Python: Taking parameter input and returning output equation

Question

I am working on a project that requires a function (I'll call it target) that will take an input named parameter_vector and will output an equation for other functions to take as an input. The output equation I am using as a test case is the standard quadratic equation:

Ax^2 + Bx + C

My target function will take a parameter_vector in that looks like:

parameter_vector = [A, B, C]

and will use these indices as coefficients in writing out my equation as an output. Here is my code so far:

def target(parameter_vector):
    pv = parameter_vector
    x = pv[0]*x**2 + pv[1]*x + pv[2]
    return x

As I'm sure you can guess, the following error pops up in my shell:

UnboundLocalError: local variable 'x' referenced before assignment

So I did a little research and came across the lambda variable that is used for place-holding a variable. I tried to input:

x = pv[0]*lambda**2 + pv[1]*x +pv[2]

But this popped up:

File "source_parameters.py", line 7
x = pv[0]*lambda**2 + pv[1]*lambda +pv[2]
               ^
SyntaxError: invalid syntax

Am I approaching this completely wrong? I'm sure even sure whether the output should be a string or not. Any help would be greatly appreciated!

Martijn Pieters · Accepted Answer · 2015-01-12 22:20:17Z

2

You misunderstood how lambdas work; they create a function from an expression. Given x as input, that'd look like this:

lambda x: pv[0]*x**2 + pv[1]*x + pv[2]

The expression produces a function object, which you can return directly from your target() function:

def target(pv):
    return lambda x: pv[0]*x**2 + pv[1]*x + pv[2]

You could unpack the parameter_vector into the three components first to make it clearer you need to have 3 values in it:

def target(parameter_vector):
    a, b, c = parameter_vector
    return lambda x: a * x ** 2 + b * x + c

or you can generalise this to any length of parameter_vector:

def target(parameter_vector):
    return lambda x: sum(p * x ** (len(parameter_vector) - i - 1)
                         for i, p in enumerate(parameter_vector))

Quick demo:

>>> lambda x: pv[0]*x**2 + pv[1]*x + pv[2]
<function <lambda> at 0x102a0c6e0>
>>> def target(parameter_vector):
...     a, b, c = parameter_vector
...     return lambda x: a * x ** 2 + b * x + c
... 
>>> new_function = target([3, 2, 5]) # 3x^2 + 2x + 5
>>> new_function
<function <lambda> at 0x102a0c7d0>
>>> new_function(2)
21
>>> new_function(3)
38
>>> new_function(4)
61
>>> def target(parameter_vector):
...     return lambda x: sum(p * x ** (len(parameter_vector) - i - 1)
...                          for i, p in enumerate(parameter_vector))
... 
>>> new_function = target([3, 2, 5])
>>> new_function(2)
21
>>> new_function = target([3, 2, 5, -2])  # 3x^3 + 2x^2 + 5x - 2
>>> new_function(2)
40

edited Jan 12, 2015 at 22:20

answered Jan 12, 2015 at 21:45

Martijn Pieters

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4 Comments

ibanez221 Over a year ago

Great! That is exactly what I am looking for. Is there a way to define an unknown number of indices in place of a,b,c for a more complex function? Let's say it has 12 coefficients but isn't known to the function prior to importing a different script. Is there a way to generalize each index but still use the unpacking step?

Martijn Pieters Over a year ago

@ibanez221: no, unpacking won't let you do that. You'd have to use a sum() of the parameters, each multiplied with x to the power len(parameter_vector) - index (where index is the index of the current parameter): lambda x: sum(p * x ** (len(pv) - i - 1) for i, p in enumerate(pv)).

Martijn Pieters Over a year ago

@ibanez221: added that option to the answer.

ibanez221 Over a year ago

Thank you so much! I never would have come up with that solution. This is very helpful, I appreciate it.

Cory Kramer · Accepted Answer · 2015-01-12 21:46:35Z

2

You're on the right track. You can return a lambda function as follows.

def target(pv):
    return lambda i : pv[0]*i*i + pv[1]*i + pv[2]

Testing that we do get a function back

>>> target([1,3,4])
<function target.<locals>.<lambda> at 0x029E8DB0>

Let's assign the function to f

>>> f = target([1,3,4])

Now we can call the function with an argument

>>> f(5)
44

answered Jan 12, 2015 at 21:46

Cory Kramer

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