pairs from a list

[Alan Isaac]

I want to generate sequential pairs from a list.
Here is a way::

from itertools import izip, islice
for x12 in izip(islice(x,0 ,None,2),islice (x,1,None,2)):
print x12

(Of course the print statement is just illustrative.)
What is the fastest way? (Ignore the import time.)

Thanks,
Alan Isaac

[Paul Rubin]

Re: pairs from a list

Alan Isaac <aisaac@america n.eduwrites:

(Of course the print statement is just illustrative.)
What is the fastest way? (Ignore the import time.)

You have to try a bunch of different ways and time them. One
idea (untested):

def pairs(seq):
while True:
yield (seq.next(), seq.next())

[George Sakkis]

Re: pairs from a list

On Jan 21, 10:20 pm, Alan Isaac <ais...@america n.eduwrote:

I want to generate sequential pairs from a list.
Here is a way::
>
from itertools import izip, islice
for x12 in izip(islice(x,0 ,None,2),islice (x,1,None,2)):
print x12
>
(Of course the print statement is just illustrative.)
What is the fastest way? (Ignore the import time.)

Look up the timeit module and test yourself the various alternatives;
that's the most reliable way to tell for sure.

George

[Paddy]

Re: pairs from a list

On Jan 22, 3:20 am, Alan Isaac <ais...@america n.eduwrote:

I want to generate sequential pairs from a list.

<<snip>>

What is the fastest way? (Ignore the import time.)

1) How fast is the method you have?
2) How much faster does it need to be for your application?
3) Are their any other bottlenecks in your application?
4) Is this the routine whose smallest % speed-up would give the
largest overall speed up of your application?

- Paddy.

[George Sakkis]

Re: pairs from a list

On Jan 22, 12:15 am, Paddy <paddy3...@goog lemail.comwrote :

On Jan 22, 3:20 am, Alan Isaac <ais...@america n.eduwrote:I want to generate sequential pairs from a list.
<<snip>>

What is the fastest way? (Ignore the import time.)

>
1) How fast is the method you have?
2) How much faster does it need to be for your application?
3) Are their any other bottlenecks in your application?
4) Is this the routine whose smallest % speed-up would give the
largest overall speed up of your application?

I believe the "what is the fastest way" question for such small well-
defined tasks is worth asking on its own, regardless of whether it
makes a difference in the application (or even if there is no
application to begin with). Just because cpu cycles are cheap these
days is not a good reason to be sloppy. Moreover, often the fastest
pure Python version happens to be among the most elegant and concise,
unlike other languages where optimization usually implies obfuscation.

George

[Arnaud Delobelle]

Re: pairs from a list

On Jan 22, 3:20 am, Alan Isaac <ais...@america n.eduwrote:

I want to generate sequential pairs from a list.
Here is a way::
>
    from itertools import izip, islice
    for x12 in izip(islice(x,0 ,None,2),islice (x,1,None,2)):
        print x12
>
(Of course the print statement is just illustrative.)
What is the fastest way? (Ignore the import time.)
>
Thanks,
Alan Isaac

Don't know the fastest, but here's a very concise way:

from itertools import izip

def ipairs(seq):
it = iter(seq)
return izip(it, it)

>>list(pairs(xr ange(10)))

[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]

>>list(pairs('h ello'))

[('h', 'e'), ('l', 'l')]

--
Arnaud

[Alan Isaac]

Re: pairs from a list

I suppose my question should have been,
is there an obviously faster way?
Anyway, of the four ways below, the
first is substantially fastest. Is
there an obvious reason why?

Thanks,
Alan Isaac

PS My understanding is that the behavior
of the last is implementation dependent
and not guaranteed.

def pairs1(x):
for x12 in izip(islice(x,0 ,None,2),islice (x,1,None,2)):
yield x12

def pairs2(x):
xiter = iter(x)
while True:
yield xiter.next(), xiter.next()

def pairs3(x):
for i in range( len(x)//2 ):
yield x[2*i], x[2*i+1],

def pairs4(x):
xiter = iter(x)
for x12 in izip(xiter,xite r):
yield x12

[Arnaud Delobelle]

Re: pairs from a list

On Jan 22, 1:19 pm, Alan Isaac <ais...@america n.eduwrote:
[...]

PS My understanding is that the behavior
of the last is implementation dependent
and not guaranteed.

[...]

def pairs4(x):
    xiter = iter(x)
    for x12 in izip(xiter,xite r):
        yield x12

According to the docs [1], izip is defined to be equivalent to:

def izip(*iterables ):
iterables = map(iter, iterables)
while iterables:
result = [it.next() for it in iterables]
yield tuple(result)

This guarantees that it.next() will be performed from left to right,
so there is no risk that e.g. pairs4([1, 2, 3, 4]) returns [(2, 1),
(4, 3)].

Is there anything else that I am overlooking?

[1] http://docs.python.org/lib/itertools-functions.html

--
Arnaud

[Arnaud Delobelle]

Re: pairs from a list

On Jan 22, 1:19 pm, Alan Isaac <ais...@america n.eduwrote:

I suppose my question should have been,
is there an obviously faster way?
Anyway, of the four ways below, the
first is substantially fastest.  Is
there an obvious reason why?

Can you post your results?

I get different ones (pairs1 and pairs2 rewritten slightly to avoid
unnecessary indirection).

====== pairs.py ===========
from itertools import *

def pairs1(x):
return izip(islice(x,0 ,None,2),islice (x,1,None,2))

def pairs2(x):
xiter = iter(x)
while True:
yield xiter.next(), xiter.next()

def pairs3(x):
for i in range( len(x)//2 ):
yield x[2*i], x[2*i+1],

def pairs4(x):
xiter = iter(x)
return izip(xiter,xite r)

def compare():
import timeit
for i in '1234':
t = timeit.Timer('l ist(pairs.pairs %s(l))' % i,
'import pairs; l=range(1000)')
print 'pairs%s: %s' % (i, t.timeit(10000) )

if __name__ == '__main__':
compare()
=============== ======

marigold:python arno$ python pairs.py
pairs1: 0.789824962616
pairs2: 4.08462786674
pairs3: 2.90438890457
pairs4: 0.536775827408

pairs4 wins.

--
Arnaud

[Alan Isaac]

Re: pairs from a list

Arnaud Delobelle wrote:

According to the docs [1], izip is defined to be equivalent to:
>
def izip(*iterables ):
iterables = map(iter, iterables)
while iterables:
result = [it.next() for it in iterables]
yield tuple(result)
>
This guarantees that it.next() will be performed from left to right,
so there is no risk that e.g. pairs4([1, 2, 3, 4]) returns [(2, 1),
(4, 3)].
>
Is there anything else that I am overlooking?
>
[1] http://docs.python.org/lib/itertools-functions.html

<URL:http://bugs.python.org/issue1121416>

fwiw,
Alan Isaac

[Arnaud Delobelle]

Re: pairs from a list

On Jan 22, 4:10 pm, Alan Isaac <ais...@america n.eduwrote:

<URL:http://bugs.python.org/issue1121416>
>
fwiw,
Alan Isaac

Thanks. So I guess I shouldn't take the code snippet I quoted as a
specification of izip but rather as an illustration.

--
Arnaud

[Paddy]

Re: pairs from a list

On Jan 22, 5:34 am, George Sakkis <george.sak...@ gmail.comwrote:

On Jan 22, 12:15 am, Paddy <paddy3...@goog lemail.comwrote :
>

On Jan 22, 3:20 am, Alan Isaac <ais...@america n.eduwrote:I want to generate sequential pairs from a list.
<<snip>>

What is the fastest way? (Ignore the import time.)

>

1) How fast is the method you have?
2) How much faster does it need to be for your application?
3) Are their any other bottlenecks in your application?
4) Is this the routine whose smallest % speed-up would give the
largest overall speed up of your application?

>
I believe the "what is the fastest way" question for such small well-
defined tasks is worth asking on its own, regardless of whether it
makes a difference in the application (or even if there is no
application to begin with).

Hi George,
You need to 'get it right' first. Micro optimizations for speed
without thought of the wider context is a bad habit to form and a time
waster.
If the routine is all that needs to be delivered and it does not
perform at an acceptable speed then find out what is acceptable and
optimise towards that goal. My questions were set to get posters to
think more about the need for speed optimizations and where they
should be applied, (if at all).

A bit of forethought might justify leaving the routine alone, or
optimising for readability instead.

- Paddy.

[Arnaud Delobelle]

Re: pairs from a list

On Jan 22, 6:34 pm, Paddy <paddy3...@goog lemail.comwrote :
[...]

Hi George,
You need to 'get it right' first. Micro optimizations for speed
without thought of the wider context is a bad habit to form and a time
waster.
If the routine is all that needs to be delivered and it does not
perform at an acceptable speed then find out what is acceptable and
optimise towards that goal. My questions were set to get posters to
think more about the need for speed optimizations and where they
should be applied, (if at all).
>
A bit of forethought might justify leaving the routine alone, or
optimising for readability instead.

But it's fun!

Some-of-us-can't-help-it'ly yours
--
Arnaud

[Alan G Isaac]

Re: pairs from a list

Steven D'Aprano wrote:

In fact, "fastest" isn't even a meaningful attribute. Does it mean:
>
* the worst-case is fastest
* the best-case is fastest
* the average-case is fastest
* fastest on typical data
* all of the above

I confess that it did not occur to me that there
might be an interesting distinction among these
cases for the question of how to get sequential
pairs from a list. How would one draw these
distinctions in this case?

Thanks,
Alan Isaac

PS Just for context, the sequential pairs were
needed in a simulation, but my question was
primarily prompted by my surprise that the
approaches I looked at differed as much as they did.

[George Sakkis]

Re: pairs from a list

On Jan 23, 4:37 am, Steven D'Aprano
<ste...@REMOVE. THIS.cybersourc e.com.auwrote:

On Tue, 22 Jan 2008 23:33:00 -0800, George Sakkis wrote:

As I mentioned already, I consider the seeking of the most efficient
solution a legitimate question, regardless of whether a "dumb" solution
is fast enough for an application. Call it a "don't be sloppy" principle
if you wish.

>
Sure, by why do you limit "efficient" and "don't be sloppy" to mean
"write the fastest executing code you can, regardless of every other
trade-off"?

I explicitly didn't limit sloppiness to inefficiency and mentioned
it's a tradeoff:

"... all else being equal or at least comparable (elegance,
conciseness, readability, etc.). Of course it's a tradeoff;
spending a week to save a few milliseconds on average is usually a
waste for most applications, but being a lazy keyboard banger writing
the first thing that pops into mind is not that good either."

But... do you write list.__len__() instead of len(list) to save a few
nanoseconds?

No, of course not, it's not worth it, but that doesn't mean that being
curious about what's faster and using timeit to find out is totally
worthless.

Another example: avoiding attribute lookups within a loop. I rarely
write
bar = foo.bar
for i in big_list: bar(i)

but it's valuable to know that it can make a difference when I really
need it. Always writing the first thing that "just works" prevents one
from even considering that there might be faster (or more elegant,
more general, etc.) alternatives.

George