Python: partial that takes three arguments

Question

I'm trying to learn Python by reading the book Data Science from Scratch by Joel Grus, and on page 94 they describe how to approximate a derivative of f = x^2 using the following code

def difference_quotient(f, x, h):
    return (f(x + h) - f(x)) / h

def square(x):
    return x * x

def derivative(x):
    return 2 * x

derivative_estimate = partial(difference_quotient, square, h=0.00001)

# plot to show they're basically the same
import matplotlib.pyplot as plt
x = range(-10,10)
plt.title("Actual Derivatives vs. Estimates")
plt.plot(x, map(derivative, x), 'rx', label='Actual')
plt.plot(x, map(derivative_estimate, x), 'b+', label='Estimate')
plt.legend(loc=9)
plt.show()

Everything works fine, but when I change the line derivative_estimate = partial(difference_quotient, square, h=0.00001) to derivative_estimate = partial(difference_quotient, f=square, h=0.00001) (because I think that is clearer to read), then I get the following error

Traceback (most recent call last):
  File "page_93.py", line 37, in <module>
    plt.plot(x, map(derivative_estimate, x), 'b+', label='Estimate')
TypeError: difference_quotient() got multiple values for keyword argument 'f'

What is going on here?

Valediction

Where the function partial comes from? Some library? I believe it is not built in. – Viquelia

@Viquelia coming from functools. – Urbana

If you're lucky, Joel himself may answer your question. :) – Bourbonism

derivative_estimate = partial(f=difference_quotient, x =square , h=0.00001) is the right way to do it. If the first argument is passed as a keyword argument then the rest of your arguments must be passed as keyword only arguments, otherwise you'll get SyntaxError: non-keyword arg after keyword arg if you run Python 3.X – Quaff

Answer 1

It was answered and perfectly explained in this topic:

• functools.partial wants to use a positional argument as a keyword argument

Which in your case implies that you should pass x as a keyword argument:

plt.plot(x, [derivative_estimate(x=item) for item in x], 'b+', label='Estimate')

Urbana

Thank you, your suggestion works and I understand it, so I have accepted your answer. But I still don't really see why the code that I suggested failed. I have read the link you provided and understand it in that example, but I don't see at what point I use the same argument twice. – Valediction

@Valediction sure, when you call map(derivative_estimate, x) - python would actually call derivative_estimate(x[0]), then derivative_estimate(x[1]) etc. x[0], x[1] would be passed as the first positional argument to difference_quotient which is f. And, since you specify f keyword argument, Python complains that f was specified multiple times. Hope that helps. – Urbana

Ahhh I see, great. Thank you!! – Valediction