Hi Gou, My understanding is that the Numerical Recipies book has an algorithm for this combinatorial caculation. Maybe you could implement it in IDL? Also, I found this page that can be helpful: https://groups.google.com/forum/#!sea... I hope Numerical Recipies can help. Cheers, fernando

Hi Gou again, Yes, I think that Kenneth Bowman seems to have the answer to your question in that IDL Google forum page: https://groups.google.com/forum/#!sea... Cheers, fernando

hi, fernando Thanks for your reply. It's true that Kenneth Bowman's method can partially solve my problem(I have used a similar method in my programming before I put forward my problem), however, as I have stated in my question that as the N(the number of arrays) is a variable,and I can only choose one element from these N arrays to get a combination of N elements at each time. What I mean is that if N = 2, I will use two for loops, if N =3 , 3 for loops will be used, if N = 6, 6 for loops will be used, and so on. It's difficult to determine how many for loops will be used in the programming.I want to adapt the for loops in programming intellectually according to the value of N. Can you give me some instructions? Thanks very much! PS: this is an example code used in my programming for N =3 ; make lookup table count = 0 FOR i = 0.0,n_elements(array1)-1 DO BEGIN FOR j = 0.0, n_elements(array2)-1 DO BEGIN FOR k = 0.0, n_elements(array3)-1 DO BEGIN lut[count,*] = [array1[i],array2[j],array3[k]] count = count+1 ENDFOR ENDFOR ENDFOR

Hi Gou, What about using a CASE statement?: http://www.exelisvis.com/docs/CASE.html How big can N become? fernando

hi, fernando Thanks for your timely reply. the use of CASE statement is not a bad idea, though it will make the programming looks like verbose. In my case, the value of N may be as large as 10.