CODEWITHSUNDEEP
pythonstatisticsregressionscipy-optimizelmfit
Jamal
Jamal

Reputation: 45

Lmfit Regression with multiple inputs but only one predictor variable

Im trying to fit a function described as below:

Y = 10*(b+5/c0+c1*x) assuming x is the predictor variable and Y is the predicted variable.

However, b is also an array with the same length as x and y, and it varies for each x value. I define the function as follows:

def func(x,b,c0,c1):
    return 10*(b+5/c0+c1*x)

where x is the predictor variable b is an array with the same length as X (each x value has a corresponding b value) c0 and C1 are model coefficients

Im trying to perform regression between y and x using lmfit Module, however, it seems that I cannot define parameters as arrays.

The code below gives the following error:

model = Model(func)
params = model.make_params(b=data["b"].values,c0=0.429, c1=0.867)
result = model.fit(Y, params, X=x,calc_covar=True)



"ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()"

I'm only interested in the relation between y and x, so is this way of doing it statistically wrong? Thanks

Upvotes: 0

Views: 186

Answers (1)

M Newville
M Newville

Reputation: 7862

In the language of lmfit, your b is "independent data", just like x in that it is a predictor, not a variable to be optimized in the fit.

By default, the first argument to a model function is assumed to be the one and only independent variable. Function arguments that are not keyword arguments with a non-numerical default value will be turned into variable parameters for the fit.

But you can specify different or other independent data as with

model = Model(func, independent_vars=['x', 'b'])
params = model.make_params(c0=0.429, c1=0.867)
result = model.fit(Y, params, X=x, b=data['b'], calc_covar=True)

FWIW, the fact that b is an array of the same length as x is actually not important to lmfit.Model: independent variables can be of any type. So, you could write your model function as

def func(predictor, c0, c1):
    return 10*(predictor['b']+5/c0+c1*predictor['x'])

and then use that as

model = Model(func)
params = model.make_params(c0=0.429, c1=0.867)
result = model.fit(Y, params, predictor={'x':x, b: data['b']}, calc_covar=True)

That may seem like overkill here, but it can be helpful for more complex problems.

Upvotes: 1

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