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permuation list with repetition and order doesn't matter

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Charles
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permuation list with repetition and order doesn't matter

Charles
I am trying to create a list of all possible permuations of 14 things with 7
things chosen at a time.
Example:
Variables: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
Output:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1  <-- case 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2 <-- case 2
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1 <-- case 3
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2
...
14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14

What I want to do is treat case 2 and case 3 above as the same case. That
is, I don't want more than 1 instance of all 1's and one 2 in the output.

I have looked through the archive and adjusted what I have found. Note> I
know the line: loop throw = 1 to 2000000. does not cover the permutations.
At this point I am just trying to refine to get the output I want.



* this program creates all possible permutations of 3 "dice" variables  .
new file.
set seed = 20081031.
input program.
loop i = 1 to 14.
loop j = 1 to 14.
loop k = 1 to 14.
loop l = 1 to 14.
loop m = 1 to 14.
loop n = 1 to 14.
loop o = 1 to 14.
leave i j k, l, m, n, o.
end case.
end loop.
end loop.
end loop.
end loop.
end loop.
end loop.
end loop.
end file.
end input program.
formats i,j,k,l,m,n,o (f1).
list.

* this program  "throws three dice" .
set seed = 5743269.
input program.
vector die (7,f1).
loop throw = 1 to 2000000.
loop #k  = 1 to 7.
compute die(#k) = rnd(rv.uniform(1,14)).
end loop.
end case.
end loop.
end file.
end input program.
formats throw (f3) die1 to die3 (f1).
list.

Again, I adjusted code from the archive that I found in an attempt to reach
my desired result.

The best analogy I can think of is using 14 decks of cards to play 7 card
stud.
I can have 7 aces in my hand.
But, if I have 6 aces and 1 king it doesn't matter there the 1 king is in
the hand.



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Art Kendall
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Re: permuation list with repetition and order doesn't matter

Art Kendall
it seems you want combinations not permutations.
try something like this
loop i = 1 to 14.
loop j = 1 to i.
loop k = 1 to j.

check the count of cases  against factorial(14) / factorial(7).



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Art Kendall
Social Research Consultants
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Rich Ulrich
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Re: permuation list with repetition and order doesn't matter

Rich Ulrich

Right - and complete if you want 3 draws, like the dice example.

(Confusing Original post. Like, the first example says "7 draws" but shows 14.)


To do  7 draws with 14 choices, there will be 7 loops, like the

(i, j, k, l, m, n, o) example, but with maximums of (j, k, ..., o)

instead of "14".


--

Rich Ulrich


From: SPSSX(r) Discussion <[hidden email]> on behalf of Art Kendall <[hidden email]>
Sent: Monday, July 30, 2018 1:10:19 PM
To: [hidden email]
Subject: Re: permuation list with repetition and order doesn't matter
 
it seems you want combinations not permutations.
try something like this
loop i = 1 to 14.
loop j = 1 to i.
loop k = 1 to j.

check the count of cases  against factorial(14) / factorial(7).



-----
Art Kendall
Social Research Consultants
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Andy W
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Re: permuation list with repetition and order doesn't matter

Andy W
In reply to this post by Charles
Below is a python program to do this. Takes about a minute and produces
77,520 rows.

*************************************************************************.
DATASET DECLARE Combo.
BEGIN PROGRAM Python.
import itertools, spss

#Function to import data into SPSS
#See
https://andrewpwheeler.wordpress.com/2014/09/19/turning-data-from-python-into-spss-data/

def SPSSData(data,vars,types,name=None):
  VarDict = zip(vars,types) #combining variables and
                            #formats into tuples
  spss.StartDataStep()
  datasetObj = spss.Dataset(name=name) #if you give a name,
                                       #needs to be declared
  #appending variables to dataset
  for i in VarDict:
    datasetObj.varlist.append(i[0],i[1])
  #now the data
  for j in data:
    datasetObj.cases.append(list(j))
  spss.EndDataStep()

#Now creating the combinations you want  
YourSet = range(1,14+1)
YourLen = 7
x = itertools.combinations_with_replacement(YourSet,YourLen)
v = ['X' + str(i+1) for i in range(YourLen)]
t = [0]*YourLen
SPSSData(data=x,vars=v,types=t,name='Combo')
END PROGRAM.
DATASET ACTIVATE Combo.
*************************************************************************.



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Andy W
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Andy W
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Re: permuation list with repetition and order doesn't matter

Andy W
Here is a pure SPSS implementation I had for a different project awhile ago
as well. Don't look too closely at the macro, you might go mad.

******************************************************************.
DATASET CLOSE ALL.
OUTPUT CLOSE ALL.

*This macro takes a number of crimes, n, in a given number of bins.
*And generates all of the possible permutations.

DEFINE !CombBins (N = !TOKENS(1)
                  /Bins = !TOKENS(1) )
INPUT PROGRAM.
VECTOR Bin(!Bins,!CONCAT("F",!LENGTH(!N),".0")).
!LET !Min = !CONCAT("#",1)
!LET !DoR = !Min
LOOP !Min = !N to 0 BY -1.
!DO !L = 2 !TO !Bins
!LET !Ind = !CONCAT("#",!L)
LOOP !Ind = (!N - SUM(!Min)) to 0 BY -1.
!LET !Min = !CONCAT(!Min,",",!Ind)
!LET !DoR = !CONCAT(!DoR," ",!Ind)
!DOEND
DO REPEAT Bin = Bin1 to !CONCAT("Bin",!Bins) /#i = !DoR.
  COMPUTE Bin = #i.
END REPEAT.
COMPUTE #c = SUM(Bin1 to !CONCAT("Bin",!Bins)).
END CASE.
END LOOP IF #c = !N.
!DO !L = 2 !TO !Bins
END LOOP.
!DOEND
END FILE.
END INPUT PROGRAM.
EXECUTE.
DATASET NAME !CONCAT("Comb",!Bins,"Bins",!N,"N")
!ENDDEFINE.

****************************.
*Example Use.
!CombBins N = 7 Bins = 14.
*Now creating dataset for individual cards in each row.
VECTOR X(7,F1.0).
COMPUTE #Ord = 0.
DO REPEAT B = Bin1 TO Bin14 /#i = 1 TO 14.
  DO IF B > 0.
    LOOP #j = 1 TO B.
      COMPUTE #Ord = #Ord + 1.
      COMPUTE X(#Ord) = #i.
  END LOOP.
  END IF.
END REPEAT.
EXECUTE.
SHOW N.
****************************.
******************************************************************.



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Andy W
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Kirill Orlov
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Re: permuation list with repetition and order doesn't matter

Kirill Orlov
In reply to this post by Charles
I can't understand the task. What are all those 1,1,1,... or 14,1,4,14,... is it "pemutations" that you duplicate an element?

30.07.2018 19:34, Charles пишет:
I am trying to create a list of all possible permuations of 14 things with 7
things chosen at a time. 
Example: 
Variables: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
Output:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1  <-- case 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2 <-- case 2
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1 <-- case 3
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2
...
14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14

What I want to do is treat case 2 and case 3 above as the same case. That
is, I don't want more than 1 instance of all 1's and one 2 in the output. 

I have looked through the archive and adjusted what I have found. Note> I
know the line: loop throw = 1 to 2000000. does not cover the permutations.
At this point I am just trying to refine to get the output I want. 



* this program creates all possible permutations of 3 "dice" variables  . 
new file. 
set seed = 20081031. 
input program. 
loop i = 1 to 14. 
loop j = 1 to 14. 
loop k = 1 to 14. 
loop l = 1 to 14.
loop m = 1 to 14.
loop n = 1 to 14. 
loop o = 1 to 14. 
leave i j k, l, m, n, o. 
end case. 
end loop. 
end loop. 
end loop. 
end loop.
end loop.
end loop.
end loop.
end file. 
end input program. 
formats i,j,k,l,m,n,o (f1). 
list. 

* this program  "throws three dice" . 
set seed = 5743269. 
input program. 
vector die (7,f1). 
loop throw = 1 to 2000000. 
loop #k  = 1 to 7. 
compute die(#k) = rnd(rv.uniform(1,14)). 
end loop. 
end case. 
end loop. 
end file. 
end input program. 
formats throw (f3) die1 to die3 (f1). 
list. 

Again, I adjusted code from the archive that I found in an attempt to reach
my desired result. 

The best analogy I can think of is using 14 decks of cards to play 7 card
stud. 
I can have 7 aces in my hand. 
But, if I have 6 aces and 1 king it doesn't matter there the 1 king is in
the hand. 



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===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD
Kirill Orlov
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Re: permuation list with repetition and order doesn't matter

Kirill Orlov
In reply to this post by Charles
--> The best analogy I can think of is using 14 decks of cards to play 7 card
stud.
I can have 7 aces in my hand.
But, if I have 6 aces and 1 king it doesn't matter there [where?] the 1 king is in
the hand.


Can that be understood such that you have 14 sets of elements, and you want to get all combinations (not permutations) by 7 elements from different sets, no more than 1 element from each set.  ?

Is that what are're after?



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