efficient idiomatic queue?
aleax at aleax.it
Tue Jan 15 09:46:51 CET 2002
"David Eppstein" <eppstein at ics.uci.edu> wrote in message
news:eppstein-27FE15.16175514012002 at news.service.uci.edu...
> What is the idiomatic way in Python of implementing the following
> operations on a list of elements:
> - return the element at the front of the list, without changing the list
> - remove the element from the front of the list
> - add a new element to the rear of the list
> My first choice would be L, del L, and L.append(), however it
> seems from some brief experiments that the del operation takes time linear
> in the length of the list, so this would be very inefficient for long
> lists. I also tried replacing the del with L=L[1:] but that wasn't any
> better. A third possibility would be to have to variables L and i, and
> implement the three operations with L[i], i=i+1, and L.append(), which is
> fast but not good in space usage when the total number of operations is
> much larger than the length of the list.
> Is there a simple Pythonic way of implementing these three operations in
> constant (amortized) time per operation and space linear in the length of
> the list?
What about (warning, untested and particularly un-profiled code):
self.data = 
self.first = 0
self.first += 1
if self.first > len(self.data)/2:
self.data = self.data[self.first:]
self.first = 0
def append(self, value):
I think this should be constant-amortized-time, and never take more
space than K+2*N where N is the number of items in the list at a
given time. It may be possible to find sequences of pop and append
operations that degrade per-operation time beyond amortized constant.
Such anomalies, in similar cases, are generally ameliorated by some
further guard against too much churning in the data structure, e.g.
if self.first>self.Konstant and self.first>len(self.data)/2:
as the guard-clause for the self.data rebinding -- any constant
value self.Konstant will not matter asymptotically in terms of
space taken, of course (it goes into the K of "K+2*N").
More information about the Python-list