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Kotlin单分配过滤器和映射

filter kotlin

-1

martinseal1987 · 技术社区 · 4 年前

我想把它变成一个单一的赋值属性

val movesWithMetaDataList: MutableList<PokemonMoveWithMetaData> = arrayListOf()
for (move in pokemonWithMovesAndMetaData.moves) {
    for (metaData in pokemonWithMovesAndMetaData.pokemonMoveMetaData) {
        if (move.name == metaData.moveName) {
            movesWithMetaDataList.add(
                PokemonMoveWithMetaData(
                    metaData, move
                )
            )
        }
    }
}
setPokemonMoves(movesWithMetaDataList.separateByGeneration())

我试过很多东西,我想我需要一个过滤器,然后一个地图,但我很难找到它,名单可以不同的大小

val movesWithMetaDataList = pokemonWithMovesAndMetaData.moves.flatMap { move ->
    pokemonWithMovesAndMetaData.pokemonMoveMetaData.map { metaData -> 
        if (metaData.moveName == move.name){
            PokemonMoveWithMetaData(
                metaData, move
            )
        }
    }
}

还有这个

val movesWithMetaDataList = pokemonWithMovesAndMetaData.moves.flatMap { move ->
    pokemonWithMovesAndMetaData.pokemonMoveMetaData.takeWhile {
        pokemonMoveMetaData -> move.name == pokemonMoveMetaData.moveName
    }.map { meta ->
        PokemonMoveWithMetaData(
            meta , move
        )
    }
}

以及其他

1 回复 | 直到 4 年前

marstran 4 年前

我想应该是这样的:

val movesWithMetaDataList = pokemonWithMovesAndMetaData.moves
    .flatMap { move -> 
        pokemonWithMovesAndMetaData.pokemonMoveMetaData
            .filter { it.moveName == move.name }
            .map { PokemonMoveWithMetaData(it, move) }
    }

filter ,然后我使用 map 创造 PokemonMoveWithMetaData 从那些里面。这个 flatMap 函数将为每个move/元数据对创建的列表展平为单个结果列表。