Write a Numerical Formula for the Is Curve, Showing Y as a Function of R Alone.
Numerical Methods using Python (scipy)¶
Overview¶
The core Python language (including the standard libraries) provide enough functionality to carry out computational research tasks. However, there are dedicated (third-party) Python libraries that provide extended functionality which
-
provide numerical tools for frequently occurring tasks
-
which are convenient to use
-
and are more efficient in terms of CPU time and memory requirements than using the code Python functionality alone.
We list three such modules in particular:
-
The
numpymodule provides a data type specialised for "number crunching" of vectors and matrices (this is thearraytype provided by "numpy" as introduced in 14-numpy.ipynb), and linear algebra tools. -
The
matplotlibpackage (also knows aspylab) provides plotting and visualisation capabilities (see 15-visualising-data.ipynb) and the -
scipypackage (SCIentific PYthon) which provides a multitude of numerical algorithms and which is introduced in this chapter.
Many of the numerical algorithms available through scipy and numpy are provided by established compiled libraries which are often written in Fortran or C. They will thus execute much faster than pure Python code (which is interpreted). As a rule of thumb, we expect compiled code to be two orders of magnitude faster than pure Python code.
You can use the help function for each numerical method to find out more about the source of the implementation.
SciPy¶
Scipy is built on numpy. All functionality from numpy seems to be available in scipy as well. For example, instead of
In [1]:
import numpy x = numpy . arange ( 0 , 10 , 0.1 ) y = numpy . sin ( x )
In [2]:
import scipy as s x = s . arange ( 0 , 10 , 0.1 ) y = s . sin ( x )
First we need to import scipy:
The scipy package provides information about its own structure when we use the help command:
The output is very long, so we're showing just a part of it here:
stats --- Statistical Functions [*] sparse --- Sparse matrix [*] lib --- Python wrappers to external libraries [*] linalg --- Linear algebra routines [*] signal --- Signal Processing Tools [*] misc --- Various utilities that don't have another home. interpolate --- Interpolation Tools [*] optimize --- Optimization Tools [*] cluster --- Vector Quantization / Kmeans [*] fftpack --- Discrete Fourier Transform algorithms [*] io --- Data input and output [*] integrate --- Integration routines [*] lib.lapack --- Wrappers to LAPACK library [*] special --- Special Functions [*] lib.blas --- Wrappers to BLAS library [*] [*] - using a package requires explicit import (see pkgload) If we are looking for an algorithm to integrate a function, we might explore the integrate package:
import scipy.integrate scipy . integrate ?
produces:
============================================= Integration and ODEs (:mod:`scipy.integrate`) ============================================= .. currentmodule:: scipy.integrate Integrating functions, given function object ============================================ .. autosummary:: :toctree: generated/ quad -- General purpose integration dblquad -- General purpose double integration tplquad -- General purpose triple integration nquad -- General purpose n-dimensional integration fixed_quad -- Integrate func(x) using Gaussian quadrature of order n quadrature -- Integrate with given tolerance using Gaussian quadrature romberg -- Integrate func using Romberg integration quad_explain -- Print information for use of quad newton_cotes -- Weights and error coefficient for Newton-Cotes integration IntegrationWarning -- Warning on issues during integration Integrating functions, given fixed samples ========================================== .. autosummary:: :toctree: generated/ trapz -- Use trapezoidal rule to compute integral. cumtrapz -- Use trapezoidal rule to cumulatively compute integral. simps -- Use Simpson's rule to compute integral from samples. romb -- Use Romberg Integration to compute integral from -- (2**k + 1) evenly-spaced samples. .. seealso:: :mod:`scipy.special` for orthogonal polynomials (special) for Gaussian quadrature roots and weights for other weighting factors and regions. Integrators of ODE systems ========================== .. autosummary:: :toctree: generated/ odeint -- General integration of ordinary differential equations. ode -- Integrate ODE using VODE and ZVODE routines. complex_ode -- Convert a complex-valued ODE to real-valued and integrate.
The following sections show examples which demonstrate how to employ the algorithms provided by scipy.
Numerical integration¶
Scientific Python provides a number of integration routines. A general purpose tool to solve integrals I of the kind
$$I=\int_a^b f(x) \mathrm{d} x$$
is provided by the quad() function of the scipy.integrate module.
It takes as input arguments the function f(x) to be integrated (the "integrand"), and the lower and upper limits a and b. It returns two values (in a tuple): the first one is the computed results and the second one is an estimation of the numerical error of that result.
Here is an example: which produces this output:
In [4]:
from math import cos , exp , pi from scipy.integrate import quad # function we want to integrate def f ( x ): return exp ( cos ( - 2 * x * pi )) + 3.2 # call quad to integrate f from -2 to 2 res , err = quad ( f , - 2 , 2 ) print ( "The numerical result is {:f} (+- {:g} )" . format ( res , err ))
The numerical result is 17.864264 (+-1.55113e-11)
Note that quad() takes optional parameters epsabs and epsrel to increase or decrease the accuracy of its computation. (Use help(quad) to learn more.) The default values are epsabs=1.5e-8 and epsrel=1.5e-8. For the next exercise, the default values are sufficient.
Exercise: integrate a function¶
-
Using scipy's
quadfunction, write a program that solves the following integral numerically: $I = \int _0^1\cos(2\pi x) dx$. -
Find the analytical integral and compare it with the numerical solution.
-
Why is it important to have an estimate of the accuracy (or the error) of the numerical integral?
Exercise: plot before you integrate¶
It is good practice to plot the integrand function to check whether it is "well behaved" before you attempt to integrate. Singularities (i.e. $x$ values where the $f(x)$ tends towards minus or plus infinity) or other irregular behaviour (such as $f(x)=\sin(\frac{1}{x}$) close to $x = 0$ are difficult to handle numerically.
- Write a function with name
plotquadwhich takes the same arguments as the quad command (i.e. $f$, $a$ and $b$) and which - (i) creates a plot of the integrand $f(x)$ and
- (ii) computes the integral numerically using the
quadfunction. The return values should be as for thequadfunction.
Solving ordinary differential equations¶
To solve an ordinary differential equation of the type $$\frac{\mathrm{d}y}{\mathrm{d}t}(t) = f(y,t)$$
with a given $y(t_0)=y_0$, we can use scipy's odeint function. Here is a (self explaining) example program (useodeint.py) to find
$$y(t) \quad \mathrm{for}\quad t\in[0,2]$$
given this differential equation: $$\frac{\mathrm{d}y}{\mathrm{d}t}(t) = -2yt \quad \mathrm{with} \quad y(0)=1.$$
In [5]:
% matplotlib inline from scipy.integrate import odeint import numpy as N def f ( y , t ): """this is the rhs of the ODE to integrate, i.e. dy/dt=f(y,t)""" return - 2 * y * t y0 = 1 # initial value a = 0 # integration limits for t b = 2 t = N . arange ( a , b , 0.01 ) # values of t for # which we require # the solution y(t) y = odeint ( f , y0 , t ) # actual computation of y(t) import pylab # plotting of results pylab . plot ( t , y ) pylab . xlabel ( 't' ); pylab . ylabel ( 'y(t)' )
Out[5]:
<matplotlib.text.Text at 0x10cad9320>
The odeint command takes a number of optional parameters to change the default error tolerance of the integration (and to trigger the production of extra debugging output). Use the help command to explore these:
help ( scipy . integrate . odeint )
will show:
Help on function odeint in module scipy.integrate.odepack: odeint(func, y0, t, args=(), Dfun=None, col_deriv=0, full_output=0, ml=None, mu=None, rtol=None, atol=None, tcrit=None, h0=0.0, hmax=0.0, hmin=0.0, ixpr=0, mxstep=0, mxhnil=0, mxordn=12, mxords=5, printmessg=0) Integrate a system of ordinary differential equations. Solve a system of ordinary differential equations using lsoda from the FORTRAN library odepack. Solves the initial value problem for stiff or non-stiff systems of first order ode-s:: dy/dt = func(y, t0, ...) where y can be a vector. *Note*: The first two arguments of ``func(y, t0, ...)`` are in the opposite order of the arguments in the system definition function used by the `scipy.integrate.ode` class. Parameters ---------- func : callable(y, t0, ...) Computes the derivative of y at t0. y0 : array Initial condition on y (can be a vector). t : array A sequence of time points for which to solve for y. The initial value point should be the first element of this sequence. args : tuple, optional Extra arguments to pass to function. Dfun : callable(y, t0, ...) Gradient (Jacobian) of `func`. col_deriv : bool, optional True if `Dfun` defines derivatives down columns (faster), otherwise `Dfun` should define derivatives across rows. full_output : bool, optional True if to return a dictionary of optional outputs as the second output printmessg : bool, optional Whether to print the convergence message Returns ------- y : array, shape (len(t), len(y0)) Array containing the value of y for each desired time in t, with the initial value `y0` in the first row. infodict : dict, only returned if full_output == True Dictionary containing additional output information ======= ============================================================ key meaning ======= ============================================================ 'hu' vector of step sizes successfully used for each time step. 'tcur' vector with the value of t reached for each time step. (will always be at least as large as the input times). 'tolsf' vector of tolerance scale factors, greater than 1.0, computed when a request for too much accuracy was detected. 'tsw' value of t at the time of the last method switch (given for each time step) 'nst' cumulative number of time steps 'nfe' cumulative number of function evaluations for each time step 'nje' cumulative number of jacobian evaluations for each time step 'nqu' a vector of method orders for each successful step. 'imxer' index of the component of largest magnitude in the weighted local error vector (e / ewt) on an error return, -1 otherwise. 'lenrw' the length of the double work array required. 'leniw' the length of integer work array required. 'mused' a vector of method indicators for each successful time step: 1: adams (nonstiff), 2: bdf (stiff) ======= ============================================================ Other Parameters ---------------- ml, mu : int, optional If either of these are not None or non-negative, then the Jacobian is assumed to be banded. These give the number of lower and upper non-zero diagonals in this banded matrix. For the banded case, `Dfun` should return a matrix whose rows contain the non-zero bands (starting with the lowest diagonal). Thus, the return matrix `jac` from `Dfun` should have shape ``(ml + mu + 1, len(y0))`` when ``ml >=0`` or ``mu >=0``. The data in `jac` must be stored such that ``jac[i - j + mu, j]`` holds the derivative of the `i`th equation with respect to the `j`th state variable. If `col_deriv` is True, the transpose of this `jac` must be returned. rtol, atol : float, optional The input parameters `rtol` and `atol` determine the error control performed by the solver. The solver will control the vector, e, of estimated local errors in y, according to an inequality of the form ``max-norm of (e / ewt) <= 1``, where ewt is a vector of positive error weights computed as ``ewt = rtol * abs(y) + atol``. rtol and atol can be either vectors the same length as y or scalars. Defaults to 1.49012e-8. tcrit : ndarray, optional Vector of critical points (e.g. singularities) where integration care should be taken. h0 : float, (0: solver-determined), optional The step size to be attempted on the first step. hmax : float, (0: solver-determined), optional The maximum absolute step size allowed. hmin : float, (0: solver-determined), optional The minimum absolute step size allowed. ixpr : bool, optional Whether to generate extra printing at method switches. mxstep : int, (0: solver-determined), optional Maximum number of (internally defined) steps allowed for each integration point in t. mxhnil : int, (0: solver-determined), optional Maximum number of messages printed. mxordn : int, (0: solver-determined), optional Maximum order to be allowed for the non-stiff (Adams) method. mxords : int, (0: solver-determined), optional Maximum order to be allowed for the stiff (BDF) method. See Also -------- ode : a more object-oriented integrator based on VODE. quad : for finding the area under a curve. Examples -------- The second order differential equation for the angle `theta` of a pendulum acted on by gravity with friction can be written:: theta''(t) + b*theta'(t) + c*sin(theta(t)) = 0 where `b` and `c` are positive constants, and a prime (') denotes a derivative. To solve this equation with `odeint`, we must first convert it to a system of first order equations. By defining the angular velocity ``omega(t) = theta'(t)``, we obtain the system:: theta'(t) = omega(t) omega'(t) = -b*omega(t) - c*sin(theta(t)) Let `y` be the vector [`theta`, `omega`]. We implement this system in python as: >>> def pend(y, t, b, c): ... theta, omega = y ... dydt = [omega, -b*omega - c*np.sin(theta)] ... return dydt ... We assume the constants are `b` = 0.25 and `c` = 5.0: >>> b = 0.25 >>> c = 5.0 For initial conditions, we assume the pendulum is nearly vertical with `theta(0)` = `pi` - 0.1, and it initially at rest, so `omega(0)` = 0. Then the vector of initial conditions is >>> y0 = [np.pi - 0.1, 0.0] We generate a solution 101 evenly spaced samples in the interval 0 <= `t` <= 10. So our array of times is: >>> t = np.linspace(0, 10, 101) Call `odeint` to generate the solution. To pass the parameters `b` and `c` to `pend`, we give them to `odeint` using the `args` argument. >>> from scipy.integrate import odeint >>> sol = odeint(pend, y0, t, args=(b, c)) The solution is an array with shape (101, 2). The first column is `theta(t)`, and the second is `omega(t)`. The following code plots both components. >>> import matplotlib.pyplot as plt >>> plt.plot(t, sol[:, 0], 'b', label='theta(t)') >>> plt.plot(t, sol[:, 1], 'g', label='omega(t)') >>> plt.legend(loc='best') >>> plt.xlabel('t') >>> plt.grid() >>> plt.show() Exercise: using odeint¶
-
Open a new file with name
testodeint.pyfile in a text editor. -
Write a program that computes the solution y(t) of this ODE using the
odeintalgorithm: $$\frac{\mathrm{d}y}{\mathrm{d}t} = -\exp(-t)(10\sin(10t)+\cos(10t))$$ from $t=0$ to $t = 10$. The initial value is $y(0)=1$. -
You should display the solution graphically at points $t=0$, $t=0.01$, $t=0.02$, ..., $t=9.99$, $t=10$.
Hint: a part of the solution $y(t)$ is shown in the figure below.
Root finding¶
If you try to find a $x$ such that $$f(x)=0$$ then this is called root finding. Note that problems like $g(x)=h(x)$ fall in this category as you can rewrite them as $f(x)=g(x)−h(x)=0$.
A number of root finding tools are available in scipy's optimize module.
Root finding using the bisection method¶
First we introduce the bisect algorithm which is (i) robust and (ii) slow but conceptually very simple.
Suppose we need to compute the roots of f(x)=x 3 − 2x 2. This function has a (double) root at x = 0 (this is trivial to see) and another root which is located between x = 1.5 (where f(1.5)= − 1.125) and x = 3 (where f(3)=9). It is pretty straightforward to see that this other root is located at x = 2. Here is a program that determines this root numerically:
In [6]:
from scipy.optimize import bisect def f ( x ): """returns f(x)=x^3-2x^2. Has roots at x=0 (double root) and x=2""" return x ** 3 - 2 * x ** 2 # main program starts here x = bisect ( f , 1.5 , 3 , xtol = 1 e - 6 ) print ( "The root x is approximately x= %14.12g , \n " "the error is less than 1e-6." % ( x )) print ( "The exact error is %g ." % ( 2 - x ))
The root x is approximately x= 2.00000023842, the error is less than 1e-6. The exact error is -2.38419e-07.
The bisect() method takes three compulsory arguments: (i) the function f(x), (ii) a lower limit a (for which we have chosen 1.5 in our example) and (ii) an upper limit b (for which we have chosen 3). The optional parameter xtol determines the maximum error of the method.
One of the requirements of the bisection method is that the interval [a,b] has to be chosen such that the function is either positive at a and negative at b, or that the function is negative at a and postive at b. In other words: a and b have to enclose a root.
Exercise: root finding using the bisect method¶
-
Write a program with name
sqrttwo.pyto determine an approximation of $\sqrt{2}$ by finding a root x of the function $f(x)=2 − x^2$ using the bisection algorithm. Choose a tolerance for the approximation of the root of 10−8. -
Document your choice of the initial bracket $[a, b]$ for the root: which values have you chosen for a and for b and why?
-
Study the results:
-
Which value for the root x does the bisection algorithm return?
-
Compute the value of $\\sqrt{2}$ using
math.sqrt(2)and compare this with the approximation of the root. How big is the absolute error of x? How does this compare withxtol?
-
Root finding using the fsolve funcion¶
A (often) better (in the sense of "more efficient") algorithm than the bisection algorithm is implemented in the general purpose fsolve() function for root finding of (multidimensional) functions. This algorithm needs only one starting point close to the suspected location of the root (but is not garanteed to converge).
Here is an example:
In [7]:
from scipy.optimize import fsolve def f ( x ): return x ** 3 - 2 * x ** 2 x = fsolve ( f , 3 ) # one root is at x=2.0 print ( "The root x is approximately x= %21.19g " % x ) print ( "The exact error is %g ." % ( 2 - x ))
The root x is approximately x= 2.000000000000006661 The exact error is -6.66134e-15.
The return value[6] of fsolve is a numpy array of length n for a root finding problem with n variables. In the example above, we have n = 1.
Interpolation¶
Given a set of N points $(x_i, y_i)$ with $i = 1, 2, …N$, we sometimes need a function $\hat{f}(x)$ which returns $y_i = f(x_i)$ where $x == x_i$, and which in addition provides some interpolation of the data $(x_i, y_i)$ for all $x$.
The function y0 = scipy.interpolate.interp1d(x,y,kind='nearest') does this interpolation based on splines of varying order. Note that the function interp1d returns a function y0 which will then interpolate the x-y data for any given $x$ when called as $y0(x)$.
The code below demonstrates this, and shows the different interpolation kinds.
In [8]:
import numpy as np import scipy.interpolate import pylab def create_data ( n ): """Given an integer n, returns n data points x and values y as a numpy.array.""" xmax = 5. x = np . linspace ( 0 , xmax , n ) y = - x ** 2 #make x-data somewhat irregular y += 1.5 * np . random . normal ( size = len ( x )) return x , y #main program n = 10 x , y = create_data ( n ) #use finer and regular mesh for plot xfine = np . linspace ( 0.1 , 4.9 , n * 100 ) #interpolate with piecewise constant function (p=0) y0 = scipy . interpolate . interp1d ( x , y , kind = 'nearest' ) #interpolate with piecewise linear func (p=1) y1 = scipy . interpolate . interp1d ( x , y , kind = 'linear' ) #interpolate with piecewise constant func (p=2) y2 = scipy . interpolate . interp1d ( x , y , kind = 'quadratic' ) pylab . plot ( x , y , 'o' , label = 'data point' ) pylab . plot ( xfine , y0 ( xfine ), label = 'nearest' ) pylab . plot ( xfine , y1 ( xfine ), label = 'linear' ) pylab . plot ( xfine , y2 ( xfine ), label = 'cubic' ) pylab . legend () pylab . xlabel ( 'x' )
Out[8]:
<matplotlib.text.Text at 0x110221358>
Curve fitting¶
We have already seen in the numpy chapter that we can fit polynomial functions through a data set using the numpy.polyfit function. Here, we introduce a more generic curve fitting algorithm.
Scipy provides a somewhat generic function (based on the Levenburg-Marquardt algorithm )through scipy.optimize.curve_fit to fit a given (Python) function to a given data set. The assumption is that we have been given a set of data with points $x_1, x_2, …x_N$ and with corresponding function values $y_i$ and a dependence of $y_i$ on $x_i$ such that $y_i=f(x_i,\vec{p})$. We want to determine the parameter vector $\vec{p}=(p_1, p_2, \ldots, p_k)$ so that $r$, the sum of the residuals, is as small as possible:
$$r = \sum\limits_{i=1}^N \left(y_i - f(x_i, \vec{p})\right)^2$$
Curve fitting is of particular use if the data is noisy: for a given $x_i$ and $y_i=f(x_i,\vec{p})$ we have a (unknown) error term $\epsilon_i$ so that $y_i=f(x_i,\vec{p})+\epsilon_i$.
We use the following example to clarify this: $$f(x,\vec{p}) = a \exp(-b x) + c, \quad\mathrm{i.e.}\quad \vec{p}=\mathtt{a,b,c}$$
In [9]:
import numpy as np from scipy.optimize import curve_fit def f ( x , a , b , c ): """Fit function y=f(x,p) with parameters p=(a,b,c). """ return a * np . exp ( - b * x ) + c #create fake data x = np . linspace ( 0 , 4 , 50 ) y = f ( x , a = 2.5 , b = 1.3 , c = 0.5 ) #add noise yi = y + 0.2 * np . random . normal ( size = len ( x )) #call curve fit function popt , pcov = curve_fit ( f , x , yi ) a , b , c = popt print ( "Optimal parameters are a= %g , b= %g , and c= %g " % ( a , b , c )) #plotting import pylab yfitted = f ( x , * popt ) # equivalent to f(x, popt[0], popt[1], popt[2]) pylab . plot ( x , yi , 'o' , label = 'data $y_i$' ) pylab . plot ( x , yfitted , '-' , label = 'fit $f(x_i)$' ) pylab . xlabel ( 'x' ) pylab . legend ()
Optimal parameters are a=2.524, b=1.04295, and c=0.378715
Out[9]:
<matplotlib.legend.Legend at 0x11056dac8>
Note that in the source code above we define the fitting function $y = f(x)$ through Python code. We can thus fit (nearly) arbitrary functions using the curve_fit method.
The curve_fit function returns a tuple popt, pcov. The first entry popt contains a tuple of the OPTimal Parameters (in the sense that these minimise equation ([eq:1]). The second entry contains the covariance matrix for all parameters. The diagonals provide the variance of the parameter estimations.
For the curve fitting process to work, the Levenburg-Marquardt algorithm needs to start the fitting process with initial guesses for the final parameters. If these are not specified (as in the example above), the value "1.0" is used for the initial guess.
If the algorithm fails to fit a function to data (even though the function describes the data reasonably), we need to give the algorithm better estimates for the initial parameters. For the example shown above, we could give the estimates to the curve_fit function by changing the line
popt , pcov = curve_fit ( f , x , yi )
to
popt , pcov = curve_fit ( f , x , yi , p0 = ( 2 , 1 , 0.6 ))
if our initial guesses would be a = 2,b = 1 and c = 0.6. Once we take the algorithm "roughly in the right area" in parameter space, the fitting usually works well.
Fourier transforms¶
In the next example, we create a signal as a superposition of a 50 Hz and 70 Hz sine wave (with a slight phase shift between them). We then Fourier transform the signal and plot the absolute value of the (complex) discrete Fourier transform coefficients against frequency, and expect to see peaks at 50Hz and 70Hz.
In [10]:
import scipy import matplotlib.pyplot as plt pi = scipy . pi signal_length = 0.5 #[seconds] sample_rate = 500 #sampling rate [Hz] dt = 1. / sample_rate #time between two samples [s] df = 1 / signal_length #frequency between points in #in frequency domain [Hz] t = scipy . arange ( 0 , signal_length , dt ) #the time vector n_t = len ( t ) #length of time vector #create signal y = scipy . sin ( 2 * pi * 50 * t ) + scipy . sin ( 2 * pi * 70 * t + pi / 4 ) #compute fourier transform f = scipy . fft ( y ) #work out meaningful frequencies in fourier transform freqs = df * scipy . arange ( 0 ,( n_t - 1 ) / 2. , dtype = 'd' ) #d=double precision float n_freq = len ( freqs ) #plot input data y against time plt . subplot ( 2 , 1 , 1 ) plt . plot ( t , y , label = 'input data' ) plt . xlabel ( 'time [s]' ) plt . ylabel ( 'signal' ) #plot frequency spectrum plt . subplot ( 2 , 1 , 2 ) plt . plot ( freqs , abs ( f [ 0 : n_freq ]), label = 'abs(fourier transform)' ) plt . xlabel ( 'frequency [Hz]' ) plt . ylabel ( 'abs(DFT(signal))' )
Out[10]:
<matplotlib.text.Text at 0x1108284e0>
The lower plot shows the discrete Fourier transform computed from the data shown in the upper plot.
Optimisation¶
Often we need to find the maximum or minimum of a particular function f(x) where f is a scalar function but x could be a vector. Typical applications are the minimisation of entities such as cost, risk and error, or the maximisation of productivity, efficiency and profit. Optimisation routines typically provide a method to minimise a given function: if we need to maximise f(x) we create a new function g(x) that reverses the sign of f, i.e. g(x)= −f(x) and we minimise g(x).
Below, we provide an example showing (i) the definition of the test function and (ii) the call of the scipy.optimize.fmin function which takes as argument a function f to minimise and an initial value x 0 from which to start the search for the minimum, and which returns the value of x for which f(x) is (locally) minimised. Typically, the search for the minimum is a local search, i.e. the algorithm follows the local gradient. We repeat the search for the minimum for two values (x 0 = 1.0 and x 0 = 2.0, respectively) to demonstrate that depending on the starting value we may find different minimar of the function f.
The majority of the commands (after the two calls to fmin) in the file fmin1.py creates the plot of the function, the start points for the searches and the minima obtained:
In [11]:
from scipy import arange , cos , exp from scipy.optimize import fmin import pylab def f ( x ): return cos ( x ) - 3 * exp ( - ( x - 0.2 ) ** 2 ) # find minima of f(x), # starting from 1.0 and 2.0 respectively minimum1 = fmin ( f , 1.0 ) print ( "Start search at x=1., minimum is" , minimum1 ) minimum2 = fmin ( f , 2.0 ) print ( "Start search at x=2., minimum is" , minimum2 ) # plot function x = arange ( - 10 , 10 , 0.1 ) y = f ( x ) pylab . plot ( x , y , label = '$\cos(x)-3e^{-(x-0.2)^2}$' ) pylab . xlabel ( 'x' ) pylab . grid () pylab . axis ([ - 5 , 5 , - 2.2 , 0.5 ]) # add minimum1 to plot pylab . plot ( minimum1 , f ( minimum1 ), 'vr' , label = 'minimum 1' ) # add start1 to plot pylab . plot ( 1.0 , f ( 1.0 ), 'or' , label = 'start 1' ) # add minimum2 to plot pylab . plot ( minimum2 , f ( minimum2 ), 'vg' ,\ label = 'minimum 2' ) # add start2 to plot pylab . plot ( 2.0 , f ( 2.0 ), 'og' , label = 'start 2' ) pylab . legend ( loc = 'lower left' )
Optimization terminated successfully. Current function value: -2.023866 Iterations: 16 Function evaluations: 32 Start search at x=1., minimum is [ 0.23964844] Optimization terminated successfully. Current function value: -1.000529 Iterations: 16 Function evaluations: 32 Start search at x=2., minimum is [ 3.13847656]
Out[11]:
<matplotlib.legend.Legend at 0x11092c588>
Calling the fmin function will produce some diagnostic output, which you can also see above.
Return value of fmin¶
Note that the return value from the fmin function is a numpy array which – for the example above – contains only one number as we have only one parameter (here x) to vary. In general, fmin can be used to find the minimum in a higher-dimensional parameter space if there are several parameters. In that case, the numpy array would contain those parameters that minimise the objective function. The objective function $f(x)$ has to return a scalar even if there are more parameters, i.e. even if $x$ is a vector as in $f(\mathbf{x})$.
Other numerical methods¶
Scientific Python and Numpy provide access to a large number of other numerical algorithms including function interpolation, Fourier transforms, optimisation, special functions (such as Bessel functions), signal processing and filters, random number generation, and more. Start to explore scipy's and numpy's capabilities using the help function and the documentation provided on the web.
scipy.io: Scipy-input output¶
Scipy provides routines to read and write Matlab mat files. Here is an example where we create a Matlab compatible file storing a (1x11) matrix, and then read this data into a numpy array from Python using the scipy Input-Output library:
First we create a mat file in Octave (Octave is [mostly] compatible with Matlab):
octave : 1 > a = - 1 : 0.5 : 4 a = Columns 1 through 6 : - 1.0000 - 0.5000 0.0000 0.5000 1.0000 1.5000 Columns 7 through 11 : 2.0000 2.5000 3.0000 3.5000 4.0000 octave : 2 > save - 6 octave_a . mat a %save as version 6
Then we load this array within python:
In [12]:
from scipy.io import loadmat mat_contents = loadmat ( 'scipy/code/octave_a.mat' )
Out[13]:
{'__globals__': [], '__header__': b'MATLAB 5.0 MAT-file Platform: posix, Created on: Mon Aug 8 12:21:36 2016', '__version__': '1.0', 'a': array([[-1. , -0.5, 0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. ]])} Out[14]:
array([[-1. , -0.5, 0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. ]])
The function loadmat returns a dictionary: the key for each item in the dictionary is a string which is the name of that array when it was saved in Matlab. The key is the actual array.
A Matlab matrix file can hold several arrays. Each of those is presented by one key-value pair in the dictionary.
Let's save two arrays from Python to demonstrate that:
In [15]:
import scipy.io import numpy as np # create two numpy arrays a = np . linspace ( 0 , 50 , 11 ) b = np . ones (( 4 , 4 )) # save as mat-file # create dictionary for savemat tmp_d = { 'a' : a , 'b' : b } scipy . io . savemat ( 'data.mat' , tmp_d )
This program creates the file data.mat, which we can subsequently read using Matlab or here Octave:
HAL47:code fangohr$ octave GNU Octave, version 3.2.4 Copyright (C) 2009 John W. Eaton and others. <snip> octave:1> whos Variables in the current scope: Attr Name Size Bytes Class ==== ==== ==== ===== ===== ans 1x11 92 cell Total is 11 elements using 92 bytes octave:2> load data.mat octave:3> whos Variables in the current scope: Attr Name Size Bytes Class ==== ==== ==== ===== ===== a 11x1 88 double ans 1x11 92 cell b 4x4 128 double Total is 38 elements using 308 bytes octave:4> a a = 0 5 10 15 20 25 30 35 40 45 50 octave:5> b b = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Note that there are other functions to read from and write to in formats as used by IDL, Netcdf and other formats in scipy.io.
More → see Scipy tutorial.
Write a Numerical Formula for the Is Curve, Showing Y as a Function of R Alone.
Source: https://www.southampton.ac.uk/~fangohr/teaching/python/book/html/16-scipy.html
0 Response to "Write a Numerical Formula for the Is Curve, Showing Y as a Function of R Alone."
Post a Comment