Convert Eigen to numpy matrix

Question

I have the following code in C++ that uses the Eigen library, need help to translate to python (numpy)

Initialization

double b = 20.0;
Eigen::Vector3d C(1.0/10.2, 1.0/10.2, 1/30);

Eigen::MatrixXd U(5200, 3);
int i = 0;
for (double x = 10.2/2.0; x < 200; x += 10) {
    for (double y = 10.2/2.0; y < 200; y += 10) {
        for (double t = 0; t <= 360; t += 30) {
            U(i, 0) = x;
            U(i, 1) = y;
            U(i, 2) = psi;
            i += 1;
        }
    }
}

Function:

Eigen::VectorXd operator()(const Eigen::VectorXd& s) {
    Eigen::VectorXd p(length());

    p(0) = s[0];
    p(1) = s[1];
    p(2) = s[2];
    p(3) = s[3];

    for (int i = 0; i < U.rows(); i++) {
        p(i + 4) = b*exp(-0.5*(s.tail(U.cols()) - U.row(i).transpose()).dot(C*(s.tail(U.cols())
                            - U.row(i).transpose())));
        if (p(i + 4) < 0.1) {
            p(i + 4) = 0;
        }
    }

    return p;
}

Python version

Initialization:

my_x = 10.2/2.0
my_y = 10.2/2.0
my_p = 0

xx = []
while my_x < 200:
    xx.append(my_x)
    my_x += 10

yy = []
while my_y < 200:
    yy.append(my_y)
    my_y += 10

pps = []
while my_psi <= 360:
    pps.append(my_p)
    my_p+=30

U =[]
for x in xx:
    for y in yy:
        for p in pps:
            U.append([x,y,p])

U = numpy.matrix(U)

C = numpy.array([1.0/10.2, 1.0/10.2, 1.0/30.0])

b = 20.0

The Function

Instead of operator() I will call the function doSomething()

def doSomething(s):   # Where s is a numpy array (1-d vector)
    p[0:4] = s[0:4]
    for i in range (U.shape[0]):
        s_dash = -0.5*(s - U[i].T)
        s_ddash = C*s
        s_dddash = s_dash.dot(s_ddash) - U[i].T
        p[i+4] = b * numpy.exp(s_dddash)

        if p[i+4] < 0.1: p[i+4] = 0

What I am confused:

1. In the C++ implementation, I think p[i+4] is supposed to be a single value
2. In my python version, I get a p[i+4] as square matrix
3. Each p[i+4] is a zero matrix.

I am unable to decipher my mistake. Please help!