Description
Write a Python program to use the scheme of DE-BPSO (combination of Differential evolution and Binary Particle Swarm Optimization) to predict a linear model for an HIV inhibitor (Pill). You can either choose a linear model such as “Multiple Linear Regression”, “Support Vector Machine”, or “Partial Least Square Regression”, or a none-linear model such “Artificial Neural Network”. The algorithm of DE-BPSO is:
Step 1:
// Create the initial velocity (call it matrix V) that is between 0 and 1 (not binary)
for (i=0; i<50; i++)
for (j=0; j<385; j++)
{
V[i,j] = random number between 0 and 1;
}
———————————————————————————
Step 2:
// Create the initial population (call it matrix X) based on the values of the initial velocity
for (i=0; i<50; i++)
for (j=0; j<385; j++)
{
if (V[i,j] <= Lambda) // Note: The value of Lambda is 0.01
X[i,j] = 1;
else
X[i, j] = 0;
}
———————————————————————————
Step 3:
// Find the fitness of each row of the 50 models in the first population
———————————————————————————
Step 4:
// Since you only have the first population at this time, your local best
// matrix (call it matrix P) become the same as the first population
// Therefore:
P = X
// The row with the best fitness in P becomes the global best. Let’s call that row Row “G”
// Therefore:
G = the row in P with the best fitness
———————————————————————————
Step 5:
// This section is going to be repeated until the end of the program
// In this section we need to find the next velocity matrix
// We are using Differential Evolution (DE) algorithm to find the new velocity matrix
for (i=0; i<50; i++)
for (j=0; j<385; j++)
{
Randomly select 3 rows from the populations and call them as r1, r2, and r3
Let r = r3 + F * (r2 – r1) // the value of F should be set to 0.7
// Do the cross mutation of row “i” and “r”
if ((random between 0 and 1) < CR) // not binary, CR = 0.7
V[i,j] = r[j]
else
V[i,j] = V[i,j] // remains unchanged
}
———————————————————————————
Step 6:
// This section creates the new population. The value of Alpha
// starts from 0.5 and is decremented to 0.33. The value of beta is 0.004
for (i=0; i<50; i++)
for (j=0; j<385; j++)
{
if ( (alpha < V[i,j]) && (V[i,j] <= 0.5*(1+alpha))
X[i, j] = P[i,j];
else if ( (0.5*(1+alpha)) < V[i,j]) && (V[i,j] <= (1-beta))
X[i,j] = G[j] // the global vector value
else if (1-beta) < V[i,j]) && (V[i,j] <=1))
X[i.j] = 1 – X[i,j]
else
X[i,j] = X[i,j]; // remains unchanged
}
———————————————————————————
Step 7:
// calculate the fitness of the new population
// update the local best matrix as you did in the previous assignment
// update the global best row if necessary
// stop if the number of iterations have reached the maximum
// otherwise decrement alpha and go to step 5
———————————————————————————