Dynamically growing numarray array.
Carl Banks
invalidemail at aerojockey.com
Thu Mar 23 02:17:50 CET 2006
Ivan Vinogradov wrote:
> Hello All,
>
> this seems like a trivial problem, but I just can't find an elegant
> solution neither by myself, nor with google's help.
>
> I'd like to be able to keep an array representing coordinates for a
> system of points.
> Since I'd like to operate on each point's coordinates individually,
> for speed and ufuncs
> numarray fits the bill perfectly, especially since system.coordinates
> [4] would return proper vector for a 5th point.
> To start, read the coordinates from a text file and add them to our
> array one by one.
> Here it gets un-elegant and probably wasteful for a large number of
> points, by converting the whole array to a list only to use append
> method and then convert it back to array(sample code below). Also,
> there is potential need to add more points later on.
>
> Any pointers would be greatly appreciated.
Very simple thing to do, actually. The following class implements a
very basic array grower for an array of 3-D points:
class ArrayGrower(object):
def __init__(self,initsize=100):
self.memarray = zeros((initsize,3),Float)
self.npoints = 0
def add_point(self,point):
if self.npoints >= len(self.memarray):
oldlen = len(self.memarray)
newmemarray = zeros((oldlen*2,3),Float)
newmemarray[:oldlen] = self.memarray
self.memarray = newmemarray
self.memarray[self.npoints] = point
self.npoints += 1
def get_current_array(self):
return self.memarray[:self.npoints]
It simply uses a slice of a larger array to hold all your points. It
keeps track of the number of points and automatically grows the array
when there's no room to add another point. (You can use a factor less
than 2 if you have to conserve memory.) Whenever you need the current
array for a calculation, use get_current_array() to get an array slice
of the proper size.
>>> x = ArrayGrower(2)
>>> x.add_point([0,1,2])
>>> x.get_current_array()
array([ [ 0., 1., 2.]])
>>> x.add_point([0,0,0])
>>> x.add_point([1,1,1])
>>> x.get_current_array()
array([[ 0., 1., 2.],
[ 0., 0., 0.],
[ 1., 1., 1.]])
Carl Banks
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