how to do scoring of random point to avoid overlapping in python
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= ang1print 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. -- पूजा गुप्ता, पीएचडी राष्ट्रीय कोशिका विज्ञान केंद्र "जीवन को खुल कर मुस्कुराने दो, खोल दो बाहे अवसर आने दो । पग बढाओ हर पथ तुम्हारा है, पथ कंटक मंजिलो का इशारा है।।"
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
Thanks Hanno
I got some idea.
How about the bin(grid)??????
On Fri, Oct 18, 2013 at 8:21 PM, 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
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-- पूजा गुप्ता, पीएचडी राष्ट्रीय कोशिका विज्ञान केंद्र "जीवन को खुल कर मुस्कुराने दो, खोल दो बाहे अवसर आने दो । पग बढाओ हर पथ तुम्हारा है, पथ कंटक मंजिलो का इशारा है।।"
participants (2)
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Hanno Klemm -
Pooja Gupta