How do I return a value from a higher-order function? - python-3.x

Question

guys how can I make it so that calling make_repeater(square, 0)(5) return 5 instead of 25? I'm guessing I would need to change the line "function_successor = h" because then I'm just getting square(5) but not sure what I need to change it to...
square = lambda x: x * x
def compose1(h, g):
"""Return a function f, such that f(x) = h(g(x))."""
def f(x):
return h(g(x))
return f
def make_repeater(h, n):
iterations = 1
function_successor = h
while iterations < n:
function_successor = compose1(h, function_successor)
iterations += 1
return function_successor
it needs to satisfy a bunch of other requirements like:
make_repeater(square, 2)(5) = square(square(5)) = 625
make_repeater(square, 4)(5) = square(square(square(square(5)))) = 152587890625

Answer 1

To achieve that, you have to use the identity function (f(x) = x) as the initial value for function_successor:
def compose1(h, g):
"""Return a function f, such that f(x) = h(g(x))."""
def f(x):
return h(g(x))
return f
IDENTITY_FUNCTION = lambda x: x
def make_repeater(function, n):
function_successor = IDENTITY_FUNCTION
# simplified loop
for i in range(n):
function_successor = compose1(function, function_successor)
return function_successor
if __name__ == "__main__":
square = lambda x: x * x
print(make_repeater(square, 0)(5))
print(make_repeater(square, 2)(5))
print(make_repeater(square, 4)(5))
and the output is
5
625
152587890625
This isn't most optimal for performance though since the identity function (which doesn't do anything useful) is always part of the composed function, so an optimized version would look like this:
def make_repeater(function, n):
if n <= 0:
return IDENTITY_FUNCTION
function_successor = function
for i in range(n - 1):
function_successor = compose1(function, function_successor)
return function_successor