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Scala随机字符串

[英]Scala Random String

Hope you are doing good, I just started scala basic programming, I want to generate some string variable with foreach or something else you think is the best method. 希望您做得很好,我刚刚开始使用scala基本编程,我想使用foreach或您认为是最佳方法的其他东西生成一些字符串变量。

How do I use scala.util.Random to generate this result: 如何使用scala.util.Random生成此结果:

Var 1A:String = random between A to J
Var 1B:String = random between A to J but not 1A value
Var 1C:String = random between A to J, not 1A and not 1B
Var 1D to 1J also should have different random value from A to J

So basically from 1A to 1J could be like this: 因此基本上从1A到1J可能是这样的:

Var 1A = "B"
Var 1B = "G"
Var 1C = "A"
Var 1D = "D"
Var 1E = "H"
and so on until 1J

The simplest way is probably to use Random.shuffle . 最简单的方法可能是使用Random.shuffle

First construct your list of characters: 首先构造您的字符列表:

scala> val chars = ('A' to 'J').toList
chars: List[Char] = List(A, B, C, D, E, F, G, H, I, J)

Then shuffle them: 然后随机播放它们:

scala> val shuffled = scala.util.Random.shuffle(chars)
shuffled: List[Char] = List(C, E, G, B, J, A, F, H, D, I)

Now you've got a list of these ten characters in a random order. 现在,您将以随机顺序获得这十个字符的列表。 If you need variables referring to them, you can do something like this: 如果需要引用它们的变量,则可以执行以下操作:

scala> val List(a, b, c, d, e, f, g, h, i, j) = shuffled
a: Char = C
b: Char = E
c: Char = G
d: Char = B
e: Char = J
f: Char = A
g: Char = F
h: Char = H
i: Char = D
j: Char = I

(Note that 1A isn't a valid Scala identifier, and Val isn't valid syntax.) (请注意, 1A不是有效的Scala标识符,而Val是无效的语法。)

Or you can refer to them by index: 或者您可以按索引引用它们:

val a = shuffled(0)
val b = shuffled(1)
...

Or you could just work with them in the list. 或者,您可以在列表中与他们一起工作。

You can use shuffle from Random package: 您可以使用Random软件包中的shuffle

scala> scala.util.Random.shuffle(('A' to 'J').toList map (_.toString))
res2: List[String] = List(C, F, D, E, G, A, B, I, H, J)

Assign the result to the list of variables if you like. 如果愿意,可将结果分配给变量列表。

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