[Numpy-discussion] how to do scoring of random point to avoid overlapping in python

[Pooja Gupta]

Thanks Hanno
I got some idea.
How about the bin(grid)??????



On Fri, Oct 18, 2013 at 8:21 PM, Hanno Klemm <klemm at phys.ethz.ch> wrote:

> On 18.10.2013 12:33, Pooja Gupta wrote:
> > I have generated random point around a object and then evaluate each
> > random point on certain criteria. But problem is that every time I am
> > getting new point. How i can resolve this problem so that my result
> > should be uniform. Is any way to evaluate the position of random
> > point.
> >
> > for arang in range(1000): # generate 1000 random point around object
> >  arang = arang + 1
> >  x,y,z = 9.251, 24.410, 64.133 # coordinates of objects (i have 500
> > object coordinates)
> >
> >  x1,y1,z1 =
> > (uniform(x-3.5,x+3.5),uniform(y-3.5,y+3.5),uniform(z-3.5,z+3.5))
> > #randompoint
> >  pacord = [x1,y1,z1] #random point coordinates
> >  dist_pap = euDist(uacoord, pacord) # check distance between object
> > and random points
> >
> >  if (dist_pap > 2.5): # if the random point far from obect
> >  dist_pap1 = dist_pap
> > vecpw = euvector(uacoord, pacord) # generate vectors b/w objject and
> > random point
> >
> > # angle between angle between object and random point
> > num1 = np.dot (vect1, vecpw)
> > denom1 = np.linalg.norm(vect1) * np.linalg.norm(vecpw)
> > ang1 = rad2deg(np.arccos(num1/denom1))
> >
> > if 140 > ang1 >100: # check angle
> >  ang2= ang1
> > print pacord
> >
> > Queries
> > every time i am getting new result (new positions of the random
> > point). How to fix it.
> > on above basis I want to score each random point and the two random
> > point should be 2.5 distance apart from each other. How I can avoid
> > overlapping of the random points.
> >
>
> I am not sure if i understand the question correctly but if you always
> want to get the same random number, every time you run the script, you
> can fix the random seed by using np.random.seed(). Regarding your second
> question, when I understand correctly, you somehow want to find two
> points that have a distance of 2.5. If you already have one of them, I
> would generate the second one using spherical coordinates and specifying
> the distance a priori. something like:
>
> def pt2(pt1, distance=2.5):
>      theta = np.random.rand()*np.pi
>      phi = np.random.rand()*2*np.pi
>      r = distance
>      x = r*np.sin(theta)*np.cos(phi)
>      y = r*np.sin(theta)*np.sin(phi)
>      z = r*np.cos(theta)
>      return pt1 + np.array((x,y,z))
>
>
> In [367]: pt1 = np.array([0,0,0])
>
> In [368]: pt2(pt1)
> Out[368]: a = array([-2.29954368, -0.57223342,  0.79664785])
>
> In [369]: np.linalg.norm(a)
> Out[369]: 2.5
>
> Hope this helps,
> Hanno
>
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