尝试在APL中使用数组时出现问题.我错过了什么? [英] Problems when trying to use arrays in APL. What have I missed?

问题描述

我显然错过了一些有关如何从APL中的数组中提取元素的事情，希望有人可以看到我错过的事情，以及应该如何做才能以有意义的方式重现预期的结果.

I obviously have missed some things about how to extract elements from arrays in APL and hope that someone can see what I have missed and how I should do to get the expected results in a way that I can reproduce in a meaningful way.

我在学习APL方面相对较新，并且更习惯于Python和C之类的语言.APL中的数据类型和数组操作工具似乎使我有些困惑.

I am relatively new in learning APL and I am more used to languages like Python and C. The data types and array manipulating tools in APL seem to confuse me, a little.

考虑以下代码，并请说明为什么预期的结果(对我而言)

Consider the following code and please tell why the expected (by me) result,

┌→─────┐
│42 666│
└~─────┘

嵌入到更复杂的内容中，并且可能是一种解决方法 这个问题. (使用Dyalog APL/S-64，16.0.30320)

got embedded in something more complex, and possibly a way around that problem. (Using Dyalog APL/S-64, 16.0.30320)

      ⎕io ← 0
      a ← 17 4711 (42 666)
      z ← a[2]

      an_expected_vector←42 666
      ]DISPLAY an_expected_vector
┌→─────┐
│42 666│
└~─────┘

      ]DISPLAY z
┌──────────┐
│ ┌→─────┐ │
│ │42 666│ │
│ └~─────┘ │
└∊─────────┘

为什么z与an_expected_vector不相同?

谢谢！/汉斯

推荐答案

2是一个标量，因此a[2]返回一个标量，它恰好是向量42 666.因此，它被嵌套在一个嵌套层中.

2 is a scalar and so a[2] returns a scalar, which happens to be the vector 42 666. It is therefore enclosed in a level of nesting.

如果使用拾取"功能(二进位⊃)，您将获得预期的结果，因为⊃将从右边的参数中选择左边的参数指示的元素:

If you use the Pick function (dyadic ⊃) you will get the expected result, as ⊃ will pick the element indicated by the left argument, from the right argument:

       ⎕io ← 0
       a ← 17 4711 (42 666)
       z ← 2⊃a
       an_expected_vector ← 42 666
       z ≡ an_expected_vector
 1

Try it online!

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