[Numpy-discussion] Is there a better way to do this?

[Jeffery D. Collins]

Hee-Seng Kye wrote:

> My question is not directly related to NumPy, but since many people 
> here deal with numbers, I was wondering if I could get some help; it 
> would be even better if there is a NumPy (or Numarray) function that 
> takes care of what I want!
>
> I'm trying to write a program that computes six-digit numbers, in 
> which the left digit is always smaller than its following digit (i.e., 
> it's always ascending).  The best I could do was to have many embedded 
> 'for' statement:
>
> c = 1
> for p0 in range(0, 7):
>   for p1 in range(1, 12):
>     for p2 in range(2, 12):
>       for p3 in range(3, 12):
>         for p4 in range(4, 12):
>           for p5 in range(5, 12):
>             if p0 < p1 < p2 < p3 < p4 < p5:
>               print repr(c).rjust(3), "\t",
>               print "%X %X %X %X %X %X" % (p0, p1, p2, p3, p4, p5)
>               c += 1
> print "...Done"
>
> This works, except that it's very slow.  I need to get it up to 
> nine-digit numbers, in which case it's significantly slow.  I was 
> wondering if there is a more efficient way to do this.
>
> I would highly appreciate it if anyone could help.

This appears to give the same results and is significantly faster.

def vers1():
    c = 1
    for p0 in range(0, 7):
        for p1 in range(p0+1, 12):
            for p2 in range(p1+1, 12):
                for p3 in range(p2+1, 12):
                    for p4 in range(p3+1, 12):
                        for p5 in range(p4+1, 12):
                            print repr(c).rjust(3), "\t",
                            print "%X %X %X %X %X %X" % (p0, p1, p2, p3, 
p4, p5)
                            c += 1
    print "...Done"


>
> Many thanks.
>
> -Kye
>
--
Jeff