Number helpers

Add number conversion and calculation methods.

The clamp method

We'll write a clamp method to keep a number within a range. clamp(n, lo, hi) returns lo if n is below lo; hi if above hi; otherwise n.

Example: clamp(15, 0, 10) → 10, because 15 is above 10. In Python we can write it in one line as max(lo, min(hi, n)). First min(hi, n) applies the upper bound, then max(lo, ...) applies the lower bound.

Bounding with max and min

min(a, b) returns the smaller, max(a, b) returns the larger. We build clamp from these: first min(hi, n) applies the upper bound, then max(lo, ...) applies the lower bound.

Example: clamp(15, 0, 10) = max(0, min(10, 15)) = max(0, 10) = 10. Read inside out. This form is shorter than if branches and still readable.

Exercise Add clamp to Utility

Add a def clamp(self, n, lo, hi): method to the Utility class. In the body write return max(lo, min(hi, n)). Then create an instance with u = Utility() and call print(u.clamp(15, 0, 10)).

Show example solution

class Utility:
    version = "1.0"
    def slugify(self, text):
        return text.lower().replace(" ", "-")
    def clamp(self, n, lo, hi):
        return max(lo, min(hi, n))

u = Utility()
print(u.clamp(15, 0, 10))

Test yourself

Match the clamp parts

clamp

Method name
max(lo, min(hi, n))

Bounding
15, 0, 10

Example input
What does clamp(15, 0, 10) return?
- 5
- 15
- 10
- 0
Show answer

10 — Since 15 is above 10, the upper bound 10 is returned.

Complete the clamp body

class Utility:
    def clamp(self, n, lo, hi):
        return max(lo, min(hi, ____))

Show answer

class Utility:
    def clamp(self, n, lo, hi):
        return max(lo, min(hi, n))

Next lesson: Date helpers

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