Re: 3 number and dot..
On Fri, 02 Nov 2007 16:39:12 +, Roberto Bonvallet wrote:
On 31 oct, 22:21, Paul Rubin http://[EMAIL PROTECTED] wrote:
def convert(n):
assert type(n) in (int,long)
I'd replace this line with n = int(n), more in the spirit of duck
typing.
Not necessarily. Something duck typing is too liberal in what it accepts.
You might want convert(math.pi) to raise an exception, although I'd
suggestion an assert is the wrong test. A better test would be an
explicit type check with raise:
if not isinstance(n, (int, long)):
raise TypeError('n not an integer')
assert is for this can never happen tests rather than type checking.
--
Steven.
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Re: 3 number and dot..
On 31 oct, 22:21, Paul Rubin http://[EMAIL PROTECTED] wrote: def convert(n): assert type(n) in (int,long) I'd replace this line with n = int(n), more in the spirit of duck typing. -- Roberto Bonvallet -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On 2 nov, 14:54, Steven D'Aprano [EMAIL PROTECTED] cybersource.com.au wrote: Not necessarily. Something duck typing is too liberal in what it accepts. You might want convert(math.pi) to raise an exception What I meant was that the function should not reject unnecessarily non- numeric things that can be converted to a number, like a string. However, you're right: numeric things that are not integers should not be treated silently as if they were. although I'd suggestion an assert is the wrong test. A better test would be an explicit type check with raise. I completely agree. Cheers, -- Roberto Bonvallet -- http://mail.python.org/mailman/listinfo/python-list
3 number and dot..
Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
Abandoned wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? Assuming that the dots are always in the 3rd and 7th position in the string: def conv(s, sep=.): l = [s[0:3], s[3:6], s[6:]] return sep.join(l) print conv(12332321) -- pkm ~ http://paulmcnett.com -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 10:18 pm, Paul McNett [EMAIL PROTECTED] wrote: Abandoned wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? Assuming that the dots are always in the 3rd and 7th position in the string: def conv(s, sep=.): l = [s[0:3], s[3:6], s[6:]] return sep.join(l) print conv(12332321) -- pkm ~http://paulmcnett.com But it's starts from the end.. print conv(12332321) 123.323.21 This is wrong it would be 12.332.321 -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 2:58 pm, Abandoned [EMAIL PROTECTED] wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? x = (12332321,) while (x[0]0): x=divmod(x[0],1000)+x[1:] ... x (0, 12, 332, 321) ..join(map(str,x[1:])) '12.332.321' -- Paul -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 10:38 pm, Paul McGuire [EMAIL PROTECTED] wrote: On Oct 31, 2:58 pm, Abandoned [EMAIL PROTECTED] wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? x = (12332321,) while (x[0]0): x=divmod(x[0],1000)+x[1:] ... x (0, 12, 332, 321) ..join(map(str,x[1:])) '12.332.321' -- Paul It's very good thank you paul! -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 2007 3:24 PM, Abandoned [EMAIL PROTECTED] wrote:
On Oct 31, 10:18 pm, Paul McNett [EMAIL PROTECTED] wrote:
Abandoned wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
Assuming that the dots are always in the 3rd and 7th position in the string:
def conv(s, sep=.):
l = [s[0:3], s[3:6], s[6:]]
return sep.join(l)
print conv(12332321)
--
pkm ~http://paulmcnett.com
But it's starts from the end..
print conv(12332321)
123.323.21
This is wrong it would be 12.332.321
If you're doing this for thousands separators, look at the locale module:
import locale
locale.setlocale(locale.LC_ALL, 'US') #the default C locale
doesn't do grouping
'English_United States.1252'
locale.format_string('%d', 1, grouping=True)
'10,000'
locale.format_string('%d', 12332321, grouping=True)
'12,332,321'
I only have the US locale available on this machine, but if you use a
locale that uses . as the thousands separator, this should work for
you.
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Re: 3 number and dot..
On 31 oct, 16:58, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
x = 12332321
'.'.join(''.join(i for n, i in g) for k, g in
groupby(enumerate(reversed(str(x))), lambda (n, i): n//3))[::-1]
'12.332.321'
--
Roberto Bonvallet
--
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Re: 3 number and dot..
On Oct 31, 10:38 pm, Paul McGuire [EMAIL PROTECTED] wrote: On Oct 31, 2:58 pm, Abandoned [EMAIL PROTECTED] wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? x = (12332321,) while (x[0]0): x=divmod(x[0],1000)+x[1:] ... x (0, 12, 332, 321) ..join(map(str,x[1:])) '12.332.321' -- Paul Hmm. When the number as 1023 the result is 1.23 :( How can i fix it ? -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 9:58 pm, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
If you want do define your own function this will work, no matter how
long the number is, or what separator you choose:
def conv(s, sep='.'):
start=len(s)%3
last=start
result=s[0:start]
for i in range(start+1,len(s)):
if (i-start)%3==0:
if last==0:
result+=s[last:i]
else:
result+=sep+(s[last:i])
last=i
if last==len(s) or last==0:
result+=s[last:len(s)]
else:
result+=sep+s[last:len(s)]
return result
print conv('123456789000')
print conv('1')
print conv('123')
print conv('1234')
1.234.567.890.000.000
1
123
1.234
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Re: 3 number and dot..
On Oct 31, 10:50 pm, Roberto Bonvallet [EMAIL PROTECTED] wrote:
On 31 oct, 16:58, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
x = 12332321
'.'.join(''.join(i for n, i in g) for k, g in
groupby(enumerate(reversed(str(x))), lambda (n, i): n//3))[::-1]
'12.332.321'
--
Roberto Bonvallet
Thank you but it give me this error:
NameError: name 'groupby' is not defined
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Re: 3 number and dot..
On Oct 31, 9:58 pm, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
Hi,
If you want to define your own function, no matter what the length of
the number is or what separator you want to choose, this will work:
def conv(s, sep='.'):
start=len(s)%3
last=start
result=s[0:start]
for i in range(start+1,len(s)):
if (i-start)%3==0:
if last==0:
result+=s[last:i]
else:
result+=sep+(s[last:i])
last=i
if last==len(s) or last==0:
result+=s[last:len(s)]
else:
result+=sep+s[last:len(s)]
return result
print conv('123456789000')
print conv('1')
print conv('123')
print conv('1234')
1.234.567.890.000.000
1
123
1.234
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Re: 3 number and dot..
On Oct 31, 10:50 pm, Roberto Bonvallet [EMAIL PROTECTED] wrote:
On 31 oct, 16:58, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
x = 12332321
'.'.join(''.join(i for n, i in g) for k, g in
groupby(enumerate(reversed(str(x))), lambda (n, i): n//3))[::-1]
'12.332.321'
--
Roberto Bonvallet
I'm sorry but it give me error no module named groupby
My python version is 2.51
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Re: 3 number and dot..
Abandoned wrote: On Oct 31, 10:18 pm, Paul McNett [EMAIL PROTECTED] wrote: Abandoned wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? Assuming that the dots are always in the 3rd and 7th position in the string: def conv(s, sep=.): l = [s[0:3], s[3:6], s[6:]] return sep.join(l) print conv(12332321) But it's starts from the end.. print conv(12332321) 123.323.21 This is wrong it would be 12.332.321 Ok, my slicing was off by one. Mea culpa. I leave it as an exercise to fix it. It is really easy, if you understand basic slicing in Python. If you don't understand slicing, you should google 'python slicing tutorial'. But, from the other messages in the thread it has become apparent that you are wanting to group by thousands (that wasn't at all clear in your initial message). Chris Mellon has given you the best response: use the locale module for this. Other people have done all the work for you, all you need to do is learn how to use locale to get what you want. It seems like you are looking for a spoonfed solution rather than looking for guidance on how to solve the problem for yourself. If I'm wrong about that assessment, I apologize in advance. Best regards and good luck with your project! Paul -- pkm ~ http://paulmcnett.com -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
Abandoned [EMAIL PROTECTED] writes:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
I'm surprised that no one has proposed a regex solution, such as:
import re
re.sub(r'\d{1,3}(?=(?:\d{3})+$)', r'\g0.', str(1234567))
'1.234.567'
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Re: 3 number and dot..
Paul McNett [EMAIL PROTECTED] writes: Chris Mellon has given you the best response: use the locale module for this. It may be the best choice as far as reuse is concerned, but it's not without serious drawbacks. For one, the locale model doesn't really allow forcing of the separators. Some locales, such as the C locale, define no thousand separators whatsoever. Since many Unix installations are set up to use the C locale because it is the OS default, this problem must be addressed. Chris addresse it by forcing the locale to US, but that is a step away from the locale model because it consciously overrides the user's locale preferences. By doing that, one forces a particular way of grouping thousands, without the possibility to using a different grouping characters or to group by a different number of digits (some locales group by tens of thousands). That is quite similar to what you get when you implement the desired grouping yourself. Setting the locale makes the code platform-dependent because different platforms have different locale names. For example, Chris's code fails for me with unsupported locale name -- apparently, my system calls the US locale is en_US.utf8. (Even specifying en_US doesn't work. It might be a Python or system problem, but it just doesn't work.). Finally, it makes the code OS-installation-dependent -- even under the same OS, different installs can and do set up different locales. If his number presentation calls for thousand separators, coding them manually is not an unreasonable implementation choice. It seems like you are looking for a spoonfed solution rather than looking for guidance on how to solve the problem for yourself. If I'm wrong about that assessment, I apologize in advance. No argument here. -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Wed, 31 Oct 2007 21:39:05 +, Abandoned wrote:
On Oct 31, 10:50 pm, Roberto Bonvallet [EMAIL PROTECTED] wrote:
On 31 oct, 16:58, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
x = 12332321
'.'.join(''.join(i for n, i in g) for k, g in
groupby(enumerate(reversed(str(x))), lambda (n, i): n//3))[::-1]
'12.332.321'
--
Roberto Bonvallet
I'm sorry but it give me error no module named groupby My python
version is 2.51
from itertools import groupby
But don't re-invent the wheel. Use the locale module like Chris Mellon
suggested.
--
Steven.
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Re: 3 number and dot..
Hrvoje Niksic:
I'm surprised that no one has proposed a regex solution, such as:
I presume many Python programmers aren't much used in using REs.
import re
re.sub(r'\d{1,3}(?=(?:\d{3})+$)', r'\g0.', str(1234567))
'1.234.567'
It works with negative numbers too. It's a very nice solution of a
simple problem that shows what you can do with REs. But I think that
RE has to be expanded splitted (and maybe even commented) with a
VERBOSE, to improve readability, maybe somethign like:
\d {1,3}
(?=# lockahead assertion
(?: \d {3} )+ $ # non-group
)
Bye,
bearophile
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Re: 3 number and dot..
On Oct 31, 7:58 pm, Abandoned [EMAIL PROTECTED] wrote: Hi.. I want to do this: for examle: 12332321 == 12.332.321 How can i do? Short without being too unreadable: def conv(x, sep='.'): x = str(x)[::-1] return sep.join(x[i:i + 3] for i in range(0, len(x), 3))[::-1] Or more simple-mindedly... def conv(x, sep='.'): x, y = str(x), [] while x: y.append(x[-3:]) x = x[:-3] return sep.join(reversed(y)) -- Paul Hankin -- http://mail.python.org/mailman/listinfo/python-list
Re: 3 number and dot..
On Oct 31, 7:32 pm, [EMAIL PROTECTED] wrote:
Hrvoje Niksic:
I'm surprised that no one has proposed a regex solution, such as:
I presume many Python programmers aren't much used in using REs.
import re
re.sub(r'\d{1,3}(?=(?:\d{3})+$)', r'\g0.', str(1234567))
'1.234.567'
It works with negative numbers too. It's a very nice solution of a
simple problem that shows what you can do with REs. But I think that
RE has to be expanded splitted (and maybe even commented) with a
VERBOSE, to improve readability, maybe somethign like:
\d {1,3}
(?=# lockahead assertion
(?: \d {3} )+ $ # non-group
)
Bye,
bearophile
That's 3 times faster on my box and works for negatives too:
def localize(num, sep='.'):
d,m = divmod(abs(num),1000)
return '-'*(num0) + (localize(d)+sep+'%03d'%m if d else str(m))
George
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Re: 3 number and dot..
Abandoned [EMAIL PROTECTED] writes: 12332321 == 12.332.321 Untested: def convert(n): assert type(n) in (int,long) if n 0: return '-%s'% convert(-n) if n 1000: return str(n) return '%s.%03d' % (convert(n//1000), n % 1000) -- http://mail.python.org/mailman/listinfo/python-list
