我有export default function UserAvatar(props) {
const {
size,
src,
name,
style // <- Need to pull from props, ListItem put it here with React.cloneElement
} = props;
return (
<Avatar
style={style} // <- Pass it onto the actual material-ui Avatar
size={size || 40}
src={src || null}
>
{src ? null : name.charAt(0)}
</Avatar>
);
}
的列表,想要在右转或左转时找到下一个方向。这是我的工作代码:
directions
enum class Turn { R, L }
enum class Direction { N, E, S, W }
val directionsInRightTurnOrder = listOf(Direction.N, Direction.E, Direction.S, Direction.W)
private fun calculateNextHeading(heading: Direction, turn: Turn): Direction {
val currentIndex = directionsInRightTurnOrder.indexOf(heading)
var nextIndex = currentIndex + if (turn == Turn.R) 1 else -1
if (nextIndex >= directionsInRightTurnOrder.size)
nextIndex = directionsInRightTurnOrder.size - nextIndex
if (nextIndex < 0)
nextIndex += directionsInRightTurnOrder.size
return directionsInRightTurnOrder.get(nextIndex)
}
列表并无限地(懒惰地)循环播放,那么这将更加简单易读。在Clojure中,我可以使用clojure.core/cycle:directionsInRightTurnOrder
另一件有用的事情是,如果我可以使用负索引在列表中查找,比如在Ruby或Python中:
(take 5 (cycle ["a" "b"]))
# ("a" "b" "a" "b" "a")
吗?答案 0 :(得分:7)
此处cycle
:
fun <T : Any> cycle(vararg xs: T): Sequence<T> {
var i = 0
return generateSequence { xs[i++ % xs.size] }
}
cycle("a", "b").take(5).toList() // ["a", "b", "a", "b", "a"]
以下是如何实施转弯申请:
enum class Turn(val step: Int) { L(-1), R(1) }
enum class Direction {
N, E, S, W;
fun turned(turn: Turn): Direction {
val mod: (Int, Int) -> Int = { n, d -> ((n % d) + d) % d }
return values()[mod(values().indexOf(this) + turn.step, values().size)]
}
}
听起来modulo
就是您正在寻找的 - 负面索引环绕。无法在Kotlin的stdlib中找到它,所以我带了自己的。
Direction.N
.turned(Turn.R) // E
.turned(Turn.R) // S
.turned(Turn.R) // W
.turned(Turn.R) // N
.turned(Turn.L) // W
Enum#values()
和Enum#valueOf(_)
可让您以编程方式访问枚举成员。
答案 1 :(得分:3)
自定义序列,无限期地重复给定的序列或列表,可以很容易地写成flatten
:
fun <T> Sequence<T>.repeatIndefinitely(): Sequence<T> =
generateSequence(this) { this }.flatten()
fun <T> List<T>.repeatIndefinitely(): Sequence<T> =
this.asSequence().repeatIndefinitely()
答案 2 :(得分:3)
您可以通过生成重复返回列表/集合然后展平它的序列来遍历Kotlin中的列表/集合。 e.g:
generateSequence { listOf("a", "b") }.flatten().take(5).toList()
// [a, b, a, b, a]
您可以定义自己的模数函数,用于将负数和正数强制转换为有效索引以访问列表中的元素(另请参阅Google Guava的IntMath.mod(int, int)
):
infix fun Int.modulo(modulus: Int): Int {
if (modulus <= 0) throw ArithmeticException("modulus $modulus must be > 0")
val remainder = this % modulus
return if (remainder >= 0) remainder else remainder + modulus
}
val list = listOf("a", "b", "c", "d")
list[-1 modulo list.size] // last element
list[-2 modulo list.size] // second to last element
list[+9 modulo list.size] // second element
list[-12 modulo list.size] // first element
答案 3 :(得分:0)
解释kotlin Slack上的讨论:
使用List#modulo
会使这更简单,但不如cycle
更优雅,因为仍然需要处理负面索引。
实施循环列表的一个选项是Sequence
。但是,需要使用Sequence
编写或生成自定义generateSequence
。我们认为这种情况有点过头了。
最终我去了:
Direction
了解next
和previous
:enum class Direction {
N, E, S, W;
private val order by lazy { listOf(N, E, S, W) }
fun add(turns: Int): Direction {
val currentIndex = order.indexOf(this)
var nextIndex = (currentIndex + turns) % order.size
return order.possiblyNegativeLookup(nextIndex)
}
fun subtract(turns: Int) = add(-1 * turns)
fun next(): Direction = add(1)
fun previous(): Direction = subtract(1)
}
List
:possiblyNegativeLookup
醇>
fun <E> List<E>.possiblyNegativeLookup(i: Int): E {
return if (i < 0) this[this.size + i] else this[i]
}
所以最终的代码变成了:
val nextHeading = if (move.turn == Turn.R) heading.next() else heading.previous()