def removedups(s):
    """Return a list of the elements in s, but without duplicates.
    Thanks to Tim Peters for fast method.
    """
    n = len(s)
    if n == 0:
        return []
    u = {}
    try:
        for x in s:
            u[x] = 1
    except TypeError:
        del u  # move on to the next method
    else:
        return list(u.keys())
    # We can't hash all the elements.  Second fastest is to sort,
    # which brings the equal elements together; then duplicates are
    # easy to weed out in a single pass.
    try:
        t = list(s)
        t.sort()
    except __HOLE__:
        del t  # move on to the next method
    else:
        assert n > 0
        last = t[0]
        lasti = i = 1
        while i < n:
            if t[i] != last:
                t[lasti] = last = t[i]
                lasti = lasti + 1
            i = i + 1
        return t[:lasti]
    # Brute force is all that's left.
    u = []
    for x in s:
        if x not in u:
            u.append(x)
    return u
dataset/ETHPy150Open kdart/pycopia/aid/pycopia/aid.py/removedups