我如何可以选择使用Scala和Spark阵列非连续的子集的元素呢? [英] How can I select a non-sequential subset elements from an array using Scala and Spark?
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问题描述
在Python中,这是我会怎么做。
In Python, this is how I would do it.
>>> x
array([10, 9, 8, 7, 6, 5, 4, 3, 2])
>>> x[np.array([3, 3, 1, 8])]
array([7, 7, 9, 2])
这不会在斯卡拉星火壳工作:
This doesn't work in the Scala Spark shell:
scala> val indices = Array(3,2,0)
indices: Array[Int] = Array(3, 2, 0)
scala> val A = Array(10,11,12,13,14,15)
A: Array[Int] = Array(10, 11, 12, 13, 14, 15)
scala> A(indices)
<console>:28: error: type mismatch;
found : Array[Int]
required: Int
A(indices)
在foreach方法不起作用或者:
The foreach method doesn't work either:
scala> indices.foreach(println(_))
3
2
0
scala> indices.foreach(A(_))
<no output>
我要的是B的结果是:
What I want is the result of B:
scala> val B = Array(A(3),A(2),A(0))
B: Array[Int] = Array(13, 12, 10)
不过,我不想硬code这样说,这是因为我不知道是指数有多长或者什么将是它。
However, I don't want to hard code it like that because I don't know how long indices is or what would be in it.
推荐答案
我能想到的最简洁的方式是翻转你的心智模式,并把指数的第一:
The most concise way I can think of is to flip your mental model and put indices first:
indices map A
和,我可能会建议使用电梯
返回一个选项
And, I would potentially suggest using lift
to return an Option
indices map A.lift
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