高阶 function 可在 python 中调用

Question

I have the following code which I struggle to understand. It defines 2 functions and then it assign the the function to h, but what are X and Y?

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more info def func_max(f: Callable[[int], int], g: Callable[[int], int])-> Callable[[int], int]:

that takes as parameters the functions f and g as above. It returns the function h(x) that is defined on integers x and the return value of h on an integer x is equal to the maximum of m and n, where m = f(x) and n = g(x).

def f(x):
  return x**2
  
def g(x):
  return 5*x

h = max_func(f,g)
X = h(3)
Y = h(6)

The exercise is asking to assemble the following in the correct order, which I will try but it does not work, here it goes:

For example I don't understand how small x enters into play now...

def max_func(f,g):
  if f(x)> g(x): #For example I don't understand how small x enters into play now...
    return f(x)```
  else:
   return f(x)
def new_function (x:int): -> int
   return new_function

Answer 1

max_func should return a function that takes an argument ( x ), applies it to f and g and then return the maximal value:

def max_func(f, g):
    def mf(x):
        return max(f(x), g(x))
    return mf

Answer 2

High order functional programming can get real weird real quickly. Since this seems to be a homework exercise, I won't give you the straight answer, but consider this:

def make_function_print_arg(f):
  def new_function(x):
    print(f"Calling function with {x}")
    return f(x)
  return new_function

This defines a function that modifies the function it gets passed. I can use it like this:

def square(x):
  return x**2

print_and_square = make_function_print_arg(square)
arg = 5
ret = print_and_square(arg) # Prints out "Calling function with 5"
print(f"{arg} squared is {ret}") # Prints out "5 squared is 25"

So, make_function_print_arg is a function which takes a function f as an argument. Within its body, it defines a new function. This new function takes a single argument, prints that argument out, then calls f with that single argument. Finally, make_function_print_arg returns the new function it just defined.

Later on, we can call make_function_print_arg with a function that we've already defined, which returns a new function that's a lot like our old function, but with some modified behaviour.

Now, in your case, you want to define a function which takes two functions as arguments, calls both of them, and returns whichever result is greater. I'm pretty certain that one of the lines you need to unscramble should read, return g(x) , so I think either you or your teacher made a typo, but working around that, see if you can use the ideas in make_function_print_arg to manage it!

For your own education, you might also want to read about how decorators work, which is quite similar to what you're learning about right now: https://book.pythontips.com/en/latest/decorators.html

Answer 3

In your code, you are calling only f(x) in both if and else statement.

You can try:

def max_func(f,g):
  if f(x)> g(x): 
    return f(x)
  else:
   return g(x)
def new_function (x:int): -> int
   return new_function