nearest neighbor in 2D

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John Hunter

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Nov 2, 2003, 11:12:47 PM11/2/03
to pytho...@python.org

I have a list of two tuples containing x and y coord

(x0, y0)
(x1, y1)
...
(xn, yn)

Given a new point x,y, I would like to find the point in the list
closest to x,y. I have to do this a lot, in an inner loop, and then I
add each new point x,y to the list. I know the range of x and y in
advance.

One solution that comes to mind is to partition to space into
quadrants and store the elements by quadrant. When a new element
comes in, identify it's quadrant and only search the appropriate
quadrant for nearest neighbor. This could be done recursively, a 2D
binary search of sorts....

Can anyone point me to some code or module that provides the
appropriate data structures and algorithms to handle this task
efficiently? The size of the list will likely be in the range of
10-1000 elements.

Thanks,
John Hunter


Isaac To

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Nov 3, 2003, 12:16:00 AM11/3/03
to
>>>>> "John" == John Hunter <jdhu...@ace.bsd.uchicago.edu> writes:

John> Given a new point x,y, I would like to find the point in the list
John> closest to x,y. I have to do this a lot, in an inner loop, and
John> then I add each new point x,y to the list. I know the range of x
John> and y in advance.

John> One solution that comes to mind is to partition to space into
John> quadrants and store the elements by quadrant. When a new element
John> comes in, identify it's quadrant and only search the appropriate
John> quadrant for nearest neighbor. This could be done recursively, a
John> 2D binary search of sorts....

By recursion your solution would work in O(log n) time. The construction
would take O(n log n) time. Unluckily, it can return the wrong point, as
the nearest point within the nearest quadrant might not be the nearest
point.

The problem is a well-studied basic computational geometry problem, although
I don't really know any Python code that actually do it. Try to look at the
web for "Voronoi diagrams" and "radial triangulation" to understand how to
solve it properly in the above mentioned (perhaps randomized) time
complexity.

Regards,
Isaac.

Noen

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Nov 3, 2003, 1:03:56 AM11/3/03
to
John Hunter wrote:

You could to a for loop, and inside that loop you will have a variable
lessest_distance. I dont know much geometric mathematics, but Im pretty
sure you can use pytagoras stuff to find the lenght from (Xn,Yn) to
(X,Y) using sinus cosinus and such.

And when the function is finished, you should return lessest_distance

Graham Lee

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Nov 3, 2003, 4:43:57 AM11/3/03
to
John Hunter wrote:

>
> One solution that comes to mind is to partition to space into
> quadrants and store the elements by quadrant. When a new element
> comes in, identify it's quadrant and only search the appropriate
> quadrant for nearest neighbor. This could be done recursively, a 2D
> binary search of sorts....
>

What happens when you put a particle in near/at the boundary of a quadrant
though? It's possible for the nearest neighbour to be in the nearest
neighbour quadrant...although you could search over these as well.
However, the number of independent areas implied by use of the word
'quadrant' suggests that this would be the same as iterating over all
space.... :-)
--
Graham Lee
Wadham College
Oxford

Anthony Briggs

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Nov 3, 2003, 6:54:30 AM11/3/03
to John Hunter, pytho...@python.org
At 10:12 PM -0600 2/11/03, John Hunter wrote:
>I have a list of two tuples containing x and y coord
>
> (x0, y0)
> (x1, y1)
> ...
> (xn, yn)
>
>Given a new point x,y, I would like to find the point in the list
>closest to x,y. I have to do this a lot, in an inner loop, and then I
>add each new point x,y to the list. I know the range of x and y in
>advance.

One method that you can use is to narrow down the list of candidates
by only considering those points within a box around your new point,
eg xn-5 < x < xn+5, yn-5 < y < yn+5. You'll still need to test using
trigonometric stuff after that, and also that there won't be a point
outside the square that'll be closer, ie. that your closest point is
< 5 away from the new point.

You might also consider sorting the list by distance from some other
point (eg. 0,0), and keeping it sorted as you add new points - it'll
take some time to do, but might make things faster overall when
searching.

Hope that helps,

Anthony
--
----------------------------------------------------
HyPEraCtiVE? HeY, WhO aRE YoU cALliNg HypERaCtIve?!
aBR...@wEStNeT.cOm.aU
----------------------------------------------------

Ron Adam

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Nov 3, 2003, 7:03:31 PM11/3/03
to


This is how I would do it. Maybe it's how you are already doing it?


import math
import random

n_points = 1000
max_x = 1000
max_y = 1000
closest_distance = 10000
closest_point = (max_x,max_y)

p = []
for i in xrange(n_points):
x = round(max_x*random.random())
y = round(max_y*random.random())
p.append((x, y))

new_point = (round(max_x*random.random()), \
round(max_y*random.random()))

for point in p:
distance = math.sqrt((new_point[0]-point[0])**2 \
+(new_point[1]-point[1])**2)
if distance < closest_distance:
closest_distance = distance
closest_point = point

print 'new_point:', new_point
print 'closest_point:', closest_point,' \
out of',n_points,'points.'


I really don't know how you can make this faster. There might be a
library that has a distance between two points function that could
speed it up.

Ronald R. Adam
rad...@tampabay.rr.com

David Eppstein

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Nov 3, 2003, 8:11:45 PM11/3/03
to
> >I have a list of two tuples containing x and y coord
> >
> > (x0, y0)
> > (x1, y1)
> > ...
> > (xn, yn)
> >
> >Given a new point x,y, I would like to find the point in the list
> >closest to x,y. I have to do this a lot, in an inner loop, and then I
> >add each new point x,y to the list. I know the range of x and y in
> >advance.

Here's some not-very-heavily-tested code for doing this using a kD-tree.

Worst case efficiency is still linear per point or quadratic total
(unlike some other more sophisticated data structures) but in practice
if your points are reasonably well behaved this should be pretty good;
e.g. I tried it with 10000 random points (each queried then added) and
it made only 302144 recursive calls to nearestNeighbor.

Also note that only the test code at the end restricts to two
dimensions, everything else works in arbitrary numbers of dimensions.

def dist2(p,q):
"""Squared distance between p and q."""
d = 0
for i in range(len(p)):
d += (p[i]-q[i])**2
return d

class kdtree:
def __init__(self,dim=2,index=0):
self.dim = dim
self.index = index
self.split = None

def addPoint(self,p):
"""Include another point in the kD-tree."""
if self.split is None:
self.split = p
self.left = kdtree(self.dim, (self.index + 1) % self.dim)
self.right = kdtree(self.dim, (self.index + 1) % self.dim)
elif self.split[self.index] < p[self.index]:
self.left.addPoint(p)
else:
self.right.addPoint(p)

def nearestNeighbor(self,q,maxdist2):
"""Find pair (d,p) where p is nearest neighbor and d is squared
distance to p. Returned distance must be within maxdist2; if
not, no point itself is returned.
"""
solution = (maxdist2+1,None)
if self.split is not None:
solution = min(solution, (dist2(self.split,q),self.split))
d2split = (self.split[self.index] - q[self.index])**2
if self.split[self.index] < p[self.index]:
solution = min(solution,
self.left.nearestNeighbor(q,solution[0]))
if d2split < solution[0]:
solution = min(solution,
self.right.nearestNeighbor(q,solution[0]))
else:
solution = min(solution,
self.right.nearestNeighbor(q,solution[0]))
if d2split < solution[0]:
solution = min(solution,
self.left.nearestNeighbor(q,solution[0]))
return solution

if __name__ == "__main__":
import math
import random

n_points = 50


max_x = 1000
max_y = 1000

max_dist2 = max_x**2 + max_y**2

k = kdtree()
for i in range(n_points):


x = round(max_x*random.random())
y = round(max_y*random.random())

p = (x,y)

if i == 0:
print 'new point',p
else:
d,q = k.nearestNeighbor(p,max_dist2)
print 'new point', p, 'has neighbor',
print q, 'at distance', math.sqrt(d)

k.addPoint(p)

--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science

Bengt Richter

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Nov 3, 2003, 9:49:10 PM11/3/03
to
On Sun, 02 Nov 2003 22:12:47 -0600, John Hunter <jdhu...@ace.bsd.uchicago.edu> wrote:

>
>I have a list of two tuples containing x and y coord
>
> (x0, y0)
> (x1, y1)
> ...
> (xn, yn)
>
>Given a new point x,y, I would like to find the point in the list
>closest to x,y. I have to do this a lot, in an inner loop, and then I
>add each new point x,y to the list. I know the range of x and y in
>advance.

Are you trying to find closest location to a mouse cursor as it moves,
and then adding a point when there's a click? I.e., your particular use case
might suggest a strategy that's different from, e.g., what you'd do if
each new point's coordinates where read from file or came from a generator,
and you had exactly one search leading to exactly one update of the set.
And also what you wanted to do with the completed set.

>
>One solution that comes to mind is to partition to space into
>quadrants and store the elements by quadrant. When a new element
>comes in, identify it's quadrant and only search the appropriate
>quadrant for nearest neighbor. This could be done recursively, a 2D
>binary search of sorts....

This might be a way of pruning, but you'd have to take into account that
nearest square doesn't guarantee nearest diagonal distance. Just blathering
off the top of my head, ... I think I would try dividing x and y into maybe a
16*16 grid of squares. A new point will fall into one of those, and then if
you find some existing points in that square, you could brute force find the
closest (remembering that comparing squared radial distances works as well as
comparing their square roots ;-) and then see if that shortest distance can
reach into any adjacent squares, and search those too if so, since there could be
a point just the other side of the border, or diagonally across an adjacent
corner that could be closer than your currently determined distance.

You could keep info about points in a square in lists or dicts (16*16 might
be sparsely populated, best in a dict of squares accessed by grid coordinates
(i.e., 4 bits apiece, maybe as tuple or combined as a single number (but then
you could use a list pre-populated with None's instead of a dict, so either way).

I guess in the extreme you could compute a complete table of nearest point
coordinates for every possible x,y point, so you'd have a raster map of voronoi
regions, with each region colored by the coordinates of its nearest point. The more
points you had, the less info would have to be updated for each new point.
I wonder when the crossover would occur ;-)


>
>Can anyone point me to some code or module that provides the
>appropriate data structures and algorithms to handle this task
>efficiently? The size of the list will likely be in the range of
>10-1000 elements.
>

What are the ranges of x and y?

Regards,
Bengt Richter

John Hunter

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Nov 3, 2003, 11:15:34 PM11/3/03
to Bengt Richter, pytho...@python.org
>>>>> "Bengt" == Bengt Richter <bo...@oz.net> writes:

Bengt> Are you trying to find closest location to a mouse cursor
Bengt> as it moves, and then adding a point when there's a click?
Bengt> I.e., your particular use case might suggest a strategy
Bengt> that's different from, e.g., what you'd do if each new
Bengt> point's coordinates where read from file or came from a
Bengt> generator, and you had exactly one search leading to
Bengt> exactly one update of the set. And also what you wanted to
Bengt> do with the completed set.

I had two use cases just yesterday. The one that prompted the
question arose in making a contour plot. I'm defining a contour as an
ordered sequence of values over a 2D MxN matrix where the values
differ from some target value by at most some tolerance. I maintain a
list of i,j indices into the matrix for a given contour value, and
follow the contour from a given i,j location by examining its
neighbors. In order to close the loop (eg, when the contour finder
has passed once around a level curve of a mountain, I want to test
whether a given point i,j is close to a previously discovered point
k,l. Since I have a list of these 2 tuple coordinates, I want to find
the nearest neighbor in the list and terminate the contour when the
nearest neighbor falls within some minimum distance

3 4 5
2 6
13 1 7
12 8
11 10 9

In the example above, I am traversing a contour identifying points in
the order 1,2,3...; as above each point represents an i,j tuple which
is an index into the matrix I am contouring. I would like to
terminate the search at 13 rather than spiral around the existing
contour 1-12. Each time I add a new point to the contour, I would like
to query the existing list (excluding the most recently added points
which are close by construction) of locations and find the minimum
distance. If I'm not too close to the preexisting contour, I add the
new point and proceed.

As I write this I realize there is an important efficiency. Since
from an existing point I add the closest neighbor, the biggest step I
can make is 1,1. If on the last nearest neighbor query I find a
minimum distance of d, it will take me d minimum steps to approach the
existing contour. So I don't need to check the distance again for at
least d steps. So the algorithm can proceed 1) obtain the distance d
from the existing contour to the most recently obtained point 2) make
d steps adding points that meet the value criteria 3) repeat.

The second use case arose with gdmodule, which can only allocate 256
colors, which I cache as a dict from rgb tuples (eg, 0.0, 0.05, 1.0)
to color. When the total number of color allocations is made, and a
new rgb request comes into the color manager, I pick the already
allocated point in rgb space closest to the requested point.

I'll try David Eppstein's approach tomorrow and see how this fares.

Thanks to all for suggestions,
John Hunter

Andrew Dalke

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Nov 4, 2003, 3:15:12 AM11/4/03
to
Ron Adam

> for point in p:
> distance = math.sqrt((new_point[0]-point[0])**2 \
> +(new_point[1]-point[1])**2)

> I really don't know how you can make this faster. There might be a


> library that has a distance between two points function that could
> speed it up.

An easy way is to move the math.sqrt call outside the loop, since
sqrt(d1) < sqrt(d2) iff d1 < d2 (when d1,d2>=0)

Andrew
da...@dalkescientific.com


G.J.Giezeman

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Nov 4, 2003, 3:32:38 AM11/4/03
to
Isaac To wrote:

A solution in C++ is using the CGAL-library (www.cgal.org). Look in the
index of the basic library and search for 'nearest'. It will point you
to Delaunay triangulations, which, together with a triangulation
hierarchy, will give O(log n) time complexity, except in pathological
cases. You can call C++ code from python.
B.t.w., there will be a new release of the CGAL library very soon
(probably this week).

Alex Martelli

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Nov 4, 2003, 10:20:39 AM11/4/03
to
Andrew Dalke wrote:

> Ron Adam
>> for point in p:
>> distance = math.sqrt((new_point[0]-point[0])**2 \
>> +(new_point[1]-point[1])**2)
>
>> I really don't know how you can make this faster. There might be a

Hmmm, that's what math.hypot is for, isn't it...?

[alex@lancelot Lib]$ timeit.py -c -s'import math; p=1.6,2.5; np=2.4,1.3'
'math.sqrt((np[0]-p[0])**2 + (np[1]-p[1])**2)'
100000 loops, best of 3: 3 usec per loop

[alex@lancelot Lib]$ timeit.py -c -s'import math; p=1.6,2.5; np=2.4,1.3'
'math.hypot(np[0]-p[0], np[1]-p[1])'
100000 loops, best of 3: 1.9 usec per loop


>> library that has a distance between two points function that could
>> speed it up.
>
> An easy way is to move the math.sqrt call outside the loop, since
> sqrt(d1) < sqrt(d2) iff d1 < d2 (when d1,d2>=0)

Yes, omitting the math.sqrt gives the same speed as calling math.hypot,
and it's the classic solution to speed up minimum-distance problems.

I vaguely suspect you could shave some further fraction of a microsecond
by saving those differences as dx and dy and then computing dx*dx+dy*dy --
since another classic tip is that a**2 is slower than a*2. Let's see...:

[alex@lancelot Lib]$ timeit.py -c -s'import math; p=1.6,2.5; np=2.4,1.3'
'dx=np[0]-p[0]; dy=np[1]-p[1]; disq=dx*dx+dy*dy'
1000000 loops, best of 3: 1.39 usec per loop

...yep, another small enhancement. Ain't measuring _FUN_?-)


Alex

Peter Otten

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Nov 4, 2003, 11:13:58 AM11/4/03
to al...@aleax.it
Alex Martelli wrote:

Finally found an application for complex numbers:

...> timeit.py -s"p= 1.6+2.5j; np=2.4+1.3j" "d=abs(p-np)"
1000000 loops, best of 3: 0.436 usec per loop

...> timeit.py -s"p= 1.6,2.5; np=2.4,1.3" "dx=np[0]-p[0];
dy=np[1]-p[1];d=dx*dx+dy*dy"
1000000 loops, best of 3: 1.15 usec per loop

This is of course all premature optimization as the most promising approach
is to try hard to reduce the number of candidate points, as David Eppstein
seems to have done. But then, he could use complex numbers, too.

Peter

David Eppstein

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Nov 4, 2003, 11:25:36 AM11/4/03
to
In article <bo8j4v$tqt$03$1...@news.t-online.com>,
Peter Otten <__pet...@web.de> wrote:

> This is of course all premature optimization as the most promising approach
> is to try hard to reduce the number of candidate points, as David Eppstein
> seems to have done. But then, he could use complex numbers, too.

Well, yes, but then my code wouldn't work very well in dimensions higher
than two...

Kirk Strauser

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Nov 4, 2003, 2:15:05 PM11/4/03
to
At 2003-11-03T04:12:47Z, John Hunter <jdhu...@ace.bsd.uchicago.edu> writes:

> One solution that comes to mind is to partition to space into quadrants
> and store the elements by quadrant. When a new element comes in, identify
> it's quadrant and only search the appropriate quadrant for nearest
> neighbor.

Erm, no. Imagine that your new point is in one corner of a quadrant. The
other point in the quadrant is in the opposite corner. There is a point in
the adjacent quadrant that is infinitessimaly close to your new point.
That's where your algorithm breaks down.
--
Kirk Strauser
The Day Companies

John Hunter

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Nov 4, 2003, 2:02:36 PM11/4/03
to Ron Adam, pytho...@python.org
>>>>> "Ron" == Ron Adam <rad...@tampabay.rr.com> writes:

Ron> I really don't know how you can make this faster. There
Ron> might be a library that has a distance between two points
Ron> function that could speed it up.

If you only had to compare one point to all the other points, then the
brute force approach -- check every point -- will work great. This is
O(N) and I don't think you can beat it. The idea is that I will be
repeatedly checking and adding points to the list, so it is worthwhile
at the outset to set up a data structure which allows a more efficient
search.

The analogy is a binary search in 1D. If you plan to repeatedly
search a (possibly growing) list of numbers to see whether it contains
some number or find the nearest neighbor to a number, it is worthwhile
at the outset to put them in a data structure that allows a binary
search. Setting up the initial data structure costs you some time,
but subsequent searches are O(log2(N)). See google for 'binary
search' and the python module bisect.

So roughly, for a list with 1,000,000 elements, your brute force
approach requires a million comparisons per search. If the data is
setup for binary search, on average only 13-14 comparisons will be
required. Well worth the effort if you need to do a lot of searches,
as I do.

John Hunter

Ron Adam

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Nov 4, 2003, 3:20:12 PM11/4/03
to
On Tue, 04 Nov 2003 08:25:36 -0800, David Eppstein
<epps...@ics.uci.edu> wrote:

>In article <bo8j4v$tqt$03$1...@news.t-online.com>,
> Peter Otten <__pet...@web.de> wrote:
>
>> This is of course all premature optimization as the most promising approach
>> is to try hard to reduce the number of candidate points, as David Eppstein
>> seems to have done. But then, he could use complex numbers, too.
>
>Well, yes, but then my code wouldn't work very well in dimensions higher
>than two...


I rearranged my first example to match the output of yours and used a
random number seed to get identical results.

Moving the square root to the return line of the find shortest
distance function increased the speed of my routine about 20%. Then
using the p*p form instead of p**2 added anouther 4%.

With printing turned there is only a very small difference. Of course
printing is the bottle neck. Turning off printing resulted in the
following. All are best of 3 runs.

1000 points:
Standard loop: 0:00:00.958192
Kdtree: 0:00:00.248096

Quite a difference. I'm not quite sure how kdtree's work. (yet) But
they can be worth while when working with large data sets.

The standard loop seems to be better for small lists of < 100 points.

100 points:
Standard loop: 0:00:00.009966
kdtree 0:00:00.015247

But for really large lists.

10000 points:
Standard loop: 0:01:39.246454
kdtree 0:00:03.424873

Hehe.... no comparison.

The number of calculations the standard loop does:

100 new points, 4950 distance calculations.
1000 new points, 499500 distance calculations.
10000 new points, 49995000 distance calculations.

And I don't know how to figure it for kdtree. But we can estimate it
by using the ratio of the speeds.

1000 points:
kdtree (3.42/99.25)*49995000 = 1722749.62 est. dist. calculations.

There's probably a better way to do that. Python is fun to do this
stuff with. Playing around like this with other languages is just too
much trouble.

_Ron

Jack Diederich

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Nov 4, 2003, 3:11:32 PM11/4/03
to pytho...@python.org, jdhu...@nitace.bsd.uchicago.edu

Breaking into the thread late, I've been busy enough to put down c.l.py
for a couple weeks.

If you only need to compare 10-1000 points, try this approach below. I
wrote it for the ICFP programming contest where I was sorting lots and
lots of points lots and lots of times. It sorts a list of points for
their manhattan distance from a particular point. I tweaked it until
it was just fast enough to do what I wanted. I won't pretend it is
optimal for all N, just that it was good enough for _my_ N. The speed
trick is that the point we are sorting around is stored in an object that
is created only once, and then we make the object's __call__ be the
comparison function.

Since it uses the list.sort() method we don't have to do anything more
clever than write our comparison function in C. I trust the Timbot
has written a better list.sort() than I ever will, so let's use it.

it is used thusly:

import icfp

l = [your big list of point tuples like (1, 10)]
p = (99, 22) # find nearest point in l to p

sort_ob = icfp.manhattan_sort(p) # creates a callable Manhat object that remembers p

l.sort(sort_ob) # sorts around p
print "closest", l[0]
print "farthest", l[-1]

The below code is from my CVS tree, so it should probably work. It was
written in a rush for a contest (JDH, you may recognize parts that are
cut-n-pasted from "probstat") so YMMV.

-jack

NB, if there is a boost or pyrex guy listening you might want to
throw in an easilly derived class that makes this [ab]use of __call__ easy.
It was a big performance win for me for very little effort.


""" icfp_module.c """

#include "Python.h"
#include <stdio.h>
#include <stdlib.h>

/*
* stats module interface
*/

static PyObject *ErrorObject;

PyObject *manhat_new(PyObject *self, PyObject *args);

static PyTypeObject Manhat_Type;

static PyObject *
heur1(PyObject *self, PyObject *args)
{
PyObject *tup1;
PyObject *tup2;
long x,y;

if (!PyArg_ParseTuple(args, "O!O!", &PyTuple_Type, &tup1, &PyTuple_Type, &tup2)) {
return NULL;
}
x = PyInt_AsLong(PyTuple_GET_ITEM(tup1,0)) - PyInt_AsLong(PyTuple_GET_ITEM(tup2,0));
y = PyInt_AsLong(PyTuple_GET_ITEM(tup1,1)) - PyInt_AsLong(PyTuple_GET_ITEM(tup2,1));
return PyInt_FromLong(labs(x * x) + labs(y * y));
}

static PyMethodDef stats_methods[] = {
{"manhattan", heur1, METH_VARARGS},
{"manhattan_sort", manhat_new, METH_VARARGS},
{NULL, NULL} /* sentinel */
};

DL_EXPORT(void)
initicfp(void)
{
PyObject *m, *d;

PyPQueue_Type.ob_type = &PyType_Type;
Manhat_Type.ob_type = &PyType_Type;

/* Create the module and add the functions */
m = Py_InitModule("icfp", stats_methods);

/* Add some symbolic constants to the module */
d = PyModule_GetDict(m);
ErrorObject = PyErr_NewException("icfp.error", NULL, NULL);
}

""" manhattan.c """
#include "Python.h"
#include <stdio.h>

#define ManhatObject_Check(v) ((v)->ob_type == &Manhat_Type)

staticforward PyTypeObject Manhat_Type;

typedef struct {
PyObject_HEAD
long x;
long y;
} ManhatObject;


static void
Manhat_dealloc(ManhatObject *self) {
PyObject_Del(self);
}

static PyObject *
Manhat_call(ManhatObject *self, PyObject *args)
{
PyObject *tup1;
PyObject *tup2;
long a;
long b;
long x;
long y;

if (!PyArg_ParseTuple(args, "O!O!", &PyTuple_Type, &tup1, &PyTuple_Type, &tup2)) {
return NULL;
}
x = PyInt_AsLong(PyTuple_GET_ITEM(tup1,0)) - self->x;
y = PyInt_AsLong(PyTuple_GET_ITEM(tup1,1)) - self->y;
a = labs(x * x) + labs(y * y);

x = PyInt_AsLong(PyTuple_GET_ITEM(tup2,0)) - self->x;
y = PyInt_AsLong(PyTuple_GET_ITEM(tup2,1)) - self->y;
b = labs(x * x) + labs(y * y);

if (a == b)
return PyInt_FromLong(0);
else if (a < b)
return PyInt_FromLong(-1);
else
return PyInt_FromLong(1);
}

static PyMethodDef Manhat_methods[] = {
{NULL, NULL} /* sentinel */
};

statichere PyTypeObject Manhat_Type = {
/* The ob_type field must be initialized in the module init function
* to be portable to Windows without using C++. */
PyObject_HEAD_INIT(&PyType_Type)
0, /*ob_size*/
"Manhat", /*tp_name*/
sizeof(ManhatObject), /*tp_basicsize*/
0, /*tp_itemsize*/
/* methods */
(destructor)Manhat_dealloc, /*tp_dealloc*/
0, /*tp_print*/
0, /*tp_getattr*/
0, //(setattrfunc)Permute_setattr, /*tp_setattr*/
0, /*tp_compare*/
0, /*tp_repr*/
0, /*tp_as_number*/
0, /*tp_as_sequence*/
0, /*tp_as_mapping*/
0, /*tp_hash*/
(ternaryfunc)Manhat_call, /*tp_call*/
0, /*tp_str*/
PyObject_GenericGetAttr, /*tp_getattro*/
0, /*tp_setattro*/
0, /*tp_as_buffer*/
Py_TPFLAGS_DEFAULT, /*tp_flags*/
0, /*tp_doc*/
0, /*tp_traverse*/
0, /*tp_clear*/
0, /*tp_richcompare*/
0, /*tp_weaklistoffset*/
0, /*tp_iter*/
0, /*tp_iternext*/
Manhat_methods, /*tp_methods*/
0, /*tp_members*/
0, /*tp_getset*/
0, /*tp_base*/
0, /*tp_dict*/
0, /*tp_descr_get*/
0, /*tp_descr_set*/
0, /*tp_dictoffset*/
0, /*tp_init*/
0, /*tp_alloc*/
0, /*tp_new*/
0, /*tp_free*/
0, /*tp_is_gc*/
};

// not static so ifcp_module.c can see it
PyObject *
manhat_new(PyObject *self, PyObject *args)
{
ManhatObject *mh;
PyObject *tup1;

if (!PyArg_ParseTuple(args, "O!", &PyTuple_Type, &tup1)) {
return NULL;
}
// call object create function and return
mh = PyObject_New(ManhatObject, &Manhat_Type);
if (mh == NULL)
return NULL;

mh->x = PyInt_AsLong(PyTuple_GET_ITEM(tup1,0));
mh->y = PyInt_AsLong(PyTuple_GET_ITEM(tup1,1));
return (PyObject *)mh;
}

""" setup.py """

from distutils.core import setup, Extension

files = [
'manhattan.c',
'icfp_module.c',
]

libraries = []
#libraries = ["efence"] # uncomment to use ElectricFence
includes = []
if (__name__ == '__main__'):
setup(name = "icfp", version = "0.2",
ext_modules = [Extension("icfp", files,
libraries = libraries,
include_dirs = includes,
)
]
)


Jim Richardson

unread,
Nov 4, 2003, 8:53:27 PM11/4/03
to


I am new to timeit.py, but this is odd.

jim@grendel:~$ /usr/lib/python2.3/timeit.py -c ' p=1.6+2.5j;np=2.4+1.3j; d=abs(p-np)'
100000 loops, best of 3: 3.1 usec per loop

vs

jim@grendel:~$ /usr/lib/python2.3/timeit.py -c -s'import math; p=1.6+2.5j;np=2.4+1.3j; d=abs(p-np)'
10000000 loops, best of 3: 0.141 usec per loop

Is it because the builtin math functions are much slower?


--
Jim Richardson http://www.eskimo.com/~warlock
Televangelists: The Pro Wrestlers of Religion

Ron Adam

unread,
Nov 4, 2003, 9:26:48 PM11/4/03
to


Ah, a contour map. Maybe something like this?


"""
Where pointA and pointB are constitutive points in a list, and pointC
is a new point from a list of new points.

For each pointC in a list of new points.
For each consecutive 2 points in a list of sequential points.
If lineAC < lineAB and lineBC < lineAB
Insert pointC between pointA and pointB

If pointC was not placed in list of sequential points.
Where pointA and pointB are the beginning and
end points of the list.

IF lineAC < lineBC
Add pointC to beginning of list.
Else add pointC to end of list.

When done copy point from beginning of list to end of list
to complete polygon.
"""


Just knowing the closest point isn't quite enough because it doesn't
tell you weather to put it in front or behind the point in the list.
Storing the distance to the next point along with each point might
make it work faster. This method has an advantage in that it doesn't
have to go through the whole list. You could start from the closest
end of the list and maybe make it quicker.

One catch is you need to know in advance that the set of points are
not divided by a hill or valley.

I'm not sure what this would do with a list of random points. Maybe a
long squiggly line where the beginning to end segment cuts across
them. I don't think you will have that problem.

_Ron


Miki Tebeka

unread,
Nov 5, 2003, 2:28:13 AM11/5/03
to
Hello John,

> Can anyone point me to some code or module that provides the
> appropriate data structures and algorithms to handle this task
> efficiently? The size of the list will likely be in the range of
> 10-1000 elements.

boost (http://www.boost.org) has a graph library and a good connection
to Python using Boost.Python.

HTH.
Miki

Peter Otten

unread,
Nov 5, 2003, 3:40:23 AM11/5/03
to
Jim Richardson wrote:

> I am new to timeit.py, but this is odd.
>
> jim@grendel:~$ /usr/lib/python2.3/timeit.py -c ' p=1.6+2.5j;np=2.4+1.3j;
> d=abs(p-np)' 100000 loops, best of 3: 3.1 usec per loop

The above is actually timed.

> jim@grendel:~$ /usr/lib/python2.3/timeit.py -c -s'import math;
> p=1.6+2.5j;np=2.4+1.3j; d=abs(p-np)' 10000000 loops, best of 3: 0.141 usec
> per loop

Here you put everything in the setup parameter. As the timed statement
defaults to pass, you are essentially timing an empty loop :-(

> Is it because the builtin math functions are much slower?

You should have used math.abs() or from math import abs to really get the
abs() function in the math module. But

>>> "abs" in dir(math)
False

Not there, so we cannot compare.

Peter


WEINHANDL Herbert

unread,
Nov 5, 2003, 6:30:24 AM11/5/03
to John Hunter
John Hunter wrote:
> I have a list of two tuples containing x and y coord
>
> (x0, y0)
> (x1, y1)
> ...
> (xn, yn)
>
> Given a new point x,y, I would like to find the point in the list
> closest to x,y. I have to do this a lot, in an inner loop, and then I
> add each new point x,y to the list. I know the range of x and y in
> advance.

> Can anyone point me to some code or module that provides the


> appropriate data structures and algorithms to handle this task
> efficiently? The size of the list will likely be in the range of
> 10-1000 elements.

The plotting-library dislin (http://www.dislin.com)
contains a really fast triangulation subroutine
(~1 hour for 700000 points on an 1.5 GHz Athlon).

For an example triangulation/nearest-neighbor-search see
the attached python-script.

Dislin is available for many platforms (Linux, Winxxx, ...)
and it can be used for free if you are using free languages like
Python, Perl, etc.

Happy pythoning

Herbert

nearest-n.py

Jim Richardson

unread,
Nov 5, 2003, 9:26:49 PM11/5/03
to
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

On Wed, 05 Nov 2003 09:40:23 +0100,
Peter Otten <__pet...@web.de> wrote:
> Jim Richardson wrote:
>
>> I am new to timeit.py, but this is odd.
>>
>> jim@grendel:~$ /usr/lib/python2.3/timeit.py -c ' p=1.6+2.5j;np=2.4+1.3j;
>> d=abs(p-np)' 100000 loops, best of 3: 3.1 usec per loop
>
> The above is actually timed.
>
>> jim@grendel:~$ /usr/lib/python2.3/timeit.py -c -s'import math;
>> p=1.6+2.5j;np=2.4+1.3j; d=abs(p-np)' 10000000 loops, best of 3: 0.141 usec
>> per loop
>
> Here you put everything in the setup parameter. As the timed statement
> defaults to pass, you are essentially timing an empty loop :-(
>


Ah! that explains it, thanks. breaking the import off, gets the loop to
3.2 usec per loop, which makes more sense.

>> Is it because the builtin math functions are much slower?
>
> You should have used math.abs() or from math import abs to really get the
> abs() function in the math module. But
>
>>>> "abs" in dir(math)
> False
>
> Not there, so we cannot compare.

Actually, I just cut'n'pasted, with a tiny alteration (timeit.py isn't
in my path) I like the idea of using the complex number there, it's a
tad easier that the other method for me to visualize.

-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.2.3 (GNU/Linux)

iD8DBQE/qbFpd90bcYOAWPYRArgzAKDiNBXw3tIpooWqglmxoqn2GebyxgCcDEKV
u60SPPopUrkJ1Kak0t4eias=
=+Guh
-----END PGP SIGNATURE-----

"Man invented language to satisfy his deep need to complain."
-- Lily Tomlin

Colin J. Williams

unread,
Nov 6, 2003, 7:48:13 AM11/6/03
to
This enquiry has generated a lot of interest. I thought it might be
useful to try numarray on the problem. numarray has a compress
function, which is used to discard points which are not of interest.

The code is below.

As would be expected, each scan over the points in the neigbourhood
discards, on average, a little more than half the points.

I have not tried it, but it should be possible, with numarray, to
generalize this from 2D to multimimensional space.

Colin W.

# Neighbour.py
''' To find the closest neighbour, in a neghbourhood P,
to a point p.
'''
import math
import numarray as N
import numarray.numerictypes as _nt
import numarray.random_array as R

trial= 0
lengthRemaining= []
def find(P, p):
''' In the 2D neighbourhood P, find the point closest to p.
The approach is based on the selection of a trial value Pt, from
P, and
discarding all values further than Pt from p.
To avoid repeated sqrt calculations the discard is based on an
enclosing square.
'''
global lengthRemaining, trial
lengthRemaining+= [[P.shape[0]]]
Pz= P - p # zero based neighbourhood
while len(Pz):
Pt= Pz[0] # trial value
Pta= N.abs(Pt)
Pz= Pz[1:]
pd= math.sqrt(N.dot(Pta, Pta)) # distance of p from the trial
value
goodCases= N.logical_and((Pz < pd),
(Pz > -pd))# use the enclosing square
goodCases= N.logical_and.reduce(goodCases, 1)
Pz= N.compress(goodCases, Pz) # discard points outside the square
if len(Pz) == 1:
Pt= Pz[0] # We have found the closest
Pz= []
lengthRemaining[trial]+= [len(Pz)]
z= 100
trial+= 1
return Pt + p

if __name__ == '__main__':
for sampleSize in range(100, 5000, 100):
P= R.random(shape= (sampleSize, 2))
for i in range(20):
p= R.random((1, 2)) # point
a= find(P, p)
## print 'Closest neighbour:', a[0]
## print 'Check - Point(p):', p[0]
## check= []
## for i in range(len(P)):
## check+= [(math.sqrt((P[i, 0]-p[0, 0])**2 + (P[i, 1]-p[0,
1])**2), P[i, 0], P[i, 1])]
## print P[i], math.sqrt((P[i, 0]-p[0, 0])**2 + (P[i, 1]-p[0, 1])**2)
## check.sort()
## print check[0]
## print check
print 'Number of scans:', sum([len(lst) for lst in lengthRemaining])
print 'Sample size:', P.shape[0], ' Average numner of scans:',
sum([len(lst) for lst in lengthRemaining])/float(len(lengthRemaining))

> ------------------------------------------------------------------------
>
> #!/usr/bin/python
>
> # (c) by H. Weinhandl 04.Nov.2003
>
> import math
> import dislin
>
>
> def dist( ia,ib ) :
> return math.sqrt( (X1[ia]-X1[ib])**2 + (Y1[ia]-Y1[ib])**2 )
>
>
> def find_nearest_neighb() :
>
> print 'extracting neighbours ... ',
>
> for i in range( nr_dat+3 ) :
> # initualize list wit the point i itself
> neighb.append( [i] )
>
> for i in range( nr_tri ) :
> # get a indes-triple, dislin seems to start indices with 1,
> # so I'm subtracting 1 to get a zero-based index
> U,V,W = I1[i]-1, I2[i]-1, I3[i]-1
>
> # add indices from all triangles, which contain the point itself
> if ( U in neighb[U] ) :
> if not (V in neighb[U] ) : neighb[U].append( V ) # do not append V if already in the list
> if not (W in neighb[U] ) : neighb[U].append( W ) # do not append W if already in the list
>
> if ( V in neighb[V] ) :
> if not (U in neighb[V] ) : neighb[V].append( U ) # do not append U if already in the list
> if not (W in neighb[V] ) : neighb[V].append( W ) # do not append W if already in the list
>
> if ( W in neighb[W] ) :
> if not (U in neighb[W] ) : neighb[W].append( U ) # do not append U if already in the list
> if not (V in neighb[W] ) : neighb[W].append( V ) # do not append V if already in the list
>
> print ' OK'
> for i in range( nr_dat ) :
> neighb[i].remove( i ) # remove the point i itself
> neighb[i].sort()
> r_mi = 9.9e9
> i_mi = i
>
> # search for the nearest neighbor of point i
> for j in neighb[i] :
> r = dist( i, j )
> if r < r_mi :
> r_mi = r
> i_mi = j
> print 'NB %2d : r_mi=%5.3f, i_mi=%2d '% (i, r_mi, i_mi), neighb[i]
>
>
>
>
> if __name__ == '__main__' :
>
> X1 = [ 1.0, 2.0, 3.0, 4.0, 5.0, 4.0, 3.0 ]
> Y1 = [ 1.0, 6.0, 3.0, 3.0, 2.0, 6.0, 0.0 ]
> nr_dat = len( X1 )
> nr_max = 2 * nr_dat + 1
> I1 = [ 0 ] * nr_max
> I2 = [ 0 ] * nr_max
> I3 = [ 0 ] * nr_max
> neighb = [ ]
>
> # padding the 2 input-lists with 3 additional elements is required by dislin
> X1.append( 0 )
> X1.append( 0 )
> X1.append( 0 )
> Y1.append( 0 )
> Y1.append( 0 )
> Y1.append( 0 )
>
>
> # delaunay triangulation of the input lists
> # I1,I2,I3 are the indices of the triangular edge-points
> nr_tri = dislin.triang( X1,Y1, nr_dat, I1,I2,I3, nr_max )
>
> print X1
> print Y1
> print nr_dat, nr_tri, nr_max
> print I1
> print I2
> print I3
>
> find_nearest_neighb()
>
> #---- end ----

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