Re: [Tutor] range function and floats?
Joel Knoll mindar...@live.com wrote I'm trying to write a simple program to give the sine of each of a range of numbers, including multiples of pi. I keep getting a syntax error, highlighting my use of 2pi as an argument in the range, range requires integers. You need to scale your floats appropriately and convert to integers. Or in your case use the multipliers of pi as the arguments to range() and then multiply by pi when getting the sine. ...this is more about learning how the range function and floats work than about writing a super-efficient program. Unfortunately they don't work together. range(0.1,0.5,0.1) - [0.1,0.2,0.3,0.4] doesn't work you need to do: for n in range(1,5): use( n/10 ) HTH, -- Alan Gauld Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] range function and floats?
Alan Gauld wrote: ...this is more about learning how the range function and floats work than about writing a super-efficient program. Unfortunately they don't work together. range(0.1,0.5,0.1) - [0.1,0.2,0.3,0.4] doesn't work you need to do: for n in range(1,5): use( n/10 ) There are pitfalls in writing a range() equivalent for floats. Here's one way that doesn't work: def new_range(start, end, step): # DON'T USE THIS!!! x = start while x end: yield x x += step Here it seems to work: list(new_range(5, 10, 0.25)) [5, 5.25, 5.5, 5.75, 6.0, 6.25, 6.5, 6.75, 7.0, 7.25, 7.5, 7.75, 8.0, 8.25, 8.5, 8.75, 9.0, 9.25, 9.5, 9.75] (Remember that the end argument is excluded!) But here it fails: L = list(new_range(5, 10, 0.1)) L[-1] == 9.9 # expect the last value to be 9.9 False L[-1] == 10.0 # maybe it's 10 then False In fact, the last value is a totally unexpected 9.9822. Such is the perils of floating point rounding errors. I've written a recipe for a float range which I hope avoids as many of these problems as possible. It isn't possible to avoid *all* rounding error when doing floating point calculation, but this should minimize them: http://code.activestate.com/recipes/577068-floating-point-range/ As a bonus, it allows you to choose whether the start and end points are included or excluded. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] syntax error that i cant spot!
Noah Hall wrote: He has no classes in there. Therefore, there is no place it should be in this code. Please remember this is Python, and not Java nor anything else. [...] It just makes life easier. Oh the irony... talking about making life easier, who are you talking to? What about? Please quote enough of the previous message to establish context -- when you are replying to the message, it is fresh in your mind. When others read your reply (possibly days later like I'm doing now), the context is anything but clear and your message comes across as merely mysterious and obscure. Thank you. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] syntax error that i cant spot!
Noah Hall wrote: Please quote enough of the previous message to establish context -- when you are replying to the message, it is fresh in your mind. When others read your reply (possibly days later like I'm doing now), the context is anything but clear and your message comes across as merely mysterious and obscure. Certainly. In fact, it was an error in that message - I had deleted accidentally that which I had quoted. Please be aware, however, rudeness does nothing for you. Please and Thank you are rude? Oh my, have you lived a sheltered life :) -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] syntax error that i cant spot!
Please and Thank you are rude? Oh my, have you lived a sheltered life :) Nej, it was your condescension that I found rude. Also, you did it again, perhaps on purpose though.. ;) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] range function and floats?
On Wed, Jan 5, 2011 at 5:14 AM, Steven D'Aprano st...@pearwood.info wrote: Alan Gauld wrote: ...this is more about learning how the range function and floats work than about writing a super-efficient program. Unfortunately they don't work together. range(0.1,0.5,0.1) - [0.1,0.2,0.3,0.4] doesn't work you need to do: for n in range(1,5): use( n/10 ) There are pitfalls in writing a range() equivalent for floats. Here's one way that doesn't work: def new_range(start, end, step): # DON'T USE THIS!!! x = start while x end: yield x x += step Here it seems to work: list(new_range(5, 10, 0.25)) [5, 5.25, 5.5, 5.75, 6.0, 6.25, 6.5, 6.75, 7.0, 7.25, 7.5, 7.75, 8.0, 8.25, 8.5, 8.75, 9.0, 9.25, 9.5, 9.75] (Remember that the end argument is excluded!) But here it fails: L = list(new_range(5, 10, 0.1)) L[-1] == 9.9 # expect the last value to be 9.9 False L[-1] == 10.0 # maybe it's 10 then False In fact, the last value is a totally unexpected 9.9822. Such is the perils of floating point rounding errors. I've written a recipe for a float range which I hope avoids as many of these problems as possible. It isn't possible to avoid *all* rounding error when doing floating point calculation, but this should minimize them: http://code.activestate.com/recipes/577068-floating-point-range/ As a bonus, it allows you to choose whether the start and end points are included or excluded. As an alternative to floating point, you can use the Decimal module: import decimal def new_range(start, stop, step): x = decimal.Decimal(str(start)) step = decimal.Decimal(str(step)) while x stop: yield x x += step x = list(new_range(5, 10, 0.1)) x[-1] == decimal.Decimal(str(9.9)) #True float(x[-1]) == 9.9 #True The decimal module allows you to get rid of those pesky floating point errors. See http://docs.python.org/library/decimal.html for more info. On a related note, if you're interested in working with rational numbers (1/23, 3/4, etc.) there is also a fraction module http://docs.python.org/library/fractions.html. HTH, Wayne ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] range function and floats?
Wayne Werner wrote: The decimal module allows you to get rid of those pesky floating point errors. See http://docs.python.org/library/decimal.html for more info. That's a myth. Decimal suffers from the same floating point issues as binary floats. It's quite easy to demonstrate the same sort of rounding errors with Decimal as for float: from decimal import Decimal as D x = D(1)/D(3) 3*x == 1 False Between 1 and 1000 inclusive, there are 354 such numbers: nums = [] for i in range(1, 1001): ... x = D(1)/D(i) ... if x*i != 1: nums.append(i) ... len(nums) 354 The problem isn't just division and multiplication, nor does it just affect fractional numbers: x = D(10)**30 x + 100 == x True Here's another nice little demonstration of the problem: def f(a): ... b = 9*a - 9 ... c = a - 1/D(9)*b ... return c # c should always equal 1 exactly ... f(D(51.1)) == 1 False Decimal and float share more things in common than differences. Both are floating point numbers. Both have a fixed precision (although you can configure what that precision is for Decimal, but not float). Both use a finite number of bits, and therefore have a finite resolution. The only differences are: * as mentioned, you can configure Decimal to use more bits and higher precision, at the cost of speed and memory; * floats use base 2 and Decimal uses base 10; * floats are done in hardware and so are much faster than Decimal; So it simply isn't correct to suggest that Decimal doesn't suffer from rounding error. -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cx_Oracle help
This guide helped me a lot with cx_Oracle
http://codingtutorials.co.uk/blog/?p=31
Greg Lindstrom-3 wrote:
Hello,
I'm trying to help out a friend and am stumped. Can you help me out?
Thanks,
--greg
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
I will briefly explain the problem I am facing.
I am using Oracle 9.2, Python 2.5 and I installed cx_Oracle-
4.3.1-win32-9i-py25 in Python.
From python I tried following :
import cx_Oracle
myDsn = cx_Oracle.makedsn('ISCN47',1521,'AUBDBS01')
CONN = cx_Oracle.connect(myusr, mypwd, myDsn)
Traceback (most recent call last):
File pyshell#4, line 1, in module
conn = cx_Oracle.Connection('scott','tiger',myDsn)
RuntimeError: Unable to acquire Oracle environment
handle
I have set the below environment variables too
NLS_LANG: snip.WE8MSWIN1252
ORACLE_HOME:D:\Tools\oracle\ora92
ORACLE_SID: AUBDBS01
PYTHON_HOME:d:\Utility\Python25
PYTHONPATH:
%PYTHON_HOME%\lib;%PYTHON_HOME%\DLLs;%PYTHON_HOME%\Lib\site-packages;%ORACLE_HOME%\bin
LD_LIBRARY_PATH: %LD_LIBRARY_PATH%;D:\Tools\oracle\ora92\lib
Not getting any idea where I am wrong?
Regards,
Kishore
___
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
--
View this message in context:
http://old.nabble.com/cx_Oracle-help-tp15293236p30596546.html
Sent from the Python - tutor mailing list archive at Nabble.com.
___
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] range function and floats?
On Wed, Jan 5, 2011 at 10:43 AM, Steven D'Aprano st...@pearwood.infowrote:
Wayne Werner wrote:
The decimal module allows you to get rid of those pesky floating point
errors. See http://docs.python.org/library/decimal.html for more info.
That's a myth. Decimal suffers from the same floating point issues as
binary floats. It's quite easy to demonstrate the same sort of rounding
errors with Decimal as for float:
from decimal import Decimal as D
x = D(1)/D(3)
3*x == 1
False
I should have clarified - when I say pesky floating point errors I mean
errors in precision that you naturally would not expect.
1/3 == .333 (repeating forever(?)). But on a computer, you're limited to a
specific precision point (sort of), and the moment you truncate
.333(repeating) to *any* finite points of precision you no longer have the
result of the mathematical operation 1/3. Yes, 1/3*3 == 1, but the error in
the Decimal module is *only* in division. It might be useful to define a
repeating flag in the Decimal module for the results of such operations as
1/3, which would get rid of the error in truncation. But this is a
fundamentally different error from the standard floating point errors.
In [106]: 0.9
Out[106]: 0.90002
These are two different numbers, mathematically. If I say x = 0.9, I
naturally assume that 0.9 is the value in x, not 0.90002. Of
course it's all in the implementation of floating points, and the fact that
Python evaluates 0.9 in a floating point context which results in the stored
value, but that ignores the fact that *naturally* one does not expect this.
And anyone who's been through 2nd or 3rd grade or whenever they teach about
equality would expect that this would evaluate to False.
In [112]: 0.90002 == 0.9
Out[112]: True
You don't get such silliness with the Decimal module:
In [125]: D('0.90002') == D('0.9')
Out[125]: False
Between 1 and 1000 inclusive, there are 354 such numbers:
nums = []
for i in range(1, 1001):
... x = D(1)/D(i)
... if x*i != 1: nums.append(i)
...
len(nums)
354
The problem isn't just division and multiplication, nor does it just affect
fractional numbers:
x = D(10)**30
x + 100 == x
True
Decimal DOES get rid of floating point errors, except in the case of
repeating (or prohibitively large precision values)
In [127]: x = D(10)**30
In [128]: x
Out[128]: Decimal('1.000E+30')
In [129]: x + 100
Out[129]: Decimal('1.000E+30')
If you reset the precision to an incredibly large number:
decimal.getcontext().prec = 1000
In [131]: x = D(10)**30
In [132]: x
Out[132]: Decimal('100')
In [133]: x + 100 == x
Out[133]: False
Voila, the error has vanished!
So it simply isn't correct to suggest that Decimal doesn't suffer from
rounding error.
I never said rounding errors - I said pesky floating point errors. When
performing the operation 1/3, I naturally expect that my computer won't hold
each of the 3's after the decimal point, and I don't categorize that as
pesky - that's just good sense if you know a little about computers. I also
expect that .333 * 3 would give me the number .999, and only .999, not
.99911 or some other wonky value. Of course it's interesting to note
that Python handles the precision properly when dealing with strings, but
not with the floating points themselves (at least on this particular trial):
In [141]: .333 * 3
Out[141]: 0.99911
In [142]: str(.333*3)
Out[142]: '0.999'
In [143]: .333 * 3 == .999
Out[143]: False
In [144]: str(.333*3) == str(.999)
Out[144]: True
Decimal and float share more things in common than differences. Both are
floating point numbers. Both have a fixed precision (although you can
configure what that precision is for Decimal, but not float). Both use a
finite number of bits, and therefore have a finite resolution. The only
differences are:
* as mentioned, you can configure Decimal to use more bits and higher
precision, at the cost of speed and memory;
* floats use base 2 and Decimal uses base 10;
* floats are done in hardware and so are much faster than Decimal;
Precisely. It's not a magic bullet (1/3 != . mathematically, after
all!), *but* it eliminates the errors that you wouldn't normally expect when
working with standard points of precision, such as the expectation that
0.333 * 3 resulting in .999, not .999011.
Hopefully a little more precise,
Wayne
___
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
[Tutor] user input help
Hi all, I'm pretty new to programming in general and figured I'd try out python. I'm working on a small program to add users to a sqlite db. The problem I'm having it dealing with the user input, I'd like to be able to repeat the function to get the input if the user doesn't accept it. here's the code I have now: def promptInput(): Get employee data from user running this program lname = raw_input(Please enter employees last name\n) fname = raw_input(Please enter employees first name\n) email = raw_input(Please enter employee email address (or press enter to \ leave blank)\n) result = (lname, fname, email) return result def getEmplyInfo(): # get the data from input result = promptInput() # print the data so the user can check and verify spelling print Is the following info correct [y/n]\n%s, %s %s % (result[1], \ result[0], result[2]) check = raw_input() #see if the user needs to make corrections to the data he entered if check == y: print this check is done so we can add user print %s, %s %s % (result[1], result[0], result[2]) else: check = promptInput() The if else loop is were I'm loosing it. If I select n it will ask for the input again but only once. If on the second time around I enter n to re-do it just exits. Thanks Jason ..·º ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] user input help
On 1/5/11, Jason Staudenmayer jas...@adventureaquarium.com wrote: Hi all, I'm pretty new to programming in general and figured I'd try out python. I'm working on a small program to add users to a sqlite db. The problem I'm having it dealing with the user input, I'd like to be able to repeat the function to get the input if the user doesn't accept it. here's the code I have now: def promptInput(): Get employee data from user running this program lname = raw_input(Please enter employees last name\n) fname = raw_input(Please enter employees first name\n) email = raw_input(Please enter employee email address (or press enter to \ leave blank)\n) result = (lname, fname, email) return result def getEmplyInfo(): # get the data from input result = promptInput() # print the data so the user can check and verify spelling print Is the following info correct [y/n]\n%s, %s %s % (result[1], \ result[0], result[2]) check = raw_input() #see if the user needs to make corrections to the data he entered if check == y: print this check is done so we can add user print %s, %s %s % (result[1], result[0], result[2]) else: check = promptInput() The if else loop is were I'm loosing it. If I select n it will ask for the input again but only once. If on the second time around I enter n to re-do it just exits. This is because the function is done once it detects the y or n, so after you enter the n, one of those if/else statements has fired, and the function has nothing else to do. You will want a while loop, something like: repeat=True while repeat: answer=raw_input(Is the data okay?) if answer==y: repeat=False else: promptInput() repeat=True Anyway, something along those lines. Look in the manual for while loops. Basically, they are a way to repeat an action until a condition is met. You will also run across for loops, which are mostly used for repeating an event a set number of times. You can use them interchangeably, but they each have situations where one works better than the other, and you want a while loop here. Thanks Jason ..·º ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor -- Have a great day, Alex (msg sent from GMail website) mehg...@gmail.com; http://www.facebook.com/mehgcap ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] wx accelerator table: one keystroke seems skipped
Hello all,
First, this is about a wx accelerator table, so if it is too off-topic
for this list, let me know.
I have a table with 23 entries, all of which work. I added another
entry last night, and it does not work. The odd thing, though, is that
I do not get an error of any kind anywhere in the program, the
keystroke simply does not call the function to which it should be
bound. I even changed the function name to something that should have
thrown an exception, but nothing at all happens. The keystroke in
question is ctrl+m and is tied to ctrlM_id and the cancelMove()
function. Below I have pasted my entire setHotkeys() function. Again,
everything works except ctrl+m, but even that keystroke does not cause
problems, it is like it is not even there.
def setHotkeys():
upArrow_id=wx.NewId()
downArrow_id=wx.NewId()
leftArrow_id=wx.NewId()
rightArrow_id=wx.NewId()
space_id=wx.NewId()
a_id=wx.NewId()
f_id=wx.NewId()
m_id=wx.NewId()
r_id=wx.NewId()
ctrlM_id=wx.NewId()
ctrlN_id=wx.NewId()
one_id=wx.NewId()
two_id=wx.NewId()
three_id=wx.NewId()
four_id=wx.NewId()
five_id=wx.NewId()
six_id=wx.NewId()
seven_id=wx.NewId()
f1_id=wx.NewId()
f2_id=wx.NewId()
f3_id=wx.NewId()
f4_id=wx.NewId()
hotkeyList=[
(wx.ACCEL_NORMAL, wx.WXK_UP, upArrow_id), #up one face-up card in
the current stack
(wx.ACCEL_NORMAL, wx.WXK_DOWN, downArrow_id), #down one face-up card
(wx.ACCEL_NORMAL, wx.WXK_LEFT, leftArrow_id), #left a stack
(wx.ACCEL_NORMAL, wx.WXK_RIGHT, rightArrow_id), #right a stack
(wx.ACCEL_NORMAL, ord('a'), a_id), #try to move last card to Ace stack
(wx.ACCEL_NORMAL, ord('1'), one_id), #jump to stack 1
(wx.ACCEL_NORMAL, ord('2'), two_id), #jump to stack 2
(wx.ACCEL_NORMAL, ord('3'), three_id), #jump to stack 3
(wx.ACCEL_NORMAL, ord('4'), four_id), #jump to stack 4
(wx.ACCEL_NORMAL, ord('5'), five_id), #jump to stack 5
(wx.ACCEL_NORMAL, ord('6'), six_id), #jump to stack 6
(wx.ACCEL_NORMAL, ord('7'), seven_id), #jump to stack 7
(wx.ACCEL_NORMAL, ord('f'), f_id), #flip up a new card
(wx.ACCEL_NORMAL, ord('r'), r_id), #read the current deck card in play
(wx.ACCEL_CTRL, ord('m'), ctrlM_id), #cancel Move mode
(wx.ACCEL_CTRL, ord('n'), ctrlN_id), #start a new hand
(wx.ACCEL_NORMAL, ord('m'), m_id), #mark stack to be moved
(wx.ACCEL_NORMAL, wx.WXK_SPACE, space_id), #move deck's facing card
to activeStack
#now f1-f4, for reviewing the four Ace stacks
(wx.ACCEL_NORMAL, wx.WXK_F1, f1_id),
(wx.ACCEL_NORMAL, wx.WXK_F2, f2_id),
(wx.ACCEL_NORMAL, wx.WXK_F3, f3_id),
(wx.ACCEL_NORMAL, wx.WXK_F4, f4_id)]
#now bind the keys to their functions, using lambda to pass args
f.Bind(wx.EVT_MENU, lambda evt, dir=1: exploreStack(evt, dir), id=upArrow_id)
f.Bind(wx.EVT_MENU, lambda evt, dir=-1: exploreStack(evt, dir),
id=downArrow_id)
f.Bind(wx.EVT_MENU, lambda evt, dir=-1: move(evt, dir), id=leftArrow_id)
f.Bind(wx.EVT_MENU, lambda evt, dir=1: move(evt, dir), id=rightArrow_id)
f.Bind(wx.EVT_MENU, lambda evt: useDeckCard(evt), id=space_id)
f.Bind(wx.EVT_MENU, lambda evt: toAcePile(evt), id=a_id)
f.Bind(wx.EVT_MENU, lambda evt: deal(evt), id=f_id)
f.Bind(wx.EVT_MENU, lambda evt: readDeckCard(evt), id=r_id)
f.Bind(wx.EVT_MENU, lambda evt, target=1, jump=True: move(evt,
target, jump), id=one_id)
f.Bind(wx.EVT_MENU, lambda evt, target=2, jump=True: move(evt,
target, jump), id=two_id)
f.Bind(wx.EVT_MENU, lambda evt, target=3, jump=True: move(evt,
target, jump), id=three_id)
f.Bind(wx.EVT_MENU, lambda evt, target=4, jump=True: move(evt,
target, jump), id=four_id)
f.Bind(wx.EVT_MENU, lambda evt, target=5, jump=True: move(evt,
target, jump), id=five_id)
f.Bind(wx.EVT_MENU, lambda evt, target=6, jump=True: move(evt,
target, jump), id=six_id)
f.Bind(wx.EVT_MENU, lambda evt, target=7, jump=True: move(evt,
target, jump), id=seven_id)
f.Bind(wx.EVT_MENU, lambda evt, sel=activeStack: selectStack(evt,
sel), id=m_id)
f.Bind(wx.EVT_MENU, lambda evt: cancelMove(evt), id=ctrlM_id)
f.Bind(wx.EVT_MENU, lambda evt: promptForNewGame(evt), id=ctrlN_id)
f.Bind(wx.EVT_MENU, lambda evt, target=0: reviewAceStack(evt,
target), id=f1_id)
f.Bind(wx.EVT_MENU, lambda evt, target=1: reviewAceStack(evt,
target), id=f2_id)
f.Bind(wx.EVT_MENU, lambda evt, target=2: reviewAceStack(evt,
target), id=f3_id)
f.Bind(wx.EVT_MENU, lambda evt, target=3: reviewAceStack(evt,
target), id=f4_id)
hotkeys=wx.AcceleratorTable(hotkeyList)
f.SetAcceleratorTable(hotkeys)
--
Have a great day,
Alex (msg sent from GMail website)
mehg...@gmail.com; http://www.facebook.com/mehgcap
___
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] range function and floats?
On 01/-10/-28163 02:59 PM, Wayne Werner wrote:
On Wed, Jan 5, 2011 at 10:43 AM, Steven D'Apranost...@pearwood.infowrote:
Wayne Werner wrote:
The decimal module allows you to get rid of those pesky floating point
errors. See http://docs.python.org/library/decimal.html for more info.
That's a myth. Decimal suffers from the same floating point issues as
binary floats. It's quite easy to demonstrate the same sort of rounding
errors with Decimal as for float:
from decimal import Decimal as D
x = D(1)/D(3)
3*x == 1
False
I should have clarified - when I say pesky floating point errors I mean
errors in precision that you naturally would not expect.
I'd expect them. Apparently you would not.
1/3 == .333 (repeating forever(?)). But on a computer, you're limited to a
specific precision point (sort of), and the moment you truncate
.333(repeating) to *any* finite points of precision you no longer have the
result of the mathematical operation 1/3. Yes, 1/3*3 == 1, but the error in
the Decimal module is *only* in division. It might be useful to define a
repeating flag in the Decimal module for the results of such operations as
1/3, which would get rid of the error in truncation.
Interesting flag. So how would it indicate that the decimal version of
the number 762/477 repeats every 476 digits?
But this is a
fundamentally different error from the standard floating point errors.
In [106]: 0.9
Out[106]: 0.90002
These errors are in the conversion between string and floating point,
and back again. The number 0.9 has a repeating bit pattern in binary
floating point, so it cannot be represented exactly. That's not an
error, it's a direct consequence of the decision to use binary floating
point. And that decision was made decades ago, and is independent of
Python. When I learned Fortran in the 60's, one of the things the book
(McCracken) emphasized was not to compare real numbers with equality,
but to use the form abs(x-y)delta.
These are two different numbers, mathematically. If I say x = 0.9, I
naturally assume that 0.9 is the value in x, not 0.90002. Of
course it's all in the implementation of floating points, and the fact that
Python evaluates 0.9 in a floating point context which results in the stored
value, but that ignores the fact that *naturally* one does not expect this.
And anyone who's been through 2nd or 3rd grade or whenever they teach about
equality would expect that this would evaluate to False.
Lots of stuff they teach in 3rd grade is oversimplified for the real world.
In [112]: 0.90002 == 0.9
Out[112]: True
You don't get such silliness with the Decimal module:
In [125]: D('0.90002') == D('0.9')
Out[125]: False
The conversion between string and float is logically a divide, where the
denominator is a power of 10. So decimal floating point doesn't have
any quantization errors for that conversion.
Even if you use the decimal package, you need to understand these
things, or you'll make the kind of error made a few days ago, where
binary floating point values were passed into a decimal constructor.
That won't fix problems that were already there. It just may mask
them for certain cases.
Between 1 and 1000 inclusive, there are 354 such numbers:
nums = []
for i in range(1, 1001):
... x = D(1)/D(i)
... if x*i != 1: nums.append(i)
...
len(nums)
354
The problem isn't just division and multiplication, nor does it just affect
fractional numbers:
x = D(10)**30
x + 100 == x
True
Decimal DOES get rid of floating point errors, except in the case of
repeating (or prohibitively large precision values)
In [127]: x = D(10)**30
In [128]: x
Out[128]: Decimal('1.000E+30')
In [129]: x + 100
Out[129]: Decimal('1.000E+30')
If you reset the precision to an incredibly large number:
decimal.getcontext().prec = 1000
In [131]: x = D(10)**30
In [132]: x
Out[132]: Decimal('100')
In [133]: x + 100 == x
Out[133]: False
Voila, the error has vanished!
So it simply isn't correct to suggest that Decimal doesn't suffer from
rounding error.
I never said rounding errors - I said pesky floating point errors. When
performing the operation 1/3, I naturally expect that my computer won't hold
each of the 3's after the decimal point, and I don't categorize that as
pesky - that's just good sense if you know a little about computers. I also
expect that .333 * 3 would give me the number .999, and only .999, not
.99911 or some other wonky value. Of course it's interesting to note
that Python handles the precision properly when dealing with strings, but
not with the floating points themselves (at least on this particular trial):
In [141]: .333 * 3
Out[141]: 0.99911
In [142]: str(.333*3)
Out[142]: '0.999'
In [143]: .333 * 3 == .999
Out[143]: False
In [144]: str(.333*3) == str(.999)
Out[144]: True
Decimal and
Re: [Tutor] range function and floats?
Wayne Werner wrote:
On Wed, Jan 5, 2011 at 10:43 AM, Steven D'Aprano st...@pearwood.infowrote:
Wayne Werner wrote:
The decimal module allows you to get rid of those pesky floating point
errors. See http://docs.python.org/library/decimal.html for more info.
That's a myth. Decimal suffers from the same floating point issues as
binary floats. It's quite easy to demonstrate the same sort of rounding
errors with Decimal as for float:
from decimal import Decimal as D
x = D(1)/D(3)
3*x == 1
False
I should have clarified - when I say pesky floating point errors I mean
errors in precision that you naturally would not expect.
Say what?
But that's the whole point -- you *must* expect them, because Decimal
floating point numbers suffer the *exact* same issues with rounding and
finite precision as binary floats. It may affect different specific
numbers, but the issues are exactly the same. Just because Decimal uses
base 10 doesn't make it immune to the same issues of rounding and
precision as base 2 floats.
1/3 == .333 (repeating forever(?)). But on a computer, you're limited to a
specific precision point (sort of), and the moment you truncate
.333(repeating) to *any* finite points of precision you no longer have the
result of the mathematical operation 1/3. Yes, 1/3*3 == 1, but the error in
the Decimal module is *only* in division. It might be useful to define a
I'm surprised you can make such an obviously wrong claim -- you even
discuss errors in addition further down. How can you do that and still
say that errors only occur with division???
Decimal includes a significant amount of code to detect when the result
of some operation is inexact, and to signal the user when it occurs.
This does *not* only happen in division:
import decimal
decimal.getcontext().traps[decimal.Inexact] = 1
a = decimal.Decimal('1e-30')
b = decimal.Decimal('11.1')
a+b
Traceback (most recent call last):
File stdin, line 1, in module
File /usr/local/lib/python3.1/decimal.py, line 1178, in __add__
ans = ans._fix(context)
File /usr/local/lib/python3.1/decimal.py, line 1653, in _fix
context._raise_error(Inexact)
File /usr/local/lib/python3.1/decimal.py, line 3812, in _raise_error
raise error(explanation)
decimal.Inexact: None
repeating flag in the Decimal module for the results of such operations as
1/3, which would get rid of the error in truncation. But this is a
You haven't thought that through very well. How do you deal with square
roots and other fractional powers without truncation errors?
What happens when you operate on two numbers with this repeating flag --
is the Decimal module supposed to include a full-blown algebra system to
generate exact answers to such things as this?
x = 1/Decimal(3) # 0.333 repeating
y = 1/Decimal(27) # 0.037037 repeating
x - y/10**9 # 0.3296296296 repeating
How do you deal with numbers like pi or e without truncation?
No, a repeating flag would just add unnecessary complexity for no real
benefit. If you want to track fractions exactly, use the fractions module.
fundamentally different error from the standard floating point errors.
But they aren't fundamentally different! That's the point! They are the
same errors. They effect different numbers, because float is base 2 and
Decimal is base 10, but they have the same fundamental cause.
In [106]: 0.9
Out[106]: 0.90002
These are two different numbers, mathematically. If I say x = 0.9, I
naturally assume that 0.9 is the value in x, not 0.90002. Of
Yes, and the exact same issue can occur in Decimal. It just affects
different numbers. 0.9 is exact in base 10, but can't be represented
exactly in base 2. Fortunately, the opposite is not the case: any finite
number of binary bits is equivalent to a finite number of decimal
digits, so unless you overflow the number of digits, you won't get an
equivalent error for binary-decimal conversion. But there are numbers
which can't be represented exactly in base 10 at any finite precision.
1/3 is one of them, so is 5/17 and an infinite number of other examples.
You can even find examples of numbers which are exact in base 2 but
inexact in Decimal at whatever finite precision you choose.
2**120
1329227995784915872903807060280344576
int( 1/( 1/Decimal(2**120) ) )
13292279957849158729038070600
Choose more digits for Decimal, and 2**120 can be stored exactly, but
other, larger, numbers can't.
course it's all in the implementation of floating points, and the fact that
Python evaluates 0.9 in a floating point context which results in the stored
value, but that ignores the fact that *naturally* one does not expect this.
And anyone who's been through 2nd or 3rd grade or whenever they teach about
equality would expect that this would evaluate to False.
Floating point numbers are not the same as the real numbers you learn
about in school. It doesn't matter whether you use binary
Re: [Tutor] user input help
Jason Staudenmayer jas...@adventureaquarium.com wrote I'm working on a small program to add users to a sqlite db. Try the database topic in my tutorial where I do a very similar thing... code follows - def promptInput(): ... result = (lname, fname, email) return result def getEmplyInfo(): # get the data from input result = promptInput() # print the data so the user can check and verify spelling print Is the following info correct [y/n]\n%s, %s %s % (result[1], \ result[0], result[2]) check = raw_input() #see if the user needs to make corrections to the data he entered if check == y: print this check is done so we can add user print %s, %s %s % (result[1], result[0], result[2]) else: check = promptInput() --- The if else loop is were I'm loosing it. if else is not a loop it is a branch. See the What is Profgramming topic in my tutorial for a description of the different structures. Then see the loops and branching topics for loop and branch structures. You need a loop. The topics give examples. HTH, -- Alan Gauld Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] wx accelerator table: one keystroke seems skipped
From: Alex Hall mehg...@gmail.com Hello all, First, this is about a wx accelerator table, so if it is too off-topic for this list, let me know. I have a table with 23 entries, all of which work. I added another entry last night, and it does not work. The odd thing, though, is that I do not get an error of any kind anywhere in the program, the keystroke simply does not call the function to which it should be bound. I even changed the function name to something that should have thrown an exception, but nothing at all happens. The keystroke in question is ctrl+m and is tied to ctrlM_id and the cancelMove() Control+M is equivalent with a enter key so this might be the problem. Octavian ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cx_Oracle help
On 05/01/2011 16:33, F1r3f1y wrote:
This guide helped me a lot with cx_Oracle
http://codingtutorials.co.uk/blog/?p=31
Greg Lindstrom-3 wrote:
Hello,
I'm trying to help out a friend and am stumped. Can you help me out?
Thanks,
--greg
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
I will briefly explain the problem I am facing.
I am using Oracle 9.2, Python 2.5 and I installed cx_Oracle-
4.3.1-win32-9i-py25 in Python.
From python I tried following :
import cx_Oracle
myDsn = cx_Oracle.makedsn('ISCN47',1521,'AUBDBS01')
CONN = cx_Oracle.connect(myusr, mypwd, myDsn)
Traceback (most recent call last):
File pyshell#4, line 1, inmodule
conn = cx_Oracle.Connection('scott','tiger',myDsn)
RuntimeError: Unable to acquire Oracle environment
handle
I have set the below environment variables too
NLS_LANG:snip.WE8MSWIN1252
ORACLE_HOME:D:\Tools\oracle\ora92
ORACLE_SID: AUBDBS01
PYTHON_HOME:d:\Utility\Python25
PYTHONPATH:
%PYTHON_HOME%\lib;%PYTHON_HOME%\DLLs;%PYTHON_HOME%\Lib\site-packages;%ORACLE_HOME%\bin
LD_LIBRARY_PATH: %LD_LIBRARY_PATH%;D:\Tools\oracle\ora92\lib
Not getting any idea where I am wrong?
Regards,
Kishore
___
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
You need to install the Runtime, 3rd product option in the Oracle
Universal Installer.
--
Kind Regards,
Christian Witts
___
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
