a) I've got an int[] which contains only 0 und 1. And I want to end with a string, containing 0 and 1. So [1,1,0,1,0,1] should become "110101". Of course I can do this with a loop and ~. But I think it should be doable with functional style, which is something I would like to understand better. Can anyone help me here? (I think a.map!(a=>to!char(a+'0')) does the trick, but is this good style or is there a better way?)

b) After having this I'd like to split this resulting string into chunks of a fixed length, e.g. length 4, so "110101" from above should become ["1101","01"]. Again, is it possible to do that with functional style? Probably chunks from std.range might help here, but I don't get it to work.

All in all, can you fill in the magic functional chain below?

> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>
> auto result = magic functional chain here :-)
>
> assert(result==["1011","1010","1111","0"]);

On Friday, 14 September 2018 at 19:44:37 UTC, berni wrote:
> a) I've got an int[] which contains only 0 und 1. And I want to end with a string, containing 0 and 1. So [1,1,0,1,0,1] should become "110101". Of course I can do this with a loop and ~. But I think it should be doable with functional style, which is something I would like to understand better. Can anyone help me here? (I think a.map!(a=>to!char(a+'0')) does the trick, but is this good style or is there a better way?)
>
> b) After having this I'd like to split this resulting string into chunks of a fixed length, e.g. length 4, so "110101" from above should become ["1101","01"]. Again, is it possible to do that with functional style? Probably chunks from std.range might help here, but I don't get it to work.
>
> All in all, can you fill in the magic functional chain below?
>
>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>>
>> auto result = magic functional chain here :-)
>>
>> assert(result==["1011","1010","1111","0"]);

What you want is std.range.chunks


auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
    a.map!(to!string)
     .join("")
     .chunks(4)
     .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
     .writeln;

On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
> What you want is std.range.chunks
>
>
> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>     a.map!(to!string)
>      .join("")
>      .chunks(4)
>      .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
>      .writeln;

It's easier if you chunk before joining:

auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];

a.map!(to!string)
 .chunks(4)
 .map!join
 .writeln;

On Saturday, 15 September 2018 at 03:25:38 UTC, Paul Backus wrote:
> On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
>> What you want is std.range.chunks
>>
>>
>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>>     a.map!(to!string)
>>      .join("")
>>      .chunks(4)
>>      .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
>>      .writeln;
>
> It's easier if you chunk before joining:
>
> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>
> a.map!(to!string)
>  .chunks(4)
>  .map!join
>  .writeln;

Oh, thanks. What I didn't know about was join. Now I wonder how I could have found out about it, without asking here? Even yet I know it's name I cannot find it, nighter in the language documentation nor in the library documentation.

On Friday, September 14, 2018 11:39:48 PM MDT berni via Digitalmars-d-learn wrote:
> On Saturday, 15 September 2018 at 03:25:38 UTC, Paul Backus wrote:
> > On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
> >> What you want is std.range.chunks
> >>
> >>
> >> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
> >>
> >>     a.map!(to!string)
> >>
> >>      .join("")
> >>      .chunks(4)
> >>      .map!(to!string) //don´t know why the chunks are not
> >>
> >> already strings at this point ;/
> >>
> >>      .writeln;
> >
> > It's easier if you chunk before joining:
> >
> > auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
> >
> > a.map!(to!string)
> >
> >  .chunks(4)
> >  .map!join
> >  .writeln;
>
> Oh, thanks. What I didn't know about was join. Now I wonder how I could have found out about it, without asking here? Even yet I know it's name I cannot find it, nighter in the language documentation nor in the library documentation.

https://dlang.org/phobos/std_array.html#join

- Jonathan M Davis

On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
>
> What you want is std.range.chunks
>
>
> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>     a.map!(to!string)
>      .join("")
>      .chunks(4)
>      .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
>      .writeln;

They're not strings, because they're now 4 ranges of integers.

Then you map each of those 4 integer ranges into 4 strings.

On Saturday, 15 September 2018 at 06:16:59 UTC, bauss wrote:
> On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
>>
>> What you want is std.range.chunks
>>
>>
>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>>     a.map!(to!string)
>>      .join("")
>>      .chunks(4)
>>      .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
>>      .writeln;
>
> They're not strings, because they're now 4 ranges of integers.
>
> Then you map each of those 4 integer ranges into 4 strings.

Oh wait I didn't see your first map!(to!string)

I take back what I just said.

On Saturday, 15 September 2018 at 05:39:48 UTC, berni wrote:
> Oh, thanks. What I didn't know about was join. Now I wonder how I could have found out about it, without asking here? Even yet I know it's name I cannot find it, nighter in the language documentation nor in the library documentation.

Probably the easiest way to find something in the documentation is to use the unofficial mirror at http://dpldocs.info/. Type "join" into the search box there, and you'll get `std.array.join` (the function I used) as the first result.

Anotherone I'm not getting to work: From some output with newlines I want to discard all lines, that start with a # and select part of the other lines with a regex. (I know the regex r".*" is quite useless, but it will be replaced by something more usefull.)

I tried the following, but non worked:

> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').matchFirst(r".*");

> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.matchFirst(r".*");

> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.map!(matchFirst(r".*"));

> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.map!(a=>matchFirst(a,r".*"));

Any ideas?

On Saturday, 15 September 2018 at 20:04:36 UTC, berni wrote:
> Anotherone I'm not getting to work: From some output with newlines I want to discard all lines, that start with a # and select part of the other lines with a regex. (I know the regex r".*" is quite useless, but it will be replaced by something more usefull.)
>
> I tried the following, but non worked:
>
>> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').matchFirst(r".*");
>
>> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.matchFirst(r".*");
>
>> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.map!(matchFirst(r".*"));
>
>> output.split("\n").filter!(a=>a.length>0 && a[0]!='#').array.map!(a=>matchFirst(a,r".*"));
>
> Any ideas?

Your last example works for me: https://run.dlang.io/is/F5n3mk

Can you post a complete, runnable example that illustrates your problem?