Fitting only one parameter of a function with many parameters in python

Question

In python I have a function which has many parameters. I want to fit this function to a data set, but using only one parameter, the rest of the parameters I want to supply o

Answer 1

Another way is to use upper and lower bounds that are identical (+ eps) as the initial value. Using the same example with initial conditions and bounds:

def func(x,a,b):
   return a*x*x + b
# free for a and b
popt,pcov = curve_fit(func, x1, x2, 
                      p0=[1,1], 
                      bounds=[(-inf,-inf),(inf,inf)])

# free for a; fixed for b  ; 
eps=1/100
popt,pcov = curve_fit(func, x1, x2, 
                      p0=[1,1], 
                      bounds=[(-inf,(1-eps)),(inf,(1+eps))])

Remember to insert an epsilon, otherwise, a and b must be the same

Answer 2

There is a simpler option if you are willing/able to edit the original function.

Redefine your function as:

def func(x,a):
    return a*x*x + b

Then you can simply put it in your loop for parameter b:

for b in xrange(10):
   popt,pcov = curve_fit(func, x1, x2)

Caveat: the function needs to be defined in the same script in which it is called for this to work.

Answer 3

You can wrap func in a lambda, as follows:

def func(x,a,b):
   return a*x*x + b

for b in xrange(10):
   popt,pcov = curve_fit(lambda x, a: func(x, a, b), x1, x2)

A lambda is an anonymous function, which in Python can only be used for simple one line functions. Basically, it's normally used to reduce the amount of code when don't need to assign a name to the function. A more detailed description is given in the official documentation: http://docs.python.org/tutorial/controlflow.html#lambda-forms

In this case, a lambda is used to fix one of the arguments of func. The newly created function accepts only two arguments: x and a, whereas b is fixed to the value taken from the local b variable. This new function is then passed into curve_fit as an argument.

Answer 4

Scipy's curve_fit takes three positional arguments, func, xdata and ydata. So an alternative approach (to using a function wrapper) is to treat 'b' as xdata (i.e. independent variable) by building a matrix that contains both your original xdata (x1) and a second column for your fixed parameter b.

Assuming x1 and x2 are arrays:

def func(xdata,a):
   x, b = xdata[:,0], xdata[:,1]  # Extract your x and b
   return a*x*x + b

for b in xrange(10): 
   xdata = np.zeros((len(x1),2))  # initialize a matrix
   xdata[:,0] = x1  # your original x-data
   xdata[:,1] = b  # your fixed parameter
   popt,pcov = curve_fit(func,xdata,x2)  # x2 is your y-data

Answer 5

A better approach would use lmfit, which provides a higher level interface to curve-fitting. Among other features, Lmfit makes fitting parameters be first-class objects that can have bounds or be explicitly fixed (among other features).

Using lmfit, this problem might be solved as:

from lmfit import Model
def func(x,a,b):
   return a*x*x + b

# create model
fmodel = Model(func)
# create parameters -- these are named from the function arguments --
# giving initial values
params = fmodel.make_params(a=1, b=0)

# fix b:
params['b'].vary = False

# fit parameters to data with various *static* values of b:
for b in range(10):
   params['b'].value = b
   result = fmodel.fit(ydata, params, x=x)
   print(": b=%f, a=%f+/-%f, chi-square=%f" % (b, result.params['a'].value, 
                                             result.params['a'].stderr,
                                             result.chisqr))

Answer 6

Instead of using the lambda function which might be less intuitive to digest I would recommend to specify the scikit curve_fit parameter bounds that will force your parameter to be searched within custom boundaries.

All you have to do is to let your variable a move between -inf and +inf and your variable b between (b - epsilon) and (b + epsilon)

In your example:

epsilon = 0.00001

def func(x,a,b):
    return a*x*x + b

for b in xrange(10):
    popt,pcov = curve_fit(func,x1,x2, bounds=((-np.inf,b-epsilon), (np.inf,b+epsilon))