CODEWITHSUNDEEP
pythonscipy
Kartik
Kartik

Reputation: 9853

Python Using Lambda Within a Function

I am trying to create a function called calc(f,a,b) where x is an equation with the variable f and I want to put this code within the function.

   def calc(f, a, b):
       limits = [a, b]
       integral = odeint(lambda y, x : f, 0, limits)

       return integral[1]

This function gets the integral using the built in odeint function. This is what I am trying to do

print calc(x**2, 0, 1)

where x^2 is the function to be integrated. My problem is that this function (x**2)needs to be passed on to the odeint function right after y, x: f where f after the semicolon is the f from the calc(f,a,b)

what I cant figure out is that how can I pass f from the calc function input to the odeint inside. It says that f isnt declared and if I put it within strings.. it doest work

When I run this function.. it doesnt work I get this error

NameError: name 'f' is not defined

I am not sure how to pass my equation to be integrated inside odeint

Thanks

Upvotes: 1

Views: 3003

Answers (4)

Adeel Zafar Soomro
Adeel Zafar Soomro

Reputation: 1522

If one were to rewrite the function calc as follows:

def calc(f, a, b):
    limits = [a, b]
    integral = odeint(lambda y, x: f(x), 0, limits)

    return integral[1][0]

Then one may use this function thus:

>>> calc(lambda x: x ** 2, 0, 1)    # Integrate x ** 2 over the interval [0, 1] (expected answer: 0.333...)
0.33333335809177234
>>> calc(lambda x: x, 0, 1)         # Integrate x over the interval [0, 1] (expected answer: 0.5)
0.50000001490120016
>>> calc(lambda x: 1, 0, 1)         # Integrate 1 over the interval [0, 1] (expected answer: 1.0)
1.0

The odeint function from the scipy.integrate module has the signature:

odeint(func, y0, t, ...)

where: func is a callable that accepts parameters y, t0, ... and returns dy/dt at the given point; y0 is a sequence representing initial condition of y; t is a sequence that represents intervals to solve for y (t0 is the first item in the sequence).

It appears that you are solving a first-order differential equation of the form dy/dx = f(x) over the interval [a, b] where y0 = 0. In such a case, when you pass f (which accepts one argument) to the function odeint, you must wrap it in a lambda so that the passed-in function accepts two arguments (y and x--the y parameter is essentially ignored since you need not use it for a first-order differential equation).

Upvotes: 2

martineau
martineau

Reputation: 123463

I'm not completely sure because odeint() is not a built-in Python function so I don't know much about it, but the first argument you're passing it in your code is not a function that computes x^2. An easy way to do something like that would be to pass a lambda function to calc that does that sort of calculation. For example:

def calc(f, a, b):
    limits = [a, b]
    integral = odeint(f, 0, limits)

    return integral[1]

print calc(lambda x: x**2, 0, 1)

Upvotes: 0

Hugh Bothwell
Hugh Bothwell

Reputation: 56644

x cannot have two values; therefore, if you need two values, one of them must be named something else. Rename one of your variables.

Edit:

(smacking forehead): In calc(x**2, 0, 1), x**2 is not a function - it is an expression, which gets evaluated before being passed to calc - therefore it complains about it needs to know what x is (in order to calculate x**2).

Try

calc(lambda x: x**2, a, b)

instead. This is equivalent to

def unnamedfunction(x):
    return x**2

calc(unnamedfunction, a, b)

Upvotes: 0

kindall
kindall

Reputation: 184171

I assume odeint is some function to which you are passing the lambda function. odeint will presumably call the lambda and needs to pass x and y to it. So the answer is, if you want odeint to call the function and pass it x and y, then you need to pass x and y to odeint as arguments, in addition to the function itself.

What exactly are you trying to do here? With more details and more code, we could probably get a better answer.

Upvotes: 1

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