Access dictionary in order?

[Diego Dainese]

On Mon, 26 Jul 1999 07:45:29 -0700, Emile van Sebille wrote:
[...]
> > But how much time requires the keylist generation?
> 
> I tried this test and found that the results varied on the number of
> items.  For example,
[snip]

You are right, my question was a bit naive...

> > That is, the keylist is made traversing all the dictionary nodes or is
> > cached inside it?
> 
> I can't help you with this one...

AFAICT (from you test program) the nodes are linked in a list and the
key-sequence is generated traversing this list. Moreover the sequence
is not cached:

  import time
  d = {}
  r = ('spam','eggs','eggs','bacon','eggs','and','spam')
  k = []
  for i in xrange(1000000):
    d[i] = i, 2*i, r
    k.append(i)
  
  test1start = time.clock()
  k = d.keys()
  test1end = test2start = time.clock()
  k = d.keys()
  test2end = time.clock()
  
  print 'test1: ',test1end - test1start
  print 'test2: ',test2end - test2start
  
The last 2 prints give always the same results (about 0.8 seconds on
my Linux box).

> > --
> > Sorry for my bad English!
> 
> Not bad at all.

Thanks :)

+++

Thanks for the reply (and to Skip Montanaro too), bye

-- 
Diego Dainese
--
To reply remove the numbers and the `x' from my address
--
Sorry for my bad English!