World: r3wp

or 16581375 rather

The dictionary will have 2 million strings... the index would be 
much smaller

index being the DB using triples as schema

Here's a question.. when a block of integers is converted to a HASH!.. 
what actually happens to it?

Hashes, lists and blocks can contain any type of value. The values 
are not affected by the container type.

A hash just has a different way of linking the items it contains 
together, so it performs faster in some operations at the expense 
of others.

If you just need to save large arrays of integers, you can use format 
used in AS3:
 The AS3 Integer can be encoded into between 1 and 5 bytes.

    *

      if the integer is between 0×00 and 0x7F then only one byte (representing 
      the integer)
    *
      if between 0×80 and 0x3FFF then 2 bytes :
          o
            (i & 0x7F) | 0×80
          o
            (i » 7)
    *
      if between 0×4000 and 0x1FFFFF then 3 bytes :
          o
            (i & 0x7F) | 0×80
          o
            (i » 7) | 0×80
          o
            (i » 14)
    *
      if between 0×200000 and 0xFFFFFFF then 4 bytes :
          o
            (i & 0x7F) | 0×80
          o
            (i » 7) | 0×80
          o
            (i » 14) | 0×80
          o
            (i » 21)
    *
      if more or equal than 0×10000000 :
          o
            (i & 0x7F) | 0×80
          o
            (i » 7) | 0×80
          o
            (i » 14) | 0×80
          o
            (i » 21) | 0×80
          o
            (i » 28)

but if jyou need to search such an array, it's not the best for your... 
especially if there is no rebcode yet:(

That's interesting, using a bit in each byte to "escape" and open 
up the next, more significant byte. So it's a variable-length encoding 
scheme. (I guess, kind of like UTF8)

But yes, I think this adds complexity and would slow down the access 
routines.

yeah... I think the conclusion is ... don't worry about the number 
of bytes (and thus mem) when using plain integers with my index block, 
as it's much smaller than the dictionary anyways.. and unless Im 
shown otherwise, the crawling of it (find/ foreach, append) should 
be about as fast as any other method, right?

It should be faster and simpler to just use integers. When you want 
to cut down the size of your 2 million integers (=30MB), you can 
then look at implementing 3-byte integers packed in a binary.

exactly

Here's a couple of conversion routines.

struct: make struct! [int [int]] none

; convert integer -> 3-byte binary

integer-to-3-byte-binary: func [integer [integer!]][
	struct/int: integer
	copy/part third struct 3
]

; convert 3-byte binary -> integer

binary-3-byte-to-integer: func [int3 [binary!]][
	struct/int: 0 ; just make sure all bytes are zero
	change third struct int3
	struct/int
]
	
; test

my-bin: integer-to-3-byte-binary 2000000
;== #{80841E}
my-int: binary-3-byte-to-integer my-bin
;== 2000000

ok.. that's cool.. thanks

Even though each integer uses 16 bytes, there's some compaction by 
using the smallest integers with the most commonly used dictionary 
strings

I'll add these functions to the dictionary.. 

dict: [{integer-to-3-byte-binary: func [integer [integer!]][
	struct/int: integer
	copy/part third struct 3
]}]


now whenever i want to use that function.. i can represent it with 
a single integer.. ie:  do pick dict 1

Terry, I would not make own binary storage as there should be new 
datatype dictionary! in R3 which is exactly what you need... if I 
understand it well

I cannot see hash! in this list http://www.rebol.net/r3blogs/0076.html
so it will be probably replaced

and that could be reduced further.. 

dict: [

{fetch: func [index][do pick dict index]}

{integer-to-3-byte-binary: func [integer [integer!]][
	struct/int: integer
	copy/part third struct 3
]}

]

do pick dict 1
fetch 2
my-bin: integer-to-3-byte-binary 2000000

The new vector! datatype will be usefull as well to store large block 
of same data type

Im happy with R2 ;)

yes... you can make it in R2 using index: [1 2 3]  and for R3 you 
just replace:  index: make vector! [integer! 24 [1 2 3]]

Now as for compressing the dictionary.. it seems that smaller strings 
grow with compression?

>> a: compress "a"
== #{789C4B04000062006201000000}

>> a: compress "aaaa"
== #{789C4B4C4C4C040003CE018504000000}
>> a: compress "aaaaa"
== #{789C4B4C04020005B401E605000000}

compress is uzing zlib compression... just insetad of checksum at 
the tail (last 4 bytes) is used size of the source string

the size is used to make a result buffer for decompress:
>> decompress rejoin [#{789C4B4C} #{04020005B401E6} #{00000000}]
** Script Error: Not enough memory
>> decompress rejoin [#{789C4B4C} #{04020005B401E6} #{FF000000}]
== "aaaaa"

Terry, your 'integer-to-3-byte-binary conversion function is dependent 
on platform endianness (third struct! gives you an endian-dependent 
result), watch out for that.

You can blame me for that, I wrote it quickly.

Terminology question;  I know I could probably RTFM, but sometimes

Ask A Friendly Human is more fun.  What is the correct terminology 
for the global

REBOL context.  I'm describing (or trying to at least) the parse 
dialect "copy" versus

the REBOL "copy".  Is there a one word term for the "no context" 
context?  Or is
the REBOL global namespace
 good enough (and not too confusing to new rebols)?

Yeah, we all call it the global context.

Thanks Anton.

Is there a function which can flat a block. For example,
>> FLAT [ 1 2 [ 3 4 ] 5 6 ]
== [ 1 2 3 4 5 6]

probably not what you want ...

>> to-block form  [ 1 2 [ 3 4 ] 5 6 ]
== [1 2 3 4 5 6]

flatten: func [block [any-block!]][

    parse block [

         any [block: any-block! (change/part block first block 1) :block | 
         skip]

    ]
 
   head block
]

and with greater speed but less accuracy:
>> load form [1 2 [3 4] 5 6]

and now I should of course have read what Graham posted...

Don't assume that the parse is slow - you may be surprised.

Yes, and I wouldn't recommend the LOAD FORM way, it looks too brittle.

does SORT/COMPARE not work in combination with /SKIP? it seems to 
happily ignore /SKIP here.

seems that using /ALL does not help.

>> b: [1 2 3 4 5  5 4 3 2 1  7 6 5 4 3  3 4 5 6 7]
== [1 2 3 4 5 5 4 3 2 1 7 6 5 4 3 3 4 5 6 7]
>> bb: sort/compare/skip copy b 2 5
== [1 2 3 4 5 5 4 3 2 1 3 4 5 6 7 7 6 5 4 3]
>> bb: sort/compare/skip copy b 4 5
== [5 4 3 2 1 1 2 3 4 5 7 6 5 4 3 3 4 5 6 7]

what if the record size is 2, consists of a binary and an object 
and you want to sort on a value in the object?

Here's one way -- though it assumes (for simplicity) that the binary 
is a string of equal length in all keys:


data: reduce ["z" make object! [key: 1] "y" make object! [key: 2] 
"z" make object! [key: 2]]
sort/all/skip/compare  data 2 func [a b][
     return (join a/1 a/2/key) < (join b/1 b/2/key)
    ]
probe data

hmm... funny, I did exactly that code, but it would never recognize 
the b object in the compare function. will have to try again...

oh wait a minute. I see now :-)

I guess I don't after all. I still can't get it to recognize 'b. 
If I try to probe 'b inside the function, sorting just stops. probing 
'a works fine.

and the block checks out fine. it is properly arranged.