STAR | Official code repository for the paper : STAR - A Sparse | 3D Animation library

Given your navigation graph, there is no way to associate fragmentA2 with your menu item fragmentA, so fragmentA is not selected when you return to fragmentA2. As per this issue:

NavigationUI has always used the current destination and what graph it is part of as the source of truth for what tab should be selected.

This can be seen by calling navigate() to go to your SecondFragment - even though you haven't used the bottom nav button, the selected tab was changed because the current destination has changed to R.id.frag_second.

So when you navigate() to R.id.frag_hint via your button in HomeFragment, NavigationUI receives a callback that the current destination has changed to R.id.frag_hint. It looks at that NavDestination and notes that there's no menu item that matches R.id.frag_hint. It then looks at the destination's parent graph - your R.id.sample element. There's no menu item that matches that ID either, so NavigationUI can't associated that destination with any menu item and therefore simply does nothing. That is true on all versions of Navigation.

So what is different when you tap on a bottom navigation item? Well, nothing different from a NavigationUI perspective in fact: the exact same code runs and the current destination and what graph it is part of is the source of truth for what tab should be selected. In the Navigation 2.3.5, there was no state saved for each tab, so it only 'worked' because selecting a tab forced the ID of the current destination to match the destination of the menu item you just tapped.

So what you're seeing in your sample app is that there's no link between R.id.frag_hint and any menu item, which means NavigationUI does nothing. If you want to link R.id.frag_hint to your Home tab, then that's exactly what a nested navigation graph can be used for.

I.e., your navigation graph should instead look like:

Series.value_counts + Series.unstack to convert to dataframe

Please find one possible solution using terra and sf libraries.

The idea is to convert the SpatRaster r into a SpatVector and then into an sf object in order to take advantage of the sf::st_join() function using the largest = TRUE argument. The rest of the code then consists of simply converting the sf object back into a SpatVector and then a SpatRaster using the terra::rasterize() function.

So, please find below a reprex that details the procedure.

Reprex

• Code

Just confirmed that the article linked in the comments below the question contains a valid clue on the issue.

Corrected code would be:

very weird definition of star according to that circle and pie are also stars ...

First simple and O(n) possibility I can think of is to render visibility map:

1. compute BBOX of the shape

2. create 2D map of the BBOX and clear it with zero

   so map 2D array (texture) to the BBOX of some resolution xs*ys

3. for each convex vertex increment visibility map

   simply by rendering "infinite" triangle/quad onto the map

   You can use winding rule to chose if vertex is convex or concave by simply checking the sign of z coordinate of the adjacent edges cross product against the winding rule of your shape.

4. scan the 2D map for cells containing number of convex vertexes

   all the cells/pixels containing number of convex vertexes are your possible Z so if any found your shape is a "star".

This is O(n*xs*ys) where n is number of (convex) vertexes and xs*ys is resolution of the visibility map. Note if your resolution is too low due to inaccuracies you might produce false negatives/positives ... if (max) resolution of the map is constant then the complexity will turn to O(n).

The rendering can be done simply for example with OpenGL and STENCIL buffer which directly has operation to increment STENCIL pixel however that one will limit the n to 255 as STENCIL is only 8 bit these days (after changes in OpenGL)... However you can workaround this by seting the BBOX to 1 and clear the exterior of the triangle/quad instead of incrementing its interrior. then the pixels holding 1 are your Z this might be used with any rendering engine no need for STENCIL

ANSWER: Underlying problem determined to exist in IDE's compilation/build/execution routine.

Reinstall and update IDE, adoption of non-EOL JDK.

Also, I did not import existing IDE settings.

When running this code from a terminal you will notice that your missing * on some runs is not missing at all but printed after your program is finished.

For better understanding you might print the PID's instead of stars and an additional line when the process is about to finish:

I recently answered a similar question, but there does seem to be a new issue introduced in 5.1.0 because of this change mentioned on the Bootstrap blog...

"Updated .bg- and .text- utilities Our new RGB values are built to help us make better use of CSS variables across the entire project. To start, our background-color and color utilities have been updated to use these new RGB values for real-time customization without recompiling Sass and on-the-fly transparency for any background or text color."

Currently in 5.1.0 you'd need to rebuild all the bg-* and text-* classes like this...

I don't know of any functions in base R that do circles for you, but you can concoct them manually.