Python/Scipy Integration array

Question

I am trying to write a program that does the following:

• Takes values of V from an array
• Passes the V values into an integral that is with respect to E
• Output integral results into an array I
• Plot I against V

The equation looks quite nasty, but everything is a constant other than V. Here is the equation. The equation isn't very important.

How should I go about this problem? My attempt (as shown below) does not calculate the integral for each value of V read from the file.

from scipy import integrate #integrate.quad
from numpy import *
import pylab
import datetime
import time
import os
import math

# import V
fn = 'cooltemp.dat'
V = loadtxt(fn,unpack=True,usecols=[1])

# variables
del1, del2, R, E, fE, fEeV = 1,2,1,2,1,1
e = 1.602176565*10**-19

# eqn = dint(abc)
a = E/( math.sqrt( E**2 - del1**2 ) )
b = ( E+ e*V )/( math.sqrt( ( E + e*V )**2) - del2**2)
c = fE-fEeV
d = 1/(e*R) # integration constant
eqn = a*b*c

# integrate 
result = quad(lambda E: eqn,-inf,inf)

# current
I = result*d

# plot IV curve
pylab.plot(V,I,'-r')

## customise graph
pylab.legend(['degree '+str(n),'degree '+str(q),'data'])
pylab.axis([0,max(x),0,max(y)])
pylab.xlabel('voltage (V)')
pylab.ylabel('current (A)')
tc = datetime.datetime.fromtimestamp(os.path.getmtime(fn))
pylab.title('IV curve\n'+fn+'\n'+str(tc)+'\n'+str(datetime.datetime.now()))
pylab.grid(True)
pylab.show()

* Updated attempt:

from scipy import integrate
from numpy import *
import pylab
import datetime
import time
import os
import math

# import V
fn = 'cooltemp.dat'
V = loadtxt(fn,unpack=True,usecols=[1])
# print V

# variables
del1, del2, R, E, fE, fEeV = 1.0,2.0,1.0,2.0,1.0,1.0
e = 1.602176565*10**-19

I=[]
for n in range(len(V)):

    constant = 1/(e*R) # integration constant
    eqn = (E/( math.sqrt( E**2 - del1**2 ) ))*(( E + e*V[n] )/( math.sqrt( ( E + e*V[n] )**2) - del2**2))*(fE-fEeV)

    # integrate 
    result,error = integrate.quad(lambda E: eqn,-inf,inf)
    print result
    # current
    I.append(result*constant)

I = array(I)

# plot IV curve
pylab.plot(V,I,'-b')

Answer 1

You have a few problems:

The "function" you pass to quad always returns eqn, which is just a precalculated number. You need to define a proper function that takes a given value for E as input and returns the integrand. This function will also need to assume a fixed value for V. Assuming the code you provided calculates the proper quantity for a given value of V and E (I haven't checked, just copy-pasting):

# import V
fn = 'cooltemp.dat'
V = loadtxt(fn,unpack=True,usecols=[1])
# print V

@np.vectorize
def result(x):
    def integrand(E):
        del1, del2, R, fE, fEeV = 1.0,2.0,1.0,1.0,1.0
        e = 1.602176565*10**-19
        a = E/( math.sqrt( E**2 - del1**2 ) )
        b = ( E+ e*x )/( math.sqrt( ( E + e*x )**2) - del2**2)
        c = fE-fEeV
        d = 1/(e*R) # integration constant
        return a * b * c
    return quad(integrand, -inf, inf)

I = result(V)

To Summarize:

• result(v) evaluates the full integral (over E) for a fixed value of v
• integrand(E) evaluates the integrand at fixed E (the integration variable) and, incidentally, fixed V (it grabs the value from outside the function, which is why the definition for integrand is nested inside the definition for result)
• the @np.vectorize trick is just a nice convenience function that allows you to pass arrays for V into result. Numpy will loop over the values for you, and return an array instead of a scalar

Answer 2

You should use np.vectorize to pass arrays into the equations and get arrays back. For example, this calculates the following equation (Comoving distance if you are curious...):

import numpy as np
from scipy.integrate import quad


spl=299792458.0    #speed of light in m/s
Mpc=3.0856E22      # Mpc in m
pc=3.0856E16       # pc in m

def HubbleTime(H0):
    return 3.0856e17/(H0/100.0)

def HubbleDist(H0):
        """returns the Hubble Distance (in Mpc) for given H_0"""
    return spl*HubbleTime(H0)/Mpc

def Integrand(z, Om, OLam):
        """ This is the E(z) function from Hogg (2000)
        Integrand(z, Om, OLam)
    """
    return ( Om*(1+z)3 + OLam)(-0.5)

def CosmComDist(z, H0=70, Om=0.30, OLam=0.70):
    """Gives the comoving distance at redshift z
    CosmComDist(z, H0=70, Om=0.30, OLam=0.70)
    """
    CMD=HubbleDist(H0)*quad(Integrand, 0, z, args=(Om, OLam))[0]
    return CMD

CosmComDist=np.vectorize(CosmComDist)
redshifts = np.linspace(0,1,100)
distances = CosmComDist(redshifts)