Java 中的 TreeSet 等效于 Python 吗？

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))

Answer 6

When you are coming with java treeset:

 import java.util.*;
class Main{
         public static void main(String args[])
          {
             TreeSet<Integer> tr=new TreeSet<>();
             tr.add(3);
             tr.add(5);
             tr.add(7);
             tr.add(6);
             tr.add(3);
             tr.add(8);

             Iterator itr=tr.iterator();
             for(int i=0;i<tr.size();i++)
            {
               System.out.print(tr.get(i)+" ");  
            } 
          }
     }

    >>>> **3 5 6 7 8**


  same AS in python:
from treeset import TreeSet
tr = TreeSet([1,2,2,7,4,3])
print(tr)
>>> [1, 2, 3, 4,7]