fifo queue

[drochom]

hi,

how would u improve this code?

class queue:
def __init__(self, size=8):
self.space = size
self.data = [None]*self.space
self.head = 0
self.tail = 0
self.len = 0
def __len__(self): return self.len
def push(self, x):
if self.len==self.space:
self.data.extend( self.data[:self.tail] )
self.data.extend( [None]* (self.space-self.tail) )
self.tail+=self.space
self.space*=2
self.data[self.tail]=x
self.tail+=1
if self.tail==self.space:
self.tail=0
self.len+=1
def pop(self):
if self.len:
elem = self.data[self.head]
self.head+=1
if self.head==self.space:
self.head=0
return elem
def top(self):
if self.len==0:
raise Exception, 'queue is empty'
return self.data[self.head]

thx in adv.

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Re: where function

[drochom]

On 18 Mar, 15:19, [EMAIL PROTECTED] wrote:
 Is there a function in Python analogous to the where function in
 IDL?

 x=[0,1,2,3,4,2,8,9]
 print where(x=2)

 output:
 [2,5]

You can try this:

print filter( lambda x: a[x]==2, range(len(a)))

However it's not the best solution...

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Re: a good algo to do this

[drochom]

hi,

try this:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/190465

cheers.

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Re: duplicate items in a list

[drochom]

http://groups.google.com/group/comp.lang.python/browse_thread/thread/32e545ebba11dd4d/49a9f0cc799cc1f1#49a9f0cc799cc1f1

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Re: Removing duplicates from a list

[drochom]

Rubinho napisal(a):
 I've a list with duplicate members and I need to make each entry
 unique.


hi,

other possibility (my newest discovery:) )

 a = [1,2,2,4,2,1,3,4]
 unique = d.fromkeys(a).keys()
 unique
[1, 2, 3, 4]

regards
przemek

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Re: Removing duplicates from a list

[drochom]

there wasn't any information about ordering...
maybe i'll find something better which don't destroy original ordering

regards
przemek

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Re: Removing duplicates from a list

[drochom]

i suppose this one is faster (but in most cases efficiency doesn't
matter)

 def stable_unique(s):
e = {}
ret = []
for x in s:
if not e.has_key(x):
e[x] = 1
ret.append(x)
return ret

cheers,
przemek

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Re: Removing duplicates from a list

[drochom]

thanks, nice job. but this benchmark is pretty deceptive:

try this:
(definition of unique2 and unique3 as above)

 import timeit
 a = range(1000)
 t = timeit.Timer('unique2(a)','from __main__ import unique2,a')
 t2 = timeit.Timer('stable_unique(a)','from __main__ import stable_unique,a')
 t2.timeit(2000)
1.8392596235778456
 t.timeit(2000)
51.52945844819817

unique2 has quadratic complexity
unique3 has amortized linear complexity
what it means?
it means that speed of your algorithm strongly depends on
len(unique2(a)). the greater distinct elements in a the greater
difference in execution time of both implementations

regards
przemek

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