is there a better way to achieve this? I have a l...

# getting-started

[leosan]

is there a better way to achieve this? I have a list and I want to compose it into 4 different lists

with(screens) {
            filter { it.showProgress }.mapTo(hasProgressScreens) { it.viewId }
            filter { it.showSkip }.mapTo(hasSkipScreens) { it.viewId }
            filter { it.showToolbar }.mapTo(hasToolbar) { it.viewId }
            filter { it.showBack }.mapTo(hasBackButton) { it.viewId }
        }

[david-wg2]

groupby?

[david-wg2]

🤔

[leosan]

not sure if I’m getting the groupBy syntax right

[leosan]

.groupBy({it.showProgress} + ???)

[david-wg2]

would have to give it an enum or something instead

[arekolek]

but groupBy would produce lists that do not intersect, right? and using four filters would produce lists that may intersect

[arekolek]

not sure which one you want

[leosan]

hmm the list may intersect, that’s right

[david-wg2]

ah, then i think you have to do what you're currently doing

[diesieben07]

You could avoid the intermediary lists with something like this:

mapNotNullTo(hasProgressScreens) { it.takeIf { it.showProgress }?.viewId }

[leosan]

I know this case the performance doesn’t matter too much because it has only 30 elements, but is there a way to achieve this with a sequence? Similar to doing with a loop?

[leosan]

now out of just curiosity

[leosan]

something like

screens.forEach {  screen ->
           if(screen.showProgress) hasProgressScreens.add(screen.viewId)
           if(screen.showSkip) hasSkipScreens.add(screen.viewId)
           if(screen.showToolbar) hasToolbar.add(screen.viewId)
           if(screen.showBack) hasBackButton.add(screen.viewId)
        }

but in a sequence/functional way

[Horv]

If you create a wrapper class for the lists you can always do it as a reduce operation, not sure if it's more readable though