Generating unique, ordered Pythagorean triplets

Question

This is a program I wrote to calculate Pythagorean triplets. When I run the program it prints each set of triplets twice because of the if statement. Is there any way I can

Answer 1

A non-numpy version of the Hall/Roberts approach is

def pythag3(limit=None, all=False):
    """generate Pythagorean triples which are primitive (default)
    or without restriction (when ``all`` is True). The elements
    returned in the tuples are sorted with the smallest first.

    Examples
    ========

    >>> list(pythag3(20))
    [(3, 4, 5), (8, 15, 17), (5, 12, 13)]
    >>> list(pythag3(20, True))
    [(3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20), (8, 15, 17), (5, 12, 13)]

    """
    if limit and limit < 5:
        return
    m = [(3,4,5)]  # primitives stored here
    while m:
        x, y, z = m.pop()
        if x > y:
            x, y = y, x
        yield (x, y, z)
        if all:
            a, b, c = x, y, z
            while 1:
                c += z
                if c > limit:
                    break
                a += x
                b += y
                yield a, b, c
        # new primitives
        a = x - 2*y + 2*z, 2*x - y + 2*z, 2*x - 2*y + 3*z
        b = x + 2*y + 2*z, 2*x + y + 2*z, 2*x + 2*y + 3*z
        c = -x + 2*y + 2*z, -2*x + y + 2*z, -2*x + 2*y + 3*z
        for d in (a, b, c):
            if d[2] <= limit:
                m.append(d)

It's slower than the numpy-coded version but the primitives with largest element less than or equal to 10^6 are generated on my slow machine in about 1.4 seconds. (And the list m never grew beyond 18 elements.)