英文:
how to print values between 2 numbers in APL?
问题
我无法打印从30到50（包括30和50）的数值在APL中。
实际上，我尝试了很多运算符，但都没有奏效。
英文:
I am not able to print values ranging from 30 to 50 both inclusive in APL.
Actually I tried so much operators but not working.
答案1
得分: 2
以下是翻译好的部分：
你的约束是什么？你想定义一个函数吗，还是只想知道如何迭代？"print" 是指返回结果还是字面上打印到某个地方（输出/文件）？
在最简单的方式中，使用 iota ⍳ 来迭代两个边界之间的距离，然后将较低的一个加到结果中。假设索引起点 IO 设置为 0：
30 + ⍳21
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
如果 IO 设置为 1，则减小较低的边界：
29 + ⍳21
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
您可以随时检查或设置 IO，使用 ⎕IO 来切换：
⎕IO ← 0
0
英文:
What are your constraints? Do you want to define a function or do you just want to know how to iterate? Does "print" mean return as result or literally print to somewhere (output/file)?
In the easiest way, use iota ⍳ to iterate through the distance between the two bounds, then add the lower one to the results. Assuming index origin IO is set to 0:
30 + ⍳21
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If IO is set to 1, lower the lower bound:
29 + ⍳21
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
You can always inspect or set IO using ⎕IO if you wanted to switch:
⎕IO ← 0
0
答案2
得分: 1
Sure, here is the translated content:
"实际上，我想要打印出在8到30之间的随机12个值。
如果⎕IO设置为1，你可以使用?23（或?1+30-8）生成大于或等于1且小于或等于23的随机整数。然后，将结果加上7（或8-1），它将在8到30之间，包括这两个值：7+?23（如果⎕IO设置为0，生成的随机数将范围从0到22，因此要达到目标范围，你需要加上8而不是7，或者如果需要独立于⎕IO的通用性，可以加上(8-⎕IO)）。
现在，要从该范围内抽取多个随机数，请在其左参数中使用?并提供你想要查看的抽取数量。结果将是一个具有相同长度的数组。
7+12?23
23 21 9 8 10 27 12 30 28 19 26 14
请注意，这些数字都是唯一的（没有重复）。如果要获得一个包含独立的随机数字的数组（值可以多次抽取），请使用⍴来相应地调整?的（右）参数，然后将其应用于整个数组：
7+?12⍴23
9 8 30 16 13 25 26 8 26 19 19 18
"
英文:
> Actually I wanted to print random 12 values between 8 to 30.
If ⎕IO is set to 1, you can generate a random integer larger or equal to 1 and smaller or equal to 23 using ?23 (or ?1+30-8). Then, add 7 (or 8-1) to the result, and it will be shifted between 8 and 30, both inclusive: 7+?23. (With ⎕IO set to 0, the generated random number will instead range from 0 to 22, so to reach the destination range you'd have to add 8 rather than 7, or (8-⎕IO) more generically if independence from ⎕IO is a concern.)
Now, to draw more than one random number from that range, provide ? in its left argument with the amount of draws you'd like to see. The result is an array of that length.
7+12?23
23 21 9 8 10 27 12 30 28 19 26 14
Note that these numbers are all unique (no duplicates). To instead get an array of independent random numbers (values can be drawn more than once), use ⍴ to accordingly reshape the (right) argument of ? which will then be applied onto the whole array:
7+?12⍴23
9 8 30 16 13 25 26 8 26 19 19 18
答案3
得分: 0
@pmf's solution是简单的，但在处理两个大但接近的值时可能效率很低，并且如果任何一个数是负数，它会失败。
如果我们将边界称为⍺和⍵，那么⍺+¯1+⍳1+⍵-⍺既高效又通用。如果⎕IO为0，则移除¯1+。您可以使用以下公式创建一个to函数：
to←{(⍺-⎕IO)+¯1+⍳1+⍵-⍺}
30 to 50
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
或者，如果您使用Dyalog APL，您可以使用dfns工作区中的to实用程序函数，如下所示：
'to' ⎕CY 'dfns'
30 to 50
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
英文:
@pmf's solution is simple, but can be very inefficient (for two large, but close values) and fails if any number is negative.
If we call the bounds ⍺ and ⍵ then ⍺+¯1+⍳1+⍵-⍺ is both efficient and general. Remove ¯1+ if ⎕IO is 0. You can create a to function using this formula as follows:
to←{(⍺-⎕IO)+¯1+⍳1+⍵-⍺}
30 to 50
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Alternatively, if you use Dyalog APL, you can use the to utility function from the dfns workspace, as follows:
'to' ⎕CY 'dfns'
30 to 50
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
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