mle | flexible terminal-based text editor | Editor library

 by   adsr C Version: v1.7.2 License: Apache-2.0

kandi X-RAY | mle Summary

kandi X-RAY | mle Summary

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

mle is a small, flexible, terminal-based text editor written in C. Runs on Linux, Windows (cygwin or WSL), FreeBSD, and macOS.
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              mle has a low active ecosystem.
              It has 706 star(s) with 43 fork(s). There are 21 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 4 open issues and 57 have been closed. On average issues are closed in 143 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of mle is v1.7.2

            kandi-Quality Quality

              mle has no bugs reported.

            kandi-Security Security

              mle has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

            kandi-License License

              mle 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

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

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            mle Key Features

            No Key Features are available at this moment for mle.

            mle Examples and Code Snippets

            No Code Snippets are available at this moment for mle.

            Community Discussions

            QUESTION

            MLE of a distribution in R: fitdistrplus (SGT distribution), how do I do it?
            Asked 2021-May-16 at 16:12

            For return data I am doing research about the importance of skewness and kurtosis for the cVaR calculation. We are comparing some distributions first, by estimating the parameters of the distribution using fitdist() in R using package "fitdistrplus". However, we want to do this for a various number of distributions (see picture: SGT, GT, SGED, GED, t, norm).

            Below is a sample code for SGT, where there is a problem: it introduced NaN's for the standard errors for parameters p and q. I also don't really know how to choose the starting values exactly.

            ...

            ANSWER

            Answered 2021-May-16 at 16:12

            The answer was using package sgt

            Source https://stackoverflow.com/questions/67457891

            QUESTION

            Error in solve.default(oout$hessian) : Lapack routine dgesv: system is exactly singular: U[1,1] = 0
            Asked 2021-May-10 at 22:48

            I am using the maximum likelihood method to estimate a set of parameters. Now I am going to use mle function from the stats4 package in R to make a profile likelihood for one of the parameters. To do that, I need to fix one of the parameters when I call the mle function. Here is the code:

            ...

            ANSWER

            Answered 2021-May-07 at 22:25

            tl;dr I'm not sure your objective function makes sense, my guess is that you have a typo. (Furthermore, if your objective function is working with mle, you don't need to set fixed explicitly: the profile method will automatically compute likelihood profiles for you ...)

            Let's start with the full model, and let's use optim() rather than stats4::mle() (I know you want to get back to mle so that you can do likelihood profiling, but it's a little bit easier to debug optim() problems since there is one less layer of code to dig through.)

            Since optim() wants an objective function that takes a vector rather than a list of arguments, write a wrapper (we could also use do.call(fr, as.list(p))):

            Source https://stackoverflow.com/questions/67360202

            QUESTION

            Discrepancy between log-likelihood returned by logLik for normal linear models and "standard" manual calculation
            Asked 2021-Apr-28 at 11:14

            The log-likelihood returned by the lm method for the logLik function does not seem to match that calculated by hand.

            To demonstrate, below I fit a standard linear model with only an intercept to a simple dataset with three points. I then extract the log-likelihood using logLik and calculate it manually by summing the logs of the pdfs evaluated at the mle.

            ...

            ANSWER

            Answered 2021-Apr-28 at 11:14

            logLik evaluates the log-likelihood at the maximum-likelihood estimates of the parameters. The maximum-likelihood estimates of the coefficients are the same as the least-squares estimates, but the denominator of the maximum-likelihood estimate of the variance is n, while summary(mod_lm)$sigma is the squared-root of the unbiased estimate of the variance, whose denominator is the degrees of freedom, here n-1. So you get the same results if you do:

            Source https://stackoverflow.com/questions/67298986

            QUESTION

            issue running valued exponential random graph model (ergm) with block diagonal constraint
            Asked 2021-Mar-24 at 02:10

            I am trying to run an exponential random graph model (ergm) on a weighted network (network_ex). This network shows the interactions between individuals in four different groups. Interactions between groups cannot occur so a blockdiagonal constraint needs to be included in the model. However when I include the blockdiagonal constraint the ergm function (that runs otherwise) returns an error (see below) and my R session either runs indefinitely or abort directly.

            Is there a way to run an ergm based on a weighted block diagonal matrix?

            ...

            ANSWER

            Answered 2021-Mar-24 at 02:10

            That does appear to be a bug. I've fixed it in the latest development version. These can be obtained by going to https://github.com/statnet/ergm and https://github.com/statnet/ergm.count and either using install_github() or downloading the binaries found in the README. Note, also that, the blockdiag() constraint has been provisionally moved to https://github.com/statnet/tergm .

            Source https://stackoverflow.com/questions/66750371

            QUESTION

            R - how to print multiple MLE output as dataframe
            Asked 2021-Mar-03 at 09:46

            I am quite new in R and I am stack in getting a table of output values from multiple MLE running for different sample sizes. Here is my code:

            ...

            ANSWER

            Answered 2021-Mar-03 at 09:46

            How about using a list of data.frames and then rbindlist them (a data.table function)

            Source https://stackoverflow.com/questions/66454143

            QUESTION

            Using mle2 function
            Asked 2021-Jan-31 at 07:14

            I would like to find the MLE for parameters epsilon and mu in such a model:

            $$X \sim \frac{1}{mu1}e^{-x/mu1}+\frac{1}{mu2}e^[-x/mu2}$$

            ...

            ANSWER

            Answered 2021-Jan-27 at 23:43

            There are a bunch of issues here. The primary one is that your likelihood expression is wrong (you can't log the components separately and then add them, you have to add the components and then take the log). Your bounds are also funny: the mixture probability should be [0,1] and the means should be [0, Inf].

            The other problem you have is that with the current simulation design (n=20, prob=0.01), you have a high probability of getting no points in the first mixture component (the probability of a point being in the second component is 1-0.01=0.99, so the probability that all of the points are in the second component is 0.99^20 = 82%). In this case the MLE will be degenerate (i.e., you're trying to fit a two-component mixture to a data set that essentially only has one component); in this case any of these solutions will give equivalent likelihoods:

            • prob=0, mu2=mean of the data, mu1=anything
            • prob=1, mu1=mean of the data, mu2=anything
            • mu1=mu2=mean of the data, prob=anything

            With all these solutions, where you end up will depend very sensitively on starting conditions and optimization algorithm.

            For this problem I would encourage you to use the built-in dmixexp2 function from the Renext package (which correctly implements the log-likelihood as log(p*Prob(X|exp1) + (1-p)*Prob(X|exp2))) and the formula interface to mle2:

            Source https://stackoverflow.com/questions/65874044

            QUESTION

            Error in fitdist function: the function mle failed to estimate the parameters, with the error code 1 (Rayleigh distribution)
            Asked 2021-Jan-27 at 12:29

            I am trying to fit my data to Rayleigh distribution by using the fitdist function from the fitdistrplus package.

            ...

            ANSWER

            Answered 2021-Jan-27 at 12:13

            There are two problems. (1) the Rayleigh distribution does not seem to be a good fit to the data (see plot output below) and (2) you need a better starting value. Since sigma is proportional to the mean for the Rayleigh distribution (see wikipedia) try that:

            Source https://stackoverflow.com/questions/65918603

            QUESTION

            Pareto distribution: R vs Python - different results
            Asked 2021-Jan-18 at 23:25

            I'm trying to replicate R's fitdist() results (reference, cannot modify R code) in Python using scipy.stats. The results are totally different. Does anyone know why? How can I replicate R's results in Python?

            ...

            ANSWER

            Answered 2021-Jan-18 at 23:25

            The discrepancy is mainly due to the differing pdfs.

            Python

            In python st.pareto.fit() uses a Pareto distribution defined via this pdf:

            Source https://stackoverflow.com/questions/65719777

            QUESTION

            Weibull: R vs Python - slightly different results
            Asked 2021-Jan-14 at 18:03

            I'm trying to replicate R's fitdist() results (reference, cannot modify R code) in Python using scipy.stats. The results are quite close but still different (difference is at not acceptable level). Does anybody know why the results are different? How can I reduce the difference between the results?

            scipy_stats.weibull_min definition (https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.weibull_min.html) seems to be the same as R's weibull (https://stat.ethz.ch/R-manual/R-devel/library/stats/html/Weibull.html.

            Data example:

            ...

            ANSWER

            Answered 2021-Jan-14 at 17:01

            The difference appears to be the result of the default relative tolerances used by the optimizers (and normal floating point imprecision). If you tighten the tolerance in the R calculation, the result is closer to the SciPy result:

            Source https://stackoverflow.com/questions/65716181

            QUESTION

            Data frame inside for loop with function for a simulation study wont work
            Asked 2020-Nov-30 at 14:46

            I am doing looping for my simulation study. But it did not give a nice output. I want to be able to call out again the output. This is the function inside the looping.

            ...

            ANSWER

            Answered 2020-Nov-30 at 14:46

            Since you want to save the data into a data frame I will assume you want jj to contain the result. As your code stands you probably want to do something like:

            Source https://stackoverflow.com/questions/65075407

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

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