Question

I am aware that numpy arrays are pointer arrays. And I know that is possible to define pointers in python. But I am wondering, if I make a variable equal to an element in a numpy vector, is it still a pointer or is it de-referenced? Is there a way I can find out or test this?

Example

    import scipy
    vec = scipy.randn(10)
    vecptr = vec # vecptr is a pointer to vec
    vecval = scipy.copy(vec) # vecval is not a pointer.

    var = vec[3] # is var pointer or is it copied by value ???

    print(type(var)) # returns numpy.float64.  does this mean its a 1x1 numpy vec and therefore  a pointer ?

The reason I ask is, what I really want to know is; will the code below this double up my memory? I am trying to create more meaningful variable names to my vector that is returned

    v = self.viewCoefs[sz][sv][sa]

    gw = v[0]
    G0 = v[1]
    G1 = v[2]
    G2 = v[3]
    alpha0 = v[4]
    alpha1 = v[5]
    alpha2 = v[6]
    beta0 = v[7]
    beta1 = v[8]
    beta2 = v[9]
    beta3 = v[10]
    gamma0 = v[11]
    gamma1 = v[12]
    gamma2 = v[12]
    gamma3 = v[12]
    gamma4 = v[13]
    delta0 = v[14]
    delta1 = v[15]
    delta2 = v[16]
    delta3 = v[17]
    delta4 = v[18]
    delta5 = v[19]
    zeta_prime_0 = v[20]
    zeta_prime_1 = v[21]
    zeta_prime_2 = v[22]
    Gamma_prime_0 = v[23]
    Gamma_prime_1 = v[24]
    Gamma_prime_2 = v[25]
    Gamma_prime_3 = v[26]

Because I have lots of these to follow

    p0 = alpha0 + alpha1*scipy.log(bfrac) + alpha2*scipy.log(bfrac)**2
    p1 = beta0 + beta1*scipy.log(bfrac) + beta2*scipy.log(bfrac)**2 + beta3*scipy.log(bfrac)**3
    p2 = gamma0 + gamma1*scipy.log(bfrac) + gamma2*scipy.log(bfrac)**2 + gamma3*scipy.log(bfrac)**3 + gamma4*scipy.log(bfrac)**4
    p3 = delta0 + delta1*scipy.log(bfrac) + delta2*scipy.log(bfrac)**2 + delta3*scipy.log(bfrac)**3 + delta4*scipy.log(bfrac)**4 + delta5*scipy.log(bfrac)**5

    subSurfRrs = g*(p0*u + p1*u**2 + p2*u**3 + p3*u**4)
    ## and lots more

So I would like meaningful variable names without doubling my memory foot print.

#

Okay, If I got it right, the solution to NOT double up my memory is :

    v = self.veiwCoefs[sz][sv][sa]

    gw = v[0:1]
    G0 = v[1:2]
    G1 = v[2:1]
    alpha0 = v[3:4]
    alpha1 = v[4:5]
    alpha2 = v[5:6]
    beta0 = v[6:7]
    beta1 = v[7:8]
    beta2 = v[8:9]
    beta3 = v[9:10]
    ## etc 

    p0 = alpha0[0] + alpha1*scipy.log(bfrac) + alpha2[0]*scipy.log(bfrac)**2
    p1 = beta0[0] + beta1[0]*scipy.log(bfrac) + beta2[0]*scipy.log(bfrac)**2 + beta3[0]*scipy.log(bfrac)**3

    ## etc

Answer 1

You almost have it, but here is how to create a view of a single element:

In [1]: import numpy as np

In [23]: v = np.arange(10)

In [24]: a = v[3:4]

In [25]: a[0] = 100

In [26]: v
Out[26]: array([  0,   1,   2, 100,   4,   5,   6,   7,   8,   9])

Here a is a view of the fourth element of v, so when you change a you change the corresponding position in v.