[Catalog-sig] Questions about efficiency.

Luis Leonel Lopez luisleonellopez@hotmail.com
Sun, 27 May 2001 22:48:51 -0000

Dear friends,

I wrote two functions which receive a sequence and return a list with 
non-duplicate elements. As per "unique" function's Tim Peters (as is in 
Python Cookbook:

def unique(s):
    """Return a list of the elements in s, but without duplicates.

    For example, unique([1,2,3,1,2,3]) is some permutation of [1,2,3],
    unique("abcabc") some permutation of ["a", "b", "c"], and
    unique(([1, 2], [2, 3], [1, 2])) some permutation of
    [[2, 3], [1, 2]].

    For best speed, all sequence elements should be hashable.  Then
    unique() will usually work in linear time.

    If not possible, the sequence elements should enjoy a total
    ordering, and if list(s).sort() doesn't raise TypeError it's
    assumed that they do enjoy a total ordering.  Then unique() will
    usually work in O(N*log2(N)) time.

    If that's not possible either, the sequence elements must support
    equality-testing.  Then unique() will usually work in quadratic

    n = len(s)
    if n == 0:
        return []

    # Try using a dict first, as that's the fastest and will usually
    # work.  If it doesn't work, it will usually fail quickly, so it
    # usually doesn't cost much to *try* it.  It requires that all the
    # sequence elements be hashable, and support equality comparison.
    u = {}
        for x in s:
            u[x] = 1
    except TypeError:
        del u  # move on to the next method
        return 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.
    # NOTE:  Python's list.sort() was designed to be efficient in the
    # presence of many duplicate elements.  This isn't true of all
    # sort functions in all languages or libraries, so this approach
    # is more effective in Python than it may be elsewhere.
        t = list(s)
    except TypeError:
        del t  # move on to the next method
        assert n > 0
        last = t[0]
        lasti = i = 1
        while i < n:
            if t[i] != last:
                t[lasti] = last = t[i]
                lasti += 1
            i += 1
        return t[:lasti]

    # Brute force is all that's left.
    u = []
    for x in s:
        if x not in u:
    return u

), my functions use the slowest method by brute force. I'd want to know why 
and which of my functions is better or efficient and why.

def onlyOne1(s)
	r = []
	for x in s:
		if x not in r:
	return r

def onlyOne2(s):
	r = []
	for x in s:
	return r

Thank you in advance!

Luis Leonel Lopez
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