How to use reduce or fold to encrypt a list of string tokens?

Question

For example I want to encrypt each token of a sentence and reduce them to a final encrypted text:

def convert(str: String) = {
  str + ":"
}

val tokens = "Hi this is a text".split("\\ ").toList

val reduce = tokens.reduce((a, b) => convert(a) + convert(b))
println(reduce)
// result is `Hi:this::is::a::text:`

val fold = tokens.fold("") {
  case (a, b) => convert(a) + convert(b)
}
println(fold)
// result is `:Hi::this::is::a::text:`

val scan = tokens.scan("") {
  case (a, b) => convert(a) + convert(b)
}
println(scan)
// result is List(, :Hi:, :Hi::this:, :Hi::this::is:, :Hi::this::is::a:, :Hi::this::is::a::text:)

Assume that convert is an encryption function. So each token should encrypt only once not twice. but fold and reduce and scan reencrypt the encrypted token. I want this desired result Hi:this:is:a:text:

In all these cases you seem to be assuming that a and b are the same. They're not. a is the accumulated result, and b the next token. So you're encrypting the token (b) and adding it to the result, and next time re-encrypting that result. So miss the convert of a out and things will look a lot better! — The Archetypal Paul, Jan 21 '14 at 09:51

Kigyo · Accepted Answer · 2014-01-21T09:36:11.243

2

Well if you want to encrypt each Token individually, map should work.

val tokens = "Hi this is a text".split("\\ ").toList

val encrypted = tokens.map(convert).mkString

println(encrypted) //prints Hi:this:is:a:text:

def convert(str: String) = {
  str + ":"
}

Edit: If you want to use a fold:

val encrypted = tokens.foldLeft("")((result, token) => result + convert(token))

edited Jan 21 '14 at 09:36

answered Jan 21 '14 at 09:23

Kigyo

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score 2 · Answer 2 · answered Jan 21 '14 at 10:03

2

One-liner specialised at this very example,

"Hi this is a text" split " " mkString("",":",":")

Or

val tokens = "Hi this is a text" split " "
val sep = ":"
val encrypted = tokens mkString("",sep,sep)

answered Jan 21 '14 at 10:03

elm

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score 2 · Answer 3 · answered Jan 21 '14 at 10:23

Note that fold or reduce will operate on two operands in every step. However you want to encrypt each of the tokens -- which is a unary operand. Therefore first you should do a map and then either a fold or a reduce:

tokens map(convert)

Reduce / Fold:

scala> tokens.map(convert).fold("")(_ + _)
res10: String = Hi:this:is:a:text:

scala> tokens.map(convert)reduce(_ + _)
res11: String = Hi:this:is:a:text:

Infact you can simply use mkString which makes it even more concise:

scala> tokens.map(convert).mkString
res12: String = Hi:this:is:a:text:

Also you can do the conversion in parallel too (using par ):

scala> tokens.par.map(convert).mkString
res13: String = Hi:this:is:a:text:

scala> tokens.par.map(convert)reduce(_ + _)
res14: String = Hi:this:is:a:text:

score 1 · Answer 4 · edited May 23 '17 at 11:44

1

I think your main problem is how reduce and fold works. You can learn from other answer

As for you question, fold can help:

"Hi this is a text".split("\\ ").fold("") { (a, b) => a + convert(b) }

edited May 23 '17 at 11:44

Community

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answered Jan 21 '14 at 09:45

Windor C

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score 1 · Answer 5 · answered Jan 21 '14 at 09:55

Here is a version with the code cleaned up and unnecessary conversions removed:

def convert(str: String) = str + :

val tokens = "Hi this is a text" split " "
val encrypted = (tokens map convert) mkString " "

mkString could be seen as a specialized Version of reduce (or fold) for Strings.

If for some reason, you don't want to use mkString the code would look like this:

def convert(str: String) = str + :

val tokens = "Hi this is a text" split " "
val encrypted = (tokens map convert) reduce (_ + _)

Or shortend with fold

val encrypted = "Hi this is a text".split(" ").foldLeft ("") { case (accum, str) => accum + convert(str) }

How to use reduce or fold to encrypt a list of string tokens?

5 Answers5