Java's TreeSet equivalent in Python?

Question

I recently came across some Java code that simply put some strings into a Java TreeSet, implemented a distance based comparator for it, and then made its merry way into the sunset to compute a given score to solve the given problem.

My questions,

• Is there an equivalent data structure available for Python?

  • The Java treeset looks basically to be an ordered dictionary that can use a comparator of some sort to achieve this ordering.

• I see there's a PEP for Py3K for an OrderedDict, but I'm using 2.6.x. There are a bunch of ordered dict implementations out there - anyone in particular that can be recommended?

PS, Just to add - I could probably import DictMixin or UserDict and implement my own sorted/ordered dictionary, AND make it happen through a comparator function - but that seems to be overkill.

Thanks.

---

Update. Thanks for the answers. To elaborate a bit, lets say I've got a compare function thats defined like, (given a particular value ln),

def mycmp(x1, y1, ln):
  a = abs(x1-ln)
  b = abs(y1-ln)
  if a<b:
    return -1
  elif a>b:
    return 1
  else:
    return 0

I'm a bit unsure about how I'd integrate this into the ordering given in the ordered dict link given here...

Something like,

OrderedDict(sorted(d.items(), cmp=mycmp(len)))

Ideas would be welcome.

Answer 1

The Python 2.7 docs for collections.OrderedDict has a link to a OrderedDict recipe that runs on Python 2.4 or better.

Edit: In regard to sorting: Use key= rather than cmp=. It tends to lead to faster code and moreover, the cmp= keyword has been eliminated in Python3.

d={5:6,7:8,100:101,1:2,3:4}
print(d.items())
# [(1, 2), (3, 4), (100, 101), (5, 6), (7, 8)]

The code you posted for mycmp doesn't make it clear what you want passed as x1. Below, I assume x1 is supposed to be the value in each key-value pair. If so, you could do something like this:

length=4
print(sorted(d.items(),key=lambda item: abs(item[1]-length) ))
# [(3, 4), (1, 2), (5, 6), (7, 8), (100, 101)]

key=... is passed a function, lambda item: abs(item[1]-length). For each item in d.items(), the lambda function returns the number abs(item[1]-length). This number acts as proxy for the item as far as sorting is concerned. See this essay for more information on sorting idioms in Python.

PS. len is a Python builtin function. So as to not clobber that len, I've changed the variable name to length.

Answer 2

I recently implemented TreeSet for Python using bisect module.

https://github.com/fukatani/TreeSet

Its usage is similar to Java's Treeset.

ex.

from treeset import TreeSet
ts = TreeSet([3,7,2,7,1,3])
print(ts)
>>> [1, 2, 3, 7]

ts.add(4)
print(ts)
>>> [1, 2, 3, 4, 7]

ts.remove(7)
print(ts)
>>> [1, 2, 3, 4]

print(ts[2])
>>> 3

Answer 3

I'd need to see some example data, but if you're just trying to do a weighted sort, then the builtin python sorted() can do it, two ways.

With well ordered tuples and a key() function:

def cost_per_page(book):
    title, pagecount, cost = book
    return float(cost)/pagecount

booklist = [
        ("Grey's Anatomy", 3000, 200),
        ('The Hobbit', 300, 7.25),
        ('Moby Dick', 4000, 4.75),
]
for book in sorted(booklist, key=cost_per_page):
    print book

or with a class with a __cmp__ operator.

class Book(object):
    def __init__(self, title, pagecount, cost):
        self.title = title
        self.pagecount = pagecount
        self.cost = cost
    def pagecost(self):
        return float(self.cost)/self.pagecount
    def __cmp__(self, other):
        'only comparable with other books'
        return cmp(self.pagecost(), other.pagecost())
    def __str__(self):
        return str((self.title, self.pagecount, self.cost))

booklist = [
        Book("Grey's Anatomy", 3000, 200),
        Book('The Hobbit', 300, 7.25),
        Book('Moby Dick', 4000, 4.75),
]
for book in sorted(booklist):
    print book

Both of these return the same output:

('Moby Dick', 4000, 4.75)
('The Hobbit', 300, 7.25)
("Grey's Anatomy", 3000, 200)

Answer 4

1. I don't think python has a built-in Sorted sets. How about something like this?

letters = ['w', 'Z', 'Q', 'B', 'C', 'A']
  for l in sorted(set(letters)):
     print l

2.Java TreeSet is an implementation of the abstraction called SortedSet. Basic types will be sorted on natural order.A TreeSet instance performs all key comparisons using its compareTo (or compare) method.So your custom keys should implement proper compareTo

Answer 5

If what you want is a set that always iterates in sorted-order, this might get you most of the way there:

def invalidate_sorted(f):
    def wrapper(self, *args, **kwargs):
        self._sort_cache = None
        return f(self, *args, **kwargs)
    return wrapper

class SortedSet(set):
    _sort_cache = None

    _invalidate_sort_methods = """
        add clear difference_update discard intersection_update
        symmetric_difference_update pop remove update
        __iand__ __ior__ __isub__ __ixor__
        """.split()

    def __iter__(self):
        if not self._sort_cache:
            self._sort_cache = sorted(set.__iter__(self))
        for item in self._sort_cache:
            yield item

    def __repr__(self):
        return '%s(%r)' % (type(self).__name__, list(self))

    for methodname in _invalidate_sort_methods:
        locals()[methodname] = invalidate_sorted(getattr(set, methodname))