Ken Oliver wrote:
>
> -----Original Message-----
>> From: Kent Johnson <[EMAIL PROTECTED]>
>> Sent: Aug 6, 2007 3:46 PM
>> To: Dick Moores <[EMAIL PROTECTED]>
>> Cc: Python Tutor List <tutor@python.org>
>> Subject: Re: [Tutor] Ingenious script (IMO)
>>
>> Dick Moores wrote:
>>> At 10:16 AM 8/6/2007, Eric Brunson wrote:
>>>
>>> Your point about efficiency is well-taken.
>>>
>>>> def makechange( amount, denominations ):
>>>>
>>>> coins = {}
>>>> for d in denominations:
>>>> coins[d] = int( amount/d )
>>>> amount = amount%d
>>>>
>>>> return coins
>>> OK, I used this this way:
>>>
>>> ============================
>>> def makechange( amount, denominations ):
>>>
>>> coins = {}
>>> for d in denominations:
>>> coins[d] = int( amount/d )
>>> amount = amount%d
>>>
>>> return coins
>>>
>>> denominations = (2000, 1000, 500, 100, 50, 25, 10, 5, 1)
>>> amount = 2218
>>> print makechange(2218, denominations)
>>> ==================================
>>>
>>> And get:
>>> {1: 3, 100: 2, 5: 1, 1000: 0, 10: 1, 2000: 1, 50: 0, 500: 0, 25: 0}
>>>
>>> That's the correct change: 3 pennies, 2 $1 bills, 1 nickel, no $10
>>> bills, 1 dime, 1 $20 bill, no half-dollars, no $5 bills, no quarters.
>>>
>>> For amount = 3288:
>>> {1: 3, 100: 2, 5: 0, 1000: 1, 10: 1, 2000: 1, 50: 1, 500: 0, 25: 1}
>>>
>>> Why those weird orders?
>> Dictionaries are not ordered. If you want to see it in order you could
>> sort the list of key, value pairs:
>> print sorted(makechange(2218, denominations).items(), reverse=True)
>>
>> Kent
>
> Do you have a clever, pythonic way to suppress the denominations with "zero"
> counts?
Sure, just use a list comprehension to filter the list of items:
non_zero_items = [(k, v) for k, v in makechange(2218,
denominations).items() if v!=0]
print sorted(non_zero_items, reverse=True)
Kent
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