The evolution of a change calculator

changeCalculator.py

######## Version 1 - brute force
 
amountOfChange = 2.30
 
numQuarters = amountOfChange//.25
remainingChange = amountOfChange%.25
numDimes = remainingChange//.1
remainingChange = remainingChange%.1
numNickels = remainingChange//.05
remainingChange = remainingChange%.05
numPennies = remainingChange
 
print "From $%s, there are %s quarters, %s dimes, %s nickels, %s pennies." %(amountOfChange, numQuarters, numDimes, numNickels, numPennies)
 
# Output: From $2.3, there are 9.0 quarters, 0.0 dimes, 0.0 nickels, 0.05 pennies.
 
######## Version 2 - wrap it in a function
 
def change_calculator(amountOfChange):
        numQuarters = amountOfChange//.25
        remainingChange = amountOfChange%.25
	numDimes = remainingChange//.1
	remainingChange = remainingChange%.1
	numNickels = remainingChange//.05
	remainingChange = remainingChange%.05
	numPennies = remainingChange
 
	print "From $%s, there are %s quarters, %s dimes, %s nickels, %s pennies." %(amountOfChange, numQuarters, numDimes, numNickels, numPennies)
 
change = 2.30
change_calculator(change)
 
# Output: From $2.3, there are 9.0 quarters, 0.0 dimes, 0.0 nickels, 0.05 pennies.
 
######## Version 3 - implement loops
 
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [quarter, dime, nickel, penny]
 
	print "The amount of change is: ", amountOfChange
	for coin in coins:
		numCoin = amountOfChange//coin
		amountOfChange = amountOfChange%coin
		print "There are %s %s pieces." %(numCoin, coin)
 
change = 2.30
change_calculator(change)
 
'''
Output: 
The amount of change is:  2.3
There are 9.0 0.25 pieces.
There are 0.0 0.1 pieces.
There are 0.0 0.05 pieces.
There are 4.0 0.01 pieces.
 
------> Notice how many pennies there are... This isn't right!
There are a number of things we can try, but as we're dealing with floating point numbers, adding a little bit (smaller than any coin we have), may fix our 'rounding' error. 
 
Increasing the amount of change ever so slightly fixes the problem ->
 
The amount of change is:  2.30001
There are 9.0 0.25 pieces.
There are 0.0 0.1 pieces.
There are 1.0 0.05 pieces.
There are 0.0 0.01 pieces.
 
But that's not the most elegant solution. We need a better approach!
'''
 
######## Version 4 - fix errors (add round(amountOfChange, 2))
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [quarter, dime, nickel, penny]
 
	print "The amount of change is: ", amountOfChange
	for coin in coins:
		numCoin = amountOfChange//coin
		amountOfChange = amountOfChange%coin
		amountOfChange = round(amountOfChange, 2)
		print "There are %s %s pieces." %(numCoin, coin)
 
change = 2.30
change_calculator(change)
 
'''
Output:
The amount of change is:  2.3
There are 9.0 0.25 pieces.
There are 0.0 0.1 pieces.
There are 1.0 0.05 pieces.
There are 0.0 0.01 pieces.
 
Hey! We fixed it!
'''
 
######## Version 5 - Make the output look nicer - add a coin name
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [("Quarter", quarter), ("Dime", dime), ("Nickel", nickel), ("Penny", penny)]
 
	print "The amount of change is: ", amountOfChange
	for name, coin in coins:
		numCoin = amountOfChange//coin
		amountOfChange = amountOfChange%coin
		amountOfChange = round(amountOfChange, 2)
		print "There are %s %s." %(numCoin, name)
 
change = 2.30
change_calculator(change)
 
'''
Output:
The amount of change is:  2.3
There are 9.0 Quarter.
There are 0.0 Dime.
There are 1.0 Nickel.
There are 0.0 Penny.
 
'''
 
######## Version 6 - Make the output look nicer - make the correct things plural
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [("Quarter", quarter), ("Dime", dime), ("Nickel", nickel), ("Penny", penny)]
 
	print "The amount of change is: ", amountOfChange
	for name, coin in coins:
		numCoin = amountOfChange//coin
		if numCoin > 0:
			amountOfChange = amountOfChange%coin
			amountOfChange = round(amountOfChange, 2)
 
			if numCoin > 1:
				# Note the special case for plural with penny
				if name[-1] == "y":
					name = name[:-1] + "ies"
				else:
					name = name+"s"
			
			print "There are %s %s." %(numCoin, name)
 
change = 2.30
change_calculator(change)
 
'''
Output:
The amount of change is:  2.3
There are 9.0 Quarters.
There are 1.0 Nickel.
 
'''
 
######## Version 7 - Make the output look nicer - take care of is/are
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [("Quarter", quarter), ("Dime", dime), ("Nickel", nickel), ("Penny", penny)]
 
	print "The amount of change is: ", amountOfChange
	for name, coin in coins:
		numCoin = amountOfChange//coin
		if numCoin > 0:
			amountOfChange = amountOfChange%coin
			amountOfChange = round(amountOfChange, 2)
 
			isAre = "is"
			if numCoin > 1:
				isAre = "are"
				# Note the special case for plural with penny
				if name[-1] == "y":
					name = name[:-1] + "ies"
				else:
					name = name+"s"
			
			print "There %s %s %s." %(isAre, numCoin, name)
 
change = 2.30
change_calculator(change)
 
'''
Output:
The amount of change is:  2.3
There are 9.0 Quarters.
There is 1.0 Nickel.
'''
 
######## Version 8 - Make the output look nicer - convert numCoin to int
def change_calculator(amountOfChange):
        quarter = .25
	dime = .1
	nickel = .05
	penny = .01
 
	coins = [("Quarter", quarter), ("Dime", dime), ("Nickel", nickel), ("Penny", penny)]
 
	print "The amount of change is: ", amountOfChange
	for name, coin in coins:
		numCoin = amountOfChange//coin
		if numCoin > 0:
			amountOfChange = amountOfChange%coin
			amountOfChange = round(amountOfChange, 2)
 
			isAre = "is"
			if numCoin > 1:
				isAre = "are"
				# Note the special case for plural with penny
				if name[-1] == "y":
					name = name[:-1] + "ies"
				else:
					name = name+"s"
			
			print "There %s %s %s." %(isAre, int(numCoin), name)
 
change = 2.30
change_calculator(change)
 
'''
Output:
The amount of change is:  2.3
There are 9 Quarters.
There is 1 Nickel.
'''