Ordering things in python...?

Question

I was under the impression that set() would order a collection much like .sort()

However it seems that it doesn't, what was peculiar to me was why it reorders the collection.

>>> h = '321'
>>> set(h)
set(['1', '3', '2'])
>>> h
'321'
>>> h = '22311'
>>> set(h)
set(['1', '3', '2'])

why doesn't it return set(['1', '2', '3']). I also seems that no matter how many instances of each number I user or in what order I use them it always return set(['1', '3', '2']). Why?

Edit:

So I have read your answers and my counter to that is this.

>>> l = [1,2,3,3]
>>> set(l)
set([1, 2, 3])
>>> l = [3,3,2,3,1,1,3,2,3]
>>> set(l)
set([1, 2, 3])

Why does it order numbers and not strings?

Also

import random
l = []
for itr in xrange(101):
 l.append(random.randint(1,101))
print set(l)

Outputs

>>> 
set([1, 2, 4, 5, 6, 8, 10, 11, 12, 14, 15, 16, 18, 19, 23, 24, 25, 26, 29, 30, 31, 32, 34, 40, 43, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 69, 70, 74, 75, 77, 79, 80, 83, 84, 85, 87, 88, 89, 90, 93, 94, 96, 97, 99, 101])

Answer 1

python set is unordered, hence there is no guarantee that the elements would be ordered in the same way as you specify them

If you want a sorted output, then call sorted:

sorted(set(h))

Responding to your edit: it comes down to the implementation of set. In CPython, it boils down to two things:

1) the set will be sorted by hash (the __hash__ function) modulo a limit

2) the limit is generally the next largest power of 2

So let's look at the int case:

x=1
type(x) # int
x.__hash__() # 1

for ints, the hash equals the original value:

[x==x.__hash__() for x in xrange(1000)].count(False) # = 0

Hence, when all the values are ints, it will use the integer hash value and everything works smoothly.

for the string representations, the hashes dont work the same way:

x='1'
type(x) 
# str
x.__hash__()
# 6272018864

To understand why the sort breaks for ['1','2','3'], look at those hash values:

[str(x).__hash__() for x in xrange(1,4)]
# [6272018864, 6400019251, 6528019634]

In our example, the mod value is 4 (3 elts, 2^1 = 2, 2^2 = 4) so

[str(x).__hash__()%4 for x in xrange(1,4)]
# [0, 3, 2]
[(str(x).__hash__()%4,str(x)) for x in xrange(1,4)]
# [(0, '1'), (3, '2'), (2, '3')]

Now if you sort this beast, you get the ordering that you see in set:

[y[1] for y in sorted([(str(x).__hash__()%4,str(x)) for x in xrange(1,4)])]
# ['1', '3', '2']

Answer 2

From the python documentation of the set type:

A set object is an unordered collection of distinct hashable objects.

This means that the set doesn't have a concept of the order of the elements in it. You should not be surprised when the elements are printed on your screen in an unusual order.

Answer 3

A set in Python tries to be a "set" in the mathematical sense of the term. No duplicates, and order shouldn't matter.