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

[Hanno Klemm]

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