mgimond.github.io | mgimond.github.io landing page | Continous Integration library

by mgimond CSS Version: Current License: No License

X-Ray Key Features Code Snippets Community Discussions(3)Vulnerabilities Install Support

kandi X-RAY | mgimond.github.io Summary

mgimond.github.io is a CSS library typically used in Devops, Continous Integration, Jekyll applications. mgimond.github.io has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

mgimond.github.io landing page

Support

Quality

Security

License

Reuse

Support

mgimond.github.io has a low active ecosystem.

It has 14 star(s) with 5 fork(s). There are 5 watchers for this library.

It had no major release in the last 6 months.

There are 1 open issues and 1 have been closed. On average issues are closed in 118 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of mgimond.github.io is current.

Quality

mgimond.github.io has no bugs reported.

Security

mgimond.github.io has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

mgimond.github.io does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

mgimond.github.io releases are not available. You will need to build from source code and install.

Top functions reviewed by kandi - BETA

kandi's functional review helps you automatically verify the functionalities of the libraries and avoid rework.
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of mgimond.github.io

Get all kandi verified functions for this library.

mgimond.github.io Key Features

No Key Features are available at this moment for mgimond.github.io.

mgimond.github.io Examples and Code Snippets

No Code Snippets are available at this moment for mgimond.github.io.

Community Discussions

Trending Discussions on mgimond.github.io

Why does as(spdf, "owin") throw an error 'spatstat.options' is not an exported object from 'namespace:spatstat'

Lay a region over all USA map in R

R - Chi Square Independence Test with same probabilites for classes

QUESTION

Why does as(spdf, "owin") throw an error 'spatstat.options' is not an exported object from 'namespace:spatstat'

Asked 2021-May-11 at 00:12

From this book here

...

ANSWER

Answered 2021-Mar-17 at 15:46

Although RStudio told me, all packages were up to date, the problem continued to exist. The solution was a full update of R and all packages. The process on Windows:

Run installr::update()from Rgui.exe (in \R\R-4.0.4\bin\x64).
Update Windows environment variable R_LIBSthe the new \R\R-4.0.4\library.
update Rprofile.site in \R\R-4.0.4\etc and make sure there is only one .libPaths(). (There has to be a line .libPaths("C:/R/R-4.0.4/library") or just add it.)
Check if there are pending package updates in RStudio

Source https://stackoverflow.com/questions/66674391

QUESTION

Lay a region over all USA map in R

Asked 2019-Apr-18 at 23:18

I want to plot a map of the following data, dt_plot, which has 2 unique counties in it.

...

ANSWER

Answered 2019-Apr-18 at 23:18

cnty <- map_data("county") # Load the county data from the maps package 
cnty2<- cnty %>% 
         mutate(polyname = paste(region, subregion,sep=",")) %>% 
         left_join(county.fips, by="polyname")

Source https://stackoverflow.com/questions/55717381

QUESTION

R - Chi Square Independence Test with same probabilites for classes

Asked 2018-Dec-07 at 09:29

Edit:

As I found later somewhere else, the Chi² test is probably not appropriate for my data here or rather does not test what I want to find out. Therefore, I conducted a generalised linear model (glm) with a Poisson distribution on my data which worked out quite nicely. So bear this in mind...

After consulting various websites on this problem (like this, this or this) and of course the official documentation of the chisq.test function, I still cannot figure out a solution to my problem.

What I want:

I want to conduct a Chi² Test of Independence on my data via the chisq.test function in R. My data is composed of 4 epiphyte species found on 4 host tree species (that means: plants of 4 species growing ON these 4 tree species). Now, I want to find out if the epiphytes are equally distributed among those trees or if maybe one tree species tends to host more epiphyte individuals as the others. The standard Chi² test I can conduct quite easily (see below). But this would then as well test if epiphyte species were equally distributed, which I don't want to be tested. So, how can I submit different probabilities for my contingency table in the cisq.test function? Namely, I want the expected matrix to be according to the number of epiphyte individuals per species while expecting them to be equally distributed among the tree species. This sounds complicated, so just have a look at my example data:

Example Data:

(I edited the data format as suggested by @paoloeusebi)

Observed data: ...

ANSWER

Answered 2018-Dec-05 at 11:06

It is better to input data as matrix instead as data frame.

Source https://stackoverflow.com/questions/53630265

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

Vulnerabilities

No vulnerabilities reported

Install mgimond.github.io

You can download it from GitHub.

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 .

Find more information at: