Re: [Tutor] Projects (fwd)
Tiger12506 wrote: Nope, if you read the code you'll see the only mapping done is up to 20 and then by tens up to 100, that's all. The same code could be used with a list, you'd only have to change the exception name. I see. There were ... in between each of the tens entries which I took to mean that big huge dictionary but left out some values in email for simplicity. I feel ashamed for not recognizing that the ellipses where there for some other purpose. ;-) Checked my mail and yes, it was not outright clear what I meant in the dictionary, sorry. Anyway if you check the rest of the code you'll see it works as I said and you don't use the intermediate values between the tens. As usual I learned something from the exercise, I thought that indexed access would be faster than list access, didn't think lists where implemented as C arrays (because they have no fixed size), now I can see tuples will be faster and will save memory because memory will not have to be reserved in chunks. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Projects (fwd)
Tiger12506 wrote: Isn't dictionary access faster than list access? Why are three lists 'much more efficient'? Oh no, no, no. Dictionaries are faster when you are *searching through* for a particular value. If you already know the index of the item in the list, lists are much faster. Dictionaries are hash based. Somewhere it has to calculate the hash of the key you give it... These three lists are more efficient in terms of size/output ratio. It appeared as if the dictionary that was presented as an example was just going to map one to one all of the values from zero to 999,999 (to match my list version capabilities). Not only is that bad programming style, it's just plain unecessary. Nope, if you read the code you'll see the only mapping done is up to 20 and then by tens up to 100, that's all. The same code could be used with a list, you'd only have to change the exception name. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Projects (fwd)
Hi Jason,
Looking back at that Java code:
static String convertDigitToEnglish(int d) {
switch ( d )
{
case 1: return one;
case 2: return two;
case 3: return three;
case 4: return four;
case 5: return five;
case 6: return six;
case 7: return seven;
case 8: return eight;
case 9: return nine;
default: return \nFatal Error!\n; // should I abort pgm?
} // end of switch
} // end of convertDigitToEnglis
Frankly, this seems silly to me. First, it ignores zero, which is a cardinal
sin. I'm being somewhat serious about this: functions should do what they say,
and that function isn't.
But the code could also be written much more tightly as:
static String digitToString(int n) {
String[] words = {zero, one, two, three, four,
five, six, seven, eight, nine};
if (0 = n n 10) {
return words[n];
}
throw new IllegalArgumentException(input not a single digit);
}
I don't mean to make this post so Java-centric; it just seems a little unfair
to compare a bad Java routine to a good Python routine. :) Writing an
equivalent in Python is also pretty straightforward:
#
## Pseudocode: just a sketch
def digitToString(n):
words = [zero, one, ...]
if 0 = n 10:
return words[n]
...
#
Like dictionaries, the list data structure works pretty well for key/value
lookup if the input key is a small number.
Good luck!
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Re: [Tutor] Projects (fwd)
Danny Yoo wrote:
Hi Jason,
Looking back at that Java code:
static String convertDigitToEnglish(int d) {
switch ( d )
{
case 1: return one;
case 2: return two;
case 3: return three;
case 4: return four;
case 5: return five;
case 6: return six;
case 7: return seven;
case 8: return eight;
case 9: return nine;
default: return \nFatal Error!\n; // should I abort pgm?
} // end of switch
} // end of convertDigitToEnglis
Frankly, this seems silly to me. First, it ignores zero, which is a cardinal
sin. I'm being somewhat serious about this: functions should do what they
say,
and that function isn't.
But the code could also be written much more tightly as:
static String digitToString(int n) {
String[] words = {zero, one, two, three, four,
five, six, seven, eight, nine};
if (0 = n n 10) {
return words[n];
}
throw new IllegalArgumentException(input not a single digit);
}
I don't mean to make this post so Java-centric; it just seems a little unfair
to compare a bad Java routine to a good Python routine. :) Writing an
equivalent in Python is also pretty straightforward:
#
## Pseudocode: just a sketch
def digitToString(n):
words = [zero, one, ...]
if 0 = n 10:
return words[n]
...
#
Like dictionaries, the list data structure works pretty well for key/value
lookup if the input key is a small number.
Good luck!
up to a thousand (not tested)
words = {0:'zero', 1:'one', 2:'two', 3:'three', ... , 10:'ten',
11:'eleven', 12:'twelve', ..., 19:'nineteen',
20:'twenty', , 90:'ninety', 100:'one hundred' }
def digitToString(n) :
try :
retStr = words[n]
except KeyError :
if n 100 :
retStr = (digitToString(n // 100)
+ ' hundred and '
+ digitToString(n % 100))
else :
retStr = (digitToString(n - (n % 10))
+ ' '
+ digitToString(n % 10))
return retStr
HTH
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Re: [Tutor] Projects (fwd)
up to a thousand (not tested)
words = {0:'zero', 1:'one', 2:'two', 3:'three', ... , 10:'ten',
11:'eleven', 12:'twelve', ..., 19:'nineteen',
20:'twenty', , 90:'ninety', 100:'one hundred' }
def digitToString(n) :
try :
retStr = words[n]
except KeyError :
if n 100 :
retStr = (digitToString(n // 100)
+ ' hundred and '
+ digitToString(n % 100))
else :
retStr = (digitToString(n - (n % 10))
+ ' '
+ digitToString(n % 10))
return retStr
This could be written much more efficiently. It can be done with only these
lists~
ones =
['zero','one','two','three','four','five','six','seven','eight','nine']
teens =
['ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen']
tens =
['','','twenty','thirty','fourty','fifty','sixty','seventy','eighty','ninety']
hundstr = 'hundred'
thousand = 'thousand'
Exercise for reader to make it work ;-) Unless of course you genuinely want
to see my code...
Current range 0 - 999,999Easily extendable to millions, trillions,
billions, etc.
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Re: [Tutor] Projects (fwd)
Tiger12506 wrote:
up to a thousand (not tested)
words = {0:'zero', 1:'one', 2:'two', 3:'three', ... , 10:'ten',
11:'eleven', 12:'twelve', ..., 19:'nineteen',
20:'twenty', , 90:'ninety', 100:'one hundred' }
def digitToString(n) :
try :
retStr = words[n]
except KeyError :
if n 100 :
retStr = (digitToString(n // 100)
+ ' hundred and '
+ digitToString(n % 100))
else :
retStr = (digitToString(n - (n % 10))
+ ' '
+ digitToString(n % 10))
return retStr
This could be written much more efficiently. It can be done with only these
lists~
ones =
['zero','one','two','three','four','five','six','seven','eight','nine']
teens =
['ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen']
tens =
['','','twenty','thirty','fourty','fifty','sixty','seventy','eighty','ninety']
hundstr = 'hundred'
thousand = 'thousand'
Isn't dictionary access faster than list access? Why are three lists
'much more efficient'?
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Re: [Tutor] Projects (fwd)
This could be written much more efficiently. It can be done with only
these
lists~
ones =
['zero','one','two','three','four','five','six','seven','eight','nine']
teens =
['ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen']
What is it with me and mistakes lately? These two lists above should be
combined into one. Saves a couple of range checks in the program.
:-{
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Re: [Tutor] Projects (fwd)
Isn't dictionary access faster than list access? Why are three lists 'much more efficient'? Oh no, no, no. Dictionaries are faster when you are *searching through* for a particular value. If you already know the index of the item in the list, lists are much faster. Dictionaries are hash based. Somewhere it has to calculate the hash of the key you give it... These three lists are more efficient in terms of size/output ratio. It appeared as if the dictionary that was presented as an example was just going to map one to one all of the values from zero to 999,999 (to match my list version capabilities). Not only is that bad programming style, it's just plain unecessary. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Projects (fwd)
On 24/01/2008, Ricardo Aráoz [EMAIL PROTECTED] wrote:
Isn't dictionary access faster than list access? Why are three lists
'much more efficient'?
Well, not necessarily.
If you want a dictionary, you could use a list of tuples:
myDict = [('a', 'one'), ('b', 'two), ('c', 'three')]
Then you could do lookup as:
def lookup(key, listDict):
for k, v in listDict:
if k == key:
return v
But, as you say, this would be a lot slower than using a dict (lookup
time is proportional to the length of the list for this naive
dictionary, whereas lookup time is essentially constant with a dict).
However, in this case, the keys are integers starting from 0 with no
holes. Instead of having to search, our lookup is just:
return digits[i]
which is also constant time.
If you really want to know which is faster, the timeit module is your
friend. Here's what my computer says:
Morpork:~ repton$ python -m timeit -s 'd = {0:zero, 1:one,
2:two, 3:three, 4:four, 5:five, 6:six, 7:seven, 8:eight,
9:nine}' 'd[5]'
1000 loops, best of 3: 0.127 usec per loop
Morpork:~ repton$ python -m timeit -s 'd = [zero, one, two,
three, four, five, six, seven, eight, nine]' 'd[5]'
1000 loops, best of 3: 0.102 usec per loop
So, if that extra 25 nanoseconds is important to you, definitely go
with the list!
--
John.
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Re: [Tutor] Projects (fwd)
Danny Yoo wrote: [snip] First, it ignores zero, which is a cardinal sin. Or is it an ordinal sin? [snip] ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
