Kotlin filtering nested lists with other list-CodePudding

I have a list like this:

val list = List<FlightRecommendationQuery>

in that:

data class FlightRecommendationQuery(
 val segments: List<Segment>
)

data class Segment(
 val stops: Int
)

I have another list with size same as segments size in FlightRecommendationQuery called filter:

val filter = List<Filter>

data class Filter(
 val stops: Int
)

I want to filter 'list' when filter's stops are equal to each segments in list. Here, size of list 'filter' and size of 'segments' of FlightRecommendationQuery are same.

CodePudding user response：

I'm not sure if there is a more efficient way to do it but I think this would do it:

val filteredList = list.filter { it.segments.map { it.stops }.equals(filter.map { it.stops }) }

CodePudding user response：

Here's a Kotlin solution if I understood your question correctly

fun main() {

    // DATA SETUP 
    
    val s1 = Segment(1)
    val s2 = Segment(2)
    val s3 = Segment(3)

    val segmentList = listOf(s1,s2,s3)
    val query = FlightRecommendationQuery(segmentList)

    val f1 = Filter(1)
    val f2 = Filter(9999)
    val f3 = Filter(3)

    val filterList = listOf(f1,f2,f3)
    // END OF DATA SETUP 
    
    // Here we do the filtration
    val filteredSegments = mutableListOf<Segment>()
    for(segmentIndex in query.segments.indices) {
        // We use indices concept from Kotlin to get the items in the same 
        // position from both lists
        val segment = query.segments[segmentIndex]
        val filter = filterList[segmentIndex]

        if(segment.stops == filter.stops) filteredSegments.add(segment)
    }

    //Filtered segments now contains the necessary data
    //Will print [Segment(stops=1), Segment(stops=3)] in this case
    //Use this list of segments to create a new FlightRecommendationQuery
    println(filteredSegments)
}

CodePudding user response：

You can do this by zipping segments and filter and checking if all of them are equal:

list.filter { query ->
    (query.segments zip filter)
        .all { it.first.stops == it.second.stops }
}

CodePudding user response：

Another way:

list.filter { query ->
    filter.indices.all { filter[it].stops == query.segments[it].stops }
}

all returns true if all the elements of list match the given predicate