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Hi,
I have to lists that I need to find the common numbers (2nd rounded to
nearest integral) and I am wondering if there is a more efficient way of
doing it. a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) ==
(l,round(m))]
[(123, 1.0), (123, 2.0), (123, 8.0)]
This works but a and b can be in the order of 30K long.
A couple of other bits of info.
 a and b are ordered smallest to largest (could bisect module be used?)
 in the future I will want to round the second number of closest 0.25
rather than whole number.
Would the sets module be more efficient?
I'm using python 2.3.
Thanks for any ideas.
Regards,
Gordon Williams  
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> A couple of other bits of info.  a and b are ordered smallest to largest (could bisect module be used?)  in the future I will want to round the second number of closest 0.25 rather than whole number.
Would the sets module be more efficient?
I'm using python 2.3.
I'd go for something that uses the rounded versions of the lists and then
iterates the first list and lets the second "cach up". Sorry, I'm to lazy
to desribe it better, so here is the code:
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)]
b= [(123, 0.9), (123, 1.9), (123, 8.0)]
a = [ (i,round(j)) for i,j in a]
b = [ (i,round(j)) for i,j in b]
res = []
pos_b = 0
try:
for i, pivot in a:
while b[pos_b][1] < pivot:
pos_b += 1
while b[pos_b][1] == pivot:
res.append(b[pos_b])
pos_b += 1
except IndexError:
# If b gets exhausted somewhere
pass
print res
While it looks more complicated, it certainly is faster, as its complexity
is in O(max(len(a), len(b))) where your code was O(len(a) * len(b))  so
usually more or less quadratic.
The speed gain comes of course from the order of the elements. And you could
factor the rounding _into_ the loops, but thats more ugly.

Regards,
Diez B. Roggisch  
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Gordon Williams wrote: I have to lists that I need to find the common numbers (2nd rounded to nearest integral) and I am wondering if there is a more efficient way of doing it.
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))] [(123, 1.0), (123, 2.0), (123, 8.0)]
[snip] Would the sets module be more efficient?
Well, in Python 2.3, I believe sets are implemented in Python while
they're implemented in C in Python 2.4. So probably not, unless you
upgrade. A 2.4 solution with sets: a = [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b = [(123, 0.9), (123, 1.9), (123, 8.0)] def roundedj(pairs_iterable):
.... return ((i, round(j)) for i, j in pairs_iterable)
.... set(roundedj(a)).intersection(set(roundedj(b)))
set([(123, 8.0), (123, 2.0), (123, 1.0)])
Steve  
P: n/a

> A couple of other bits of info.  a and b are ordered smallest to largest (could bisect module be used?)  in the future I will want to round the second number of closest 0.25 rather than whole number.
Would the sets module be more efficient?
I'm using python 2.3.
I'd go for something that uses the rounded versions of the lists and then
iterates the first list and lets the second "cach up". Sorry, I'm to lazy
to desribe it better, so here is the code:
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)]
b= [(123, 0.9), (123, 1.9), (123, 8.0)]
a = [ (i,round(j)) for i,j in a]
b = [ (i,round(j)) for i,j in b]
res = []
pos_b = 0
try:
for i, pivot in a:
while b[pos_b][1] < pivot:
pos_b += 1
while b[pos_b][1] == pivot:
res.append(b[pos_b])
pos_b += 1
except IndexError:
# If b gets exhausted somewhere
pass
print res
While it looks more complicated, it certainly is faster, as its complexity
is in O(max(len(a), len(b))) where your code was O(len(a) * len(b))  so
usually more or less quadratic.
The speed gain comes of course from the order of the elements. And you could
factor the rounding _into_ the loops, but thats more ugly.

Regards,
Diez B. Roggisch  
P: n/a

Gordon Williams wrote: I have to lists that I need to find the common numbers (2nd rounded to nearest integral) and I am wondering if there is a more efficient way of doing it.
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))] [(123, 1.0), (123, 2.0), (123, 8.0)]
[snip] Would the sets module be more efficient?
Well, in Python 2.3, I believe sets are implemented in Python while
they're implemented in C in Python 2.4. So probably not, unless you
upgrade. A 2.4 solution with sets: a = [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b = [(123, 0.9), (123, 1.9), (123, 8.0)] def roundedj(pairs_iterable):
.... return ((i, round(j)) for i, j in pairs_iterable)
.... set(roundedj(a)).intersection(set(roundedj(b)))
set([(123, 8.0), (123, 2.0), (123, 1.0)])
Steve  
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Steven Bethard wrote: Well, in Python 2.3, I believe sets are implemented in Python while they're implemented in C in Python 2.4.
I think the Python 2.3 Sets implementation is likely to be quicker than
whatever listmanipulation answer you come up with instead. But there's
only one way to find out ;)

Michael Hoffman  
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Steven Bethard wrote: Well, in Python 2.3, I believe sets are implemented in Python while they're implemented in C in Python 2.4.
I think the Python 2.3 Sets implementation is likely to be quicker than
whatever listmanipulation answer you come up with instead. But there's
only one way to find out ;)

Michael Hoffman  
P: n/a

Gordon Williams wrote: a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))]
d = {}
for (l, m) in b:
d[l, round(m)] = 1
result = []
for (i, j) in a:
t = (i, round(j))
if t in d:
result.append(t)
 in the future I will want to round the second number of closest 0.25 rather than whole number.
I would do that by multiplying by 4 and rounding to
an integer to derive the dictionary key. That will
avoid any floatrepresentation problems you might have
by trying to round to a fraction.
Would the sets module be more efficient?
As another poster said, sets are implemented as dicts
in 2.3, so it comes down to much the same thing. Using
sets might be a bit faster than the above code in 2.4,
but probably not greatly so. By far the biggest
improvement will come from using an O(n) algorithm
instead of an O(n**2) one.

Greg Ewing, Computer Science Dept,
University of Canterbury,
Christchurch, New Zealand http://www.cosc.canterbury.ac.nz/~greg  
P: n/a

Gordon Williams wrote: a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))]
d = {}
for (l, m) in b:
d[l, round(m)] = 1
result = []
for (i, j) in a:
t = (i, round(j))
if t in d:
result.append(t)
 in the future I will want to round the second number of closest 0.25 rather than whole number.
I would do that by multiplying by 4 and rounding to
an integer to derive the dictionary key. That will
avoid any floatrepresentation problems you might have
by trying to round to a fraction.
Would the sets module be more efficient?
As another poster said, sets are implemented as dicts
in 2.3, so it comes down to much the same thing. Using
sets might be a bit faster than the above code in 2.4,
but probably not greatly so. By far the biggest
improvement will come from using an O(n) algorithm
instead of an O(n**2) one.

Greg Ewing, Computer Science Dept,
University of Canterbury,
Christchurch, New Zealand http://www.cosc.canterbury.ac.nz/~greg  
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On Thu, 20041202 at 22:16, Greg Ewing wrote: Gordon Williams wrote:>a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] >b= [(123, 0.9), (123, 1.9), (123, 8.0)] >[ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))]
d = {} for (l, m) in b: d[l, round(m)] = 1
result = [] for (i, j) in a: t = (i, round(j)) if t in d: result.append(t)
 in the future I will want to round the second number of closest 0.25 rather than whole number.
I would do that by multiplying by 4 and rounding to an integer to derive the dictionary key. That will avoid any floatrepresentation problems you might have by trying to round to a fraction.
Would the sets module be more efficient?
As another poster said, sets are implemented as dicts in 2.3, so it comes down to much the same thing. Using sets might be a bit faster than the above code in 2.4, but probably not greatly so. By far the biggest improvement will come from using an O(n) algorithm instead of an O(n**2) one.
Of course a low Ofactor is important; you should avoid however
confusing the statement of what you want to do with the statement of how
you want to do it. One of the benefits of a HLL like Python is you can
merely state *what* you want without worrying about how to compute it.
In the original example above you are computing a set intersection 
python's set object has an intersection method. Use it. Not only is it
faster than your O**2 solution, but it is a good deal clearer. from sets import Set set_a = Set( [(i,round(j)) for i,j in a] ) set_b = Set( [(i,round(j)) for i,j in b] ) set_a.intersection( set_b )
Set([(123, 2.0), (123, 1.0), (123, 8.0)])
Or you could say ...
set_a, set_b = [[Set((i,round(j))) for i,j in s] for s in (a,b )]
Adam DePrince  
P: n/a

On Thu, 20041202 at 22:16, Greg Ewing wrote: Gordon Williams wrote:>a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] >b= [(123, 0.9), (123, 1.9), (123, 8.0)] >[ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))]
d = {} for (l, m) in b: d[l, round(m)] = 1
result = [] for (i, j) in a: t = (i, round(j)) if t in d: result.append(t)
 in the future I will want to round the second number of closest 0.25 rather than whole number.
I would do that by multiplying by 4 and rounding to an integer to derive the dictionary key. That will avoid any floatrepresentation problems you might have by trying to round to a fraction.
Would the sets module be more efficient?
As another poster said, sets are implemented as dicts in 2.3, so it comes down to much the same thing. Using sets might be a bit faster than the above code in 2.4, but probably not greatly so. By far the biggest improvement will come from using an O(n) algorithm instead of an O(n**2) one.
Of course a low Ofactor is important; you should avoid however
confusing the statement of what you want to do with the statement of how
you want to do it. One of the benefits of a HLL like Python is you can
merely state *what* you want without worrying about how to compute it.
In the original example above you are computing a set intersection 
python's set object has an intersection method. Use it. Not only is it
faster than your O**2 solution, but it is a good deal clearer. from sets import Set set_a = Set( [(i,round(j)) for i,j in a] ) set_b = Set( [(i,round(j)) for i,j in b] ) set_a.intersection( set_b )
Set([(123, 2.0), (123, 1.0), (123, 8.0)])
Or you could say ...
set_a, set_b = [[Set((i,round(j))) for i,j in s] for s in (a,b )]
Adam DePrince  
P: n/a

Michael Hoffman wrote: Steven Bethard wrote:
Well, in Python 2.3, I believe sets are implemented in Python while they're implemented in C in Python 2.4. I think the Python 2.3 Sets implementation is likely to be quicker than whatever listmanipulation answer you come up with instead. But there's only one way to find out ;)
Yeah, almost certainly since he's looking at lists 3K long. If they
were small, you never know since the list comprehension gets the Ccode
speedup, while sets.Set is Python code:
python m timeit s "a = [(123,1.3),(123,2.4),(123,7.8),(123,10.2)];
b = [(123, 0.9), (123, 1.9), (123, 8.0)]" "[ (i,round(j)) for i,j in a
for l,m in b if (i,round(j)) == (l,round(m))]"
10000 loops, best of 3: 27.5 usec per loop
python m timeit s "import sets; a =
[(123,1.3),(123,2.4),(123,7.8),(123,10.2)]; b = [(123, 0.9), (123, 1.9
), (123, 8.0)]" "sets.Set([(i,round(j)) for i,j in
a]).intersection(sets.Set([(i, round(j)) for i, j in b]))"
10000 loops, best of 3: 47.7 usec per loop
In the case given, the O(n**2) list comprehension is faster than the
O(n) set intersection. Of course, this is not likely to be true with
any reasonable sized data. But it's something worth keeping in mind.
Steve  
P: n/a

Michael Hoffman wrote: Steven Bethard wrote:
Well, in Python 2.3, I believe sets are implemented in Python while they're implemented in C in Python 2.4. I think the Python 2.3 Sets implementation is likely to be quicker than whatever listmanipulation answer you come up with instead. But there's only one way to find out ;)
Yeah, almost certainly since he's looking at lists 3K long. If they
were small, you never know since the list comprehension gets the Ccode
speedup, while sets.Set is Python code:
python m timeit s "a = [(123,1.3),(123,2.4),(123,7.8),(123,10.2)];
b = [(123, 0.9), (123, 1.9), (123, 8.0)]" "[ (i,round(j)) for i,j in a
for l,m in b if (i,round(j)) == (l,round(m))]"
10000 loops, best of 3: 27.5 usec per loop
python m timeit s "import sets; a =
[(123,1.3),(123,2.4),(123,7.8),(123,10.2)]; b = [(123, 0.9), (123, 1.9
), (123, 8.0)]" "sets.Set([(i,round(j)) for i,j in
a]).intersection(sets.Set([(i, round(j)) for i, j in b]))"
10000 loops, best of 3: 47.7 usec per loop
In the case given, the O(n**2) list comprehension is faster than the
O(n) set intersection. Of course, this is not likely to be true with
any reasonable sized data. But it's something worth keeping in mind.
Steve  
P: n/a

"Gordon Williams" <g_****@cyberus.ca> wrote in message
news:ma**************************************@pyth on.org... Hi,
I have to lists that I need to find the common numbers (2nd rounded to nearest integral) and I am wondering if there is a more efficient way of doing it.
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))] [(123, 1.0), (123, 2.0), (123, 8.0)] This works but a and b can be in the order of 30K long.
A couple of other bits of info.  a and b are ordered smallest to largest (could bisect module be used?)  in the future I will want to round the second number of closest 0.25 rather than whole number.
Would the sets module be more efficient?
Yes: set((x,round(y)) for x,y in a) & set((x,round(y)) for x,y in b)
set([(123, 8.0), (123, 2.0), (123, 1.0)])
I'm using python 2.3.
from sets import Set as set set([(x,round(y)) for x,y in a]) & set([(x,round(y)) for x,y in b])
set([(123, 8.0), (123, 2.0), (123, 1.0)])
Raymond Hettinger  
P: n/a

"Gordon Williams" <g_****@cyberus.ca> wrote in message
news:ma**************************************@pyth on.org... Hi,
I have to lists that I need to find the common numbers (2nd rounded to nearest integral) and I am wondering if there is a more efficient way of doing it.
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) == (l,round(m))] [(123, 1.0), (123, 2.0), (123, 8.0)] This works but a and b can be in the order of 30K long.
A couple of other bits of info.  a and b are ordered smallest to largest (could bisect module be used?)  in the future I will want to round the second number of closest 0.25 rather than whole number.
Would the sets module be more efficient?
Yes: set((x,round(y)) for x,y in a) & set((x,round(y)) for x,y in b)
set([(123, 8.0), (123, 2.0), (123, 1.0)])
I'm using python 2.3.
from sets import Set as set set([(x,round(y)) for x,y in a]) & set([(x,round(y)) for x,y in b])
set([(123, 8.0), (123, 2.0), (123, 1.0)])
Raymond Hettinger  
P: n/a

Steven Bethard wrote: Yeah, almost certainly since he's looking at lists 3K long. If they were small, you never know since the list comprehension gets the Ccode speedup, while sets.Set is Python code:
[list comprehension] 10000 loops, best of 3: 27.5 usec per loop
[Python 2.3 Set] 10000 loops, best of 3: 47.7 usec per loop
In the case given, the O(n**2) list comprehension is faster than the O(n) set intersection. Of course, this is not likely to be true with any reasonable sized data. But it's something worth keeping in mind.
Of course if you're working with a dataset that small, it probably
doesn't really matter which of these implementations you use.
The exception would be if this were in an inner loop in the actual
program and *were* being run 10000 times or more.

Michael Hoffman  
P: n/a

Steven Bethard wrote: Yeah, almost certainly since he's looking at lists 3K long. If they were small, you never know since the list comprehension gets the Ccode speedup, while sets.Set is Python code:
[list comprehension] 10000 loops, best of 3: 27.5 usec per loop
[Python 2.3 Set] 10000 loops, best of 3: 47.7 usec per loop
In the case given, the O(n**2) list comprehension is faster than the O(n) set intersection. Of course, this is not likely to be true with any reasonable sized data. But it's something worth keeping in mind.
Of course if you're working with a dataset that small, it probably
doesn't really matter which of these implementations you use.
The exception would be if this were in an inner loop in the actual
program and *were* being run 10000 times or more.

Michael Hoffman  
P: n/a
 Hi,
I have to lists that I need to find the common numbers (2nd rounded to nearest integral) and I am wondering if there is a more efficient way of doing it.
a= [(123,1.3),(123,2.4),(123,7.8),(123,10.2)] b= [(123, 0.9), (123, 1.9), (123, 8.0)] [ (i,round(j)) for i,j in a for l,m in b if (i,round(j)) ==
(l,round(m))] [(123, 1.0), (123, 2.0), (123, 8.0)]
Thanks for all your suggestions. I've tried each one with lists of 1K, 10K
and 30K long and tabulated the results below. Run with profile on W2K,
python 2.3.2, 1GHz Athlon.
1K, 10K and 30K long (seconds per call)
t1= 0.009, 0.148, 0.563
t2= 0.015, 0.217, 0.777
t3= 0.008, 0.108, 0.487
t4= 0.016, 0.190, 0.749
t5= 0.015, 0.224, 0.773
The nonset algorithims (t1,t3) came out the winners (maybe due to the
conversion of the set to a sorted list. I didn't look into it any farther.)
Regards,
Gordon Williams

from sets import Set
import random
size = 1000
a = [(123,i+random.choice([.2,.1,.1,.2])) for i in range(size)]
b = [(123, 1+i+random.choice([.2,.1,.1,.2])) for i in range(size)]
def t1():
#Diez B. Roggisch <de*********@web.de>
ra = [ (i,round(j)) for i,j in a]
rb = [ (i,round(j)) for i,j in b]
res = []
pos_b = 0
try:
for i, pivot in ra:
while rb[pos_b][1] < pivot:
pos_b += 1
while rb[pos_b][1] == pivot:
res.append(rb[pos_b])
pos_b += 1
except IndexError:
# If b gets exhausted somewhere
pass
return res
def t2():
#Steven Bethard <st************@gmail.com>
def roundedj(pairs_iterable):
return [(i, round(j)) for i, j in pairs_iterable]
l=list(Set(roundedj(a)).intersection(Set(roundedj( b))))
l.sort()
return l
def t3():
#Greg Ewing <gr**@cosc.canterbury.ac.nz>
d = {}
for (l, m) in b:
d[l, round(m)] = 1
result = []
for (i, j) in a:
t = (i, round(j))
if t in d:
result.append(t)
return result
def t4():
#Adam DePrince <ad**@cognitcorp.com>
set_a = Set( [(i,round(j)) for i,j in a] )
set_b = Set( [(i,round(j)) for i,j in b] )
l= list(set_a.intersection( set_b ))
l.sort()
return l
def t5():
#Raymond Hettinger <vz******@verizon.net>
l= list(Set([(x,round(y)) for x,y in a]) & Set([(x,round(y)) for x,y in
b]))
l.sort()
return l
def test():
r1= t1()
r2= t2()
r3= t3()
r4= t4()
r5= t5()   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 2333
 replies: 17
 date asked: Jul 18 '05
