使用嵌套槽 (#) [英] Using nested slots (#)

问题描述

假设我想构造类似

Array[#1^#2 == 3 &, {3, 3}] 

现在我想用一个变量替换3".我可以做，例如:

And now I want to replace the "3" with a variable. I can do, for example:

f[x_] := Array[#1^#2 == x &, {x, x}]  

问题是:有没有办法使用仅插槽和 &作为功​​能符号?

The question is: Is there a way using only slots and & as the functional notation?

推荐答案

怎么样

Map[Last, #] & /@ Array[#1^#2 == #3 &, {#, #, #}] &[3]

非常丑陋的元素提取，非常有趣的是 Map[Last, #]& 给了我与 Last/@ 不同的结果.这是因为 Map 与 & 具有不同的属性吗?

Horrendously ugly element extraction, and very interestingly Map[Last, #]& gives me a different result than Last /@. Is this due to the fact that Map has different attributes than &?

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