My goal is to find the probability density function for a certain distribution, using a given algorithm.
This algorithm requires that I search in which interval a float is placed in. Even though the code runs perfectly, it takes too long. I was looking for a way of optimizing my code, but none came to mind.
In each iteration I check if the float is in the interval: if that's the case, I'd like to had a unity to the position I'm considering, in array p.
This is my code:
import numpy as np
import pylab as plt
import random as rd
n = [10,100,1000]
N = [10**6]
dy = 0.005
k_max = int(1/dy-1)
y = np.array([(j+0.5)*dy for j in range(k_max+1)])
intervals = np.linspace(0,1,k_max+2)
def p(y,n,N):
p = np.zeros(len(y))
Y = np.array([sum(np.array([rd.random() for k in range(n)]))/n for j in range(N)])
z = np.array([sum(np.array([rd.random() for k in range(n)])) for l in range(N)])
for j in Y:
for i in range(len(y)-1):
if intervals[i] <= j < intervals[i+1]:
p[i] += 1
return(p/(dy*N))
for a in n:
pi = p(y,a,N[0])
plt.plot(y,pi,label = 'n = ' + str(a))
plt.title('Probability Density Function')
plt.xlabel('x')
plt.ylabel('p(x)')
plt.show()
Edit: I've added the full code, as requested. Edit 2: Fixed an error intervals.
A quick and simple optimization can be made here:
for j in Y:
for i in range(len(y)-1):
if intervals[i] <= j < intervals[i+1]:
p[i] += 1
Since intervals
consists of len(y)
evenly spaced numbers over the interval [0, 1], which is also the range of Y
values, we need not search the position of j
in intervals
, but rather we can compute it.
for j in Y: p[int(j*(len(y)-1))] += 1
Also we can remove the unused
z = np.array([sum(np.array([rd.random() for k in range(n)])) for l in range(N)])
The greatest part of the remaining execution time is taken by
Y = np.array([sum(np.array([rd.random() for k in range(n)]))/n for j in range(N)])
Here the inner conversions to np.array
are very time consuming; better leave them all out:
Y = [sum([rd.random() for k in range(n)])/n for j in range(N)]