Thanks Hanno
I got some idea.
How about the bin(grid)??
On Fri, Oct 18, 2013 at 8:21 PM, Hanno Klemm kl...@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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