Hi from italy...so sorry for my bad english :)

I use this code :

Application.Run "ATPVBAEN.XLA!Histogram",
Sheets("Archivio").Range("$A$2:$H$5000") _
, ActiveSheet.Range("$D$46:$E$77"), ActiveSheet.Range("$A$1:$A$32"),
False _
, False, False, False

to generate a serial of casual number.

My question :

is possibile to *avoid* that a number is present 2 o 3 time in a row ?

example to avoid :

9 10 11 11 12 15 18 30
6 11 11 15 23 23 25 31



TIA

I am not sure how a histogram generates a serial of casual number, but if
you mean randombetween, then it does not provide unique numbers.

--
Regards,
Tom Ogilvy

"Aristotele64" wrote in message
...
Hi from italy...so sorry for my bad english :)

I use this code :

Application.Run "ATPVBAEN.XLA!Histogram",
Sheets("Archivio").Range("$A$2:$H$5000") _
, ActiveSheet.Range("$D$46:$E$77"),
ActiveSheet.Range("$A$1:$A$32"),
False _
, False, False, False

to generate a serial of casual number.

My question :

is possibile to *avoid* that a number is present 2 o 3 time in a row ?

example to avoid :

9 10 11 11 12 15 18 30
6 11 11 15 23 23 25 31



TIA

As Tom has noted, random integers returned by RANDBETWEEN are not
constrained to be unique. The usual approach to generate unique random
integers between say 1 and 50, would be to put all the integers 1..50 in
50 rows of a column; put =RAND() in the corresponding rows of the
adjacent column; sort both columns by the random column; then use
however many values you want, in order, from the first column.

Jerry

Aristotele64 wrote:

Hi from italy...so sorry for my bad english :)

I use this code :

Application.Run "ATPVBAEN.XLA!Histogram",
Sheets("Archivio").Range("$A$2:$H$5000") _
, ActiveSheet.Range("$D$46:$E$77"), ActiveSheet.Range("$A$1:$A$32"),
False _
, False, False, False

to generate a serial of casual number.

My question :

is possibile to *avoid* that a number is present 2 o 3 time in a row ?

example to avoid :

9 10 11 11 12 15 18 30
6 11 11 15 23 23 25 31

"Tom Ogilvy" ha scritto nel messaggio
...
I am not sure how a histogram generates a serial of casual number, but if
you mean randombetween, then it does not provide unique numbers.


sorry i have wrong to copy cody , this is my code :

Application.Run "ATPVBAEN.XLA!Random", ActiveSheet.Range("$P$2:$W$5000"), 8
_
, 5000, 7, , ActiveSheet.Range("$G$1:$H$32")

Is possible to write a function to generate a similar ATPVBAEN.XLA
option ?

tk

"Jerry W. Lewis" ha scritto nel messaggio
...
As Tom has noted, random integers returned by RANDBETWEEN are not
constrained to be unique. The usual approach to generate unique random
integers between say 1 and 50, would be to put all the integers 1..50 in
50 rows of a column; put =RAND() in the corresponding rows of the
adjacent column; sort both columns by the random column; then use
however many values you want, in order, from the first column.

many thanks i don't know this.........

but my problem is that i want generate casual number with a column of
number
and respective column of a probability and
ATPVBAEN XLA was perfect , only things bad is that a row have number not
unique


Example i have this :

Num Prob

1 0,035181345
2 0,037730718
3 0,045967154
4 0,0300826
5 0,03953489
6 0,038750467
7 0,025493729
8 0,032122098
9 0,025689834
10 0,030592474
11 0,027533227
12 0,02765089
13 0,030514032
14 0,03184755
15 0,028945187
16 0,025493729
17 0,031925992
18 0,0300826
19 0,032122098
20 0,032292056
21 0,0292668
22 0,032088106
23 0,030116591
24 0,034263571
25 0,028552976
26 0,028077093
27 0,024643938
28 0,026173561
29 0,029980625
30 0,032835922
31 0,032088106
32 0,032360039


this code :

Application.Run "ATPVBAEN.XLA!Random", ActiveSheet.Range("$P$2:$W$5000"), 8
_
, 5000, 7, , ActiveSheet.Range("$G$1:$H$32")


working but on 5000 rows i have 3000 with a repeat number.....

is there on web page where speaking : how to write ATPVBAEN.XLA!Random
function ?


TIA

My guess is that you want to generate 5000 random samples of 8
observations from a custom specified discrete distribution. Each sample
of 8 is to be drawn without replacement.

A very painful way to do this would be to remove the final argument from
your ATPVBAEN.XLA!Random call. That would return 5000 random samples of
8 from the Uniform(0,1) distribution.

For each random sample [row of U(0,1) random variables] proceed as follows:


If the first uniform random number <0,035181345 then the first random
number from your distribution is 1. If the first uniform random number
=0,035181345 but <(0,035181345+0,037730718) then the first random
number from your distribution is 2 ...
Create a new custom distribution for only 31 values (excluding the
one already drawn) by rescaling the remaining probabilities to sum to 1.
Map the second uniform random number to this reduced custom distribution.
Continue reducing and choosing until all 8 random numbers have been
drawn.
Repeat for the next sample.


You could automate this process with a macro, but I do not immediately
see an elegant approach.

Jerry

Aristotele64 wrote:
....

but my problem is that i want generate casual number with a column of
number
and respective column of a probability and
ATPVBAEN XLA was perfect , only things bad is that a row have number not
unique


Example i have this :

Num Prob

1 0,035181345
2 0,037730718
3 0,045967154
4 0,0300826
5 0,03953489
6 0,038750467
7 0,025493729
8 0,032122098
9 0,025689834
10 0,030592474
11 0,027533227
12 0,02765089
13 0,030514032
14 0,03184755
15 0,028945187
16 0,025493729
17 0,031925992
18 0,0300826
19 0,032122098
20 0,032292056
21 0,0292668
22 0,032088106
23 0,030116591
24 0,034263571
25 0,028552976
26 0,028077093
27 0,024643938
28 0,026173561
29 0,029980625
30 0,032835922
31 0,032088106
32 0,032360039


this code :

Application.Run "ATPVBAEN.XLA!Random", ActiveSheet.Range("$P$2:$W$5000"), 8
_
, 5000, 7, , ActiveSheet.Range("$G$1:$H$32")


working but on 5000 rows i have 3000 with a repeat number.....

is there on web page where speaking : how to write ATPVBAEN.XLA!Random
function ?