jimjilbang3.py

# Challenge one: What does this function do? How do you use it?

def welcome(name):
	return 'Hello ' + name

# Solution:

### When called and a name passed, this function returns a greeting string for that name. Importantly, this function doesn't print anything to the console, it just creates a string that we can then print later, or do other things with.

person = 'Graham'

greeting = welcome(person)

print(greeting)
## Outputs "Hello Graham"

# or:

print(welcome('Graham'))
## This also outputs "Hello Graham". Here, we're passing the name directly into the function, and passing the function's output directly to the print function.

# Challenge two: Write a function to double a number. Test it.

# Lee & Andrew W's solutions:
def calc(number):
  return 2 * number

# Usage
print(calc(8))
## As above, we're passing the result of calc(8) directly into the print function. It'll print out 16 to the console.

# Multiple parameters

def total_wheels(bikes, cars, trucks):
	bike_wheels = bikes * 2
	car_wheels = cars * 4
	truck_wheels = trucks * 18
	return bike_wheels + car_wheels + truck_wheels

print(total_wheels(3, 1, 0))
## This prints the number 10, which is the total number of wheels for 3 bikes and a car.

# Named parameters

print(total_wheels(trucks=1, bikes=2, cars=0))
## Notice that the parameters are passed in a different order to which they're declared. As long as they're named, this is fine.

# Conditions

def calculate_pay(hours_worked, hourly_rate):
	# Working weeks are 40 hours and overtime is paid at time and a half.
	if hours_worked <= 40:
		return hours_worked * hourly_rate
	else:
		overtime = 40 - hours_worked
		return (40 + 1.5 * overtime) * hourly_rate
 
# Else clause isn't required:

def calculate_pay(hours_worked, hourly_rate):
	# Working weeks are 40 hours and overtime is paid at time and a half.
	if hours_worked <= 40:
	    return hours_worked * hourly_rate
	overtime = 40 - hours_worked
	return (40 + 1.5 * overtime) * hourly_rate

# Challenge three: Write a function that takes two numbers and returns the largest.

# Andrew C

def highest(numb1, number2):
	if numb1 > numb2:
		return numb1
	if numb2 > numb1:
		return numb2

print(highest(5, 10))

## Andrew's taken my poor requirements very literally. If numb1 and numb2 are the same, the function won't hit a return statement at all. In that case, the function will return a special Python type 'None'. Let this be a lesson to me about specifying what I want!

# Lee
def largest(number1, number2):
  if number1 >= number2:
    return number1
  elif number1 <= number2:
    return number2

print largest(10, 20)

## Lee's solution will always hit a return statement as he uses >=, which is 'greater than or equal to'. In the case that the numbers are the same, it'll return the number that they both are.

# Andrew W

def high_number(number1, number2):
	if number1 > number2:
		return number1	
	else:
		return number2

print(high_number(40, 5))

## Andrew's solution is very elegant. After one comparison, we don't need to check the reverse so we can assume number2 >= number1.

# Challenge four: Write a function that takes three numbers and returns the largest.

# 'and'

if number1 > number2 and number1 > number3: # Good
if number1 > number2 and number3: # Bad

## Python uses 'and' and 'or' to combine logical statements. The Good example here works, but watch out for the Bad example, which will behave strangely.

# Lee
def largest(number1, number2, number3):
        if number1 > number2 and number1 > number3:
                return number1
        elif number2 > number1 and number2 > number3:
                return number2
        elif number3 > number1 and number3 > number2:
                return number3

print largest(1, 2, 3)

#Sonal
def largest(num1, num2, num3):
	if num1 > num2 and num1 > num3:
		return num1
	elif num2 > num1 and num2 > num3:
		return num2
	elif num3 > num1 and num3 > num2:
		return num3
		
print (largest(10, 20, 30))


#andrew w

def high_number(number1, number2, number3):
	if number1 > number2 and number1 > number3:
		return number1	
	elif number2 > number3:
		return number2
	else: return number3

print(high_number(45, 41, 66))

## Here, Andrew W uses the information gained from earlier comparisons to reduce the number of checks to make later on.

# Andrew C

def highest(numb1, numb2):
	if numb1 > numb2:
		return numb1
	if numb2 > numb1:
		return numb2

def higher(numba, numbb, numbc):
	highestnumb = highest(numba, numbb)
	highestnumb = highest(highestnumb, numbc)
	
print(higher(5, 3, 60))

## Andrew C reuses our earlier function, which finds the highest of two numbers, to help perform the work for the 'highest of three' function. This is good practice in any sort of programming.