How to optimise fit for function where a parameter depends on independant variable [duplicate]

Question

I have some data with three columns, x, y, and b. Here, b has associated values for each x value. (N.b. x always increases)

That is, the data looks something like:

x y b 1 4 7 2 5 8 3 6 9

Say this data is fit by some model y = ax + math.log(b). I could compute each value by doing something like:

def findB(x): for i in range(0, len(xData)): if xData[i] >= x: return bData[i] def f(x, a): b = findB(x) result = a*x + math.log(b) return result

Now say I want to optimise this fit for f(x,a) to find the parameter a using scipy.optimise.curve_fit.

I have attempted to do so like this:

popt, pcov = scipy.optimize.curve_fit(f, xData, yData)

but doing so results in the error:

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

This is due to the curve_fit function passing all the x data as one array resulting in the if xData[i] >= x: statement being rendered ambiguous.

I can't figure out a away around this. I have tried creating a paired list of xData and bData and passing this to curve_fit to no avail.

This works if I simply hardcode b to some pre-determined value so I believe I am calling the curve_fit function correctly. Advice welcome.

Thanks, Leporello, I hadn't spotted that question and the solution there seems much more elegant,. — Samidamaru
– Samidamaru, Commented Jul 3, 2019 at 11:58
Can you describe what findB is supposed to do in words? It's evolved from a single loop to a nested loop now, and frankly, neither one makes sense. — Mad Physicist
– Mad Physicist, Commented Jul 3, 2019 at 12:00
The problem is that curve_fit expects an array function, and if cannot be used on arrays, for that you use a mask. Same problem in math.log (it expects scalars, not vectors: use np.log instead). In your particular case, since all you are doing is the same as an excel VLOOKUP, and assuming your array is sorted (as in your example) you can get the i index using np.searchsorted: def findB(x): return bData[np.searchsorted(xData, x)] — Tarifazo
– Tarifazo, Commented Jul 3, 2019 at 12:07

Samidamaru · Accepted Answer · 2019-07-03 11:54:21Z

In discussion with some colleagues, I think we found the solution.

Using a numpy array for the values of b and replacing math.log with np.log seems to solve this:

def findB(x): tempArray = np.zeros(len(x)) for value in x: for i in range(0, len(xData)): if xData[i] >= value: tempArray[i] = bData[i] return tempArray

def f(x, a): b = findB(x) result = a*x + np.log(b) return result

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