Produce a dummy string N characters long with java 8 streams

Question

There are numerous ways to do this, but using Java 8 streams (likely IntStream), how can I produce a dummy string that is N characters long?

I've seen examples using IntStream.range(), and the various aggregator functions (sum, average), but I don't see a way to do this.

My first random guess looks like this:

IntStream.range(1, 110).map(i -> "x").collect(Collectors.joining());

But that's wrong in a couple of different ways.

I think that the fastest way, would be to do a `Array.fill` here, to produce a `char[]` and than simply do `new String(char[])`... — Eugene, Mar 19 '18 at 19:41
@Eugene Most code sucks in parallel. Not sure what your point is. — shmosel, Mar 19 '18 at 19:45
@shmosel something `InstStream` or `Array` band that can be splitter better would be much more suited for parallel processing, nothing more — Eugene, Mar 19 '18 at 19:47

davidxxx · Accepted Answer · 2018-03-19T19:52:12.527

7

You need to use mapToObj() and not map() as you actually use an IntStream and IntStream.map() takes as parameter an IntUnaryOperator, that is an (int->int) function.

For same character dummy (for example "x") :

collect = IntStream.range(1, 110)
                   .mapToObj(i ->"x")
                   .collect(Collectors.joining());

Form random dummy :

You could use Random.ints(long streamSize, int randomNumberOrigin, int randomNumberBound).

Returns a stream producing the given streamSize number of pseudorandom int values, each conforming to the given origin (inclusive) and bound (exclusive).

To generate a String containing 10 random characters between the 65 and 100 ASCII code :

public static void main(String[] args) {
    String collect = new Random().ints(10, 65, 101)
                                 .mapToObj(i -> String.valueOf((char) i))
                                 .collect(Collectors.joining());

    System.out.println(collect);

}

edited Mar 19 '18 at 19:52

answered Mar 19 '18 at 19:41

davidxxx

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it's not random... – Eugene Mar 19 '18 at 19:43
@Eugene Can you detail ? – davidxxx Mar 19 '18 at 19:46
1

the Op wants `AAAAAA`.... not `AGCVT` – Eugene Mar 19 '18 at 19:47
@Eugene Thanks ! – davidxxx Mar 19 '18 at 19:55
1

oh wow! :) really? this is what OP was missing? – Eugene Mar 19 '18 at 20:12

score 4 · Answer 2 · answered Mar 19 '18 at 19:38

4

If you really want to use a Stream for this, you can utilize Stream#generate, and limit it to n characters:

Stream.generate(() -> "x").limit(110).collect(Collectors.joining());

answered Mar 19 '18 at 19:38

Jacob G.

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Alexandre Dupriez · Answer 3 · 2018-03-19T19:42:59.297

You are actually almost there:

String s = IntStream.range(40, 110)
                    .mapToObj(i -> Character.toString((char)i))
                    .collect(Collectors.joining());

System.out.println(s);

Produces:

()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklm

If you want random ordering, with N = 60 for instance:

Random r = new Random();    

IntStream.generate(() -> 40 + r.nextInt(70))
         .limit(60)
         .mapToObj(i -> Character.toString((char)i))
         .collect(Collectors.joining()));

Produces

Z>fA+5OY@:HfP;(L:^WKDU21T(*1//@V,F9O-SA2;+),A+V/mLjm<eaE56CH

Produce a dummy string N characters long with java 8 streams

3 Answers3