            ...

            ANSWER

            Answered 2022-Jan-09 at 06:43

            Here is a solution that involving create a custom n-grams function

            Setup

            Source https://stackoverflow.com/questions/70634673

            QUESTION

            How to upload file from a browser dialog using RSelenium
            Asked 2021-Nov-17 at 15:27

            I have the following URL which I want to post the form:

            ...

            ANSWER

            Answered 2021-Nov-13 at 14:16

            The RSelenium package enables to remote control the comon web browsers from R. The following code opens a remote browser session in mozilla firefox(chrome should work as well) and works the login screen - with the correct usr and pw the next page will open. As I do not have access to the closed part I can not try nor debug my code after the login screen click so I just showed one upload:

            Source https://stackoverflow.com/questions/69953031

            QUESTION

            bnlearn Error: Wrong number of conditional probability distributions
            Asked 2021-Oct-10 at 19:29

            I am learning to work with bnlearn and I keep running into the following error in the last line of my code below:

            Error in custom.fit(dag, cpt) : wrong number of conditional probability distributions

            What am I doing wrong?

            ...

            ANSWER

            Answered 2021-Oct-10 at 19:29

            You have several errors in your CPT definitions. Primarily, you need to make sure that:

            • the number of probabilities supplied are equal to the product of the number of states in the child and parent nodes,
            • that the number of dimensions of the matrix/array is equal to the number of parent nodes plus one, for the child node,
            • the child node should be given in the first dimension when the node dimension is greater than one.
            • the names given in the dimnames arguments (e.g. the names in dimnames=list(ThisName = ...)) should match the names that were defined in the DAG, in your case with modelstring and in my answer with model2network. (So my earlier suggestion of using dimnames=list(cptNBLW = ...) should be dimnames=list(nblw = ...) to match how node nblw was declared in the model string)

            You also did not add node f into your cpt list.

            Below is your code with comments where things have been changed. (I have commented out the offending lines and added ones straight after)

            Source https://stackoverflow.com/questions/69508863

            QUESTION

            R : Function is returning Unused Arguments
            Asked 2021-Jul-08 at 07:50

            I am working with R. I am learning about optimization and trying to follow the instructions from the following references: https://www.rdocumentation.org/packages/pso/versions/1.0.3/topics/psoptim and https://rpubs.com/Argaadya/intro-PSO

            For this example, I first generate some random data:

            ...

            ANSWER

            Answered 2021-Jul-08 at 07:50

            As pointed out in the comments, if you name arguments, you need to name them according to the function definition. Also, the lower and upper bounds are not expressions, but must be constants (i.e. numbers, not names such as random_1).

            Such constraints could be handled for instance through penalties: in the objective function, compute whether a variable is outside its range; if it is, subtract a penalty from the objective function (if you maximise) or add a penalty (if you minimise).

            There are optimisation methods in which you could handle such constraints directly when creating new solutions. One such method is Local Search. Here is a (rough) example, in which I assume that you want to maximise. I use the implementation in package NMOF (which I maintain). The algorithm expects minimisation, but that is no problem: just multiply the objective function value by -1. (Note that many optimisation algorithms expect minimisation models.)

            The key ingredient to Local Search is the neighbourhood function. It takes a given solution as input and produces a slightly changed solution (a neighbour solution). Local Search then takes a random walk through the space of possible solutions (with steps as defined in the neighbourhood function), accepting better solutions, but rejecting solutions that lead to worse objective function values.

            Source https://stackoverflow.com/questions/68293973

            QUESTION

            R: subscript is out of bounds
            Asked 2021-Jul-07 at 23:48

            I am working with R. I am trying to follow this tutorial over here on function optimization: https://rpubs.com/Argaadya/bayesian-optimization

            For this example, I first generate some random data:

            ...

            ANSWER

            Answered 2021-Jul-07 at 23:48

            There appear to be a few bugs in your code, e.g. I don't think your fitness function was returning data in the required format and some of your vectors were being used before they were defined.

            I made some changes so your code was more inline with the tutorial, and it seems to complete without error, but I can't say whether the outcome is "correct" or whether it will be suitable for your use-case:

            Source https://stackoverflow.com/questions/68293872

            QUESTION

            "tibble::enframe" code no longer runs, is there a way to replicate what it does?
            Asked 2021-Jun-30 at 03:51

            I'm running a code to format data, which goes into a rake weighting analysis. I started to run the lines but when I get to the parts "tibble::enframe" they don't run. What does tibble do in this code? The strange thing was it did run once upon a time in past but no longer does.

            the code

            ...

            ANSWER

            Answered 2021-Jun-30 at 00:36

            It appears that you have not loaded the library for select. After adding library(tidyverse) before your code, it worked on my machine.

            Source https://stackoverflow.com/questions/68187064

            QUESTION

            (R) Error in Xi - Xj : non-numeric argument to binary operator
            Asked 2021-Jun-04 at 21:55

            I am working with the R programming language. I am trying to recreate the graphs shown in this tutorial over here : https://www.rpubs.com/cboettig/greta-gp

            This tutorial shows how to make a special type of regression model for 2 variables. I am able to copy and paste the code from this tutorial and successfully make the desired graphs:

            ...

            ANSWER

            Answered 2021-Jun-04 at 21:55

            I think I got the problem. First of all below is the way by which we can reproduce the error & the way you have proceed :

            Source https://stackoverflow.com/questions/67764577

            QUESTION

            Vertical 95% Confidence Interval plot 2 groups comparison
            Asked 2021-May-17 at 13:59

            So my end goal is to have a plot with multiple 95% confidence intervals plotted vertically in 2 groups like this example:

            I have found this code: https://rpubs.com/nati147/703463

            But how can I add groupwise comparison in the plot?

            Write a function ‘CI_95’ that takes input a vector of sample values, and outputs the 95% confidence interval for this sample. You can use the ‘margin_error_95’ function.

            ...

            ANSWER

            Answered 2021-May-17 at 13:59

            Here is a solution, although it might require some further refinement based on your preferences. I kept the general structure of your plot_CI_95 function, but added a loop over the different groups. This meant that the mu and sig variables now must have multiple values, one for each group if you are going to show group-wise differences. There are also some colors and other graphical parameters. The result is shown below.

            To avoid the intervals for the two groups overlapping, there are a few parameters to tweak. 1) figure height in png (increasing value makes more space between groups, 2) offset parameter (can increase up to about 0.3), or 3) lwd in the segments functions (smaller values mean thinner lines). Using png or similar function to directly save the figure will allow fine tuning of the desired look.

            Source https://stackoverflow.com/questions/67504569

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            Install rpub

            Rpub is distributed as a Ruby gem, which should be installed on most Macs and Linux systems. Once you have ensured you have a working installation of Ruby and Ruby gems, install the gem as follows from the command line:.

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