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
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
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
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
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http://mail.python.org/mailman/listinfo/python-list
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 ~
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:]]
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
--
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)
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
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
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)
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
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
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,
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
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],
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})+$)', r'\g0.', str(1234567))
'1.234.567'
--
Paul McNett [EMAIL PROTECTED] writes:
Chris Mellon has given you the best response: use the locale module
for this.
It may be the best choice as far as reuse is concerned, but it's not
without serious drawbacks.
For one, the locale model doesn't really allow forcing of the
separators. Some
On Wed, 31 Oct 2007 21:39:05 +, Abandoned wrote:
On Oct 31, 10:50 pm, Roberto Bonvallet [EMAIL PROTECTED] wrote:
On 31 oct, 16:58, Abandoned [EMAIL PROTECTED] wrote:
Hi..
I want to do this:
for examle:
12332321 == 12.332.321
How can i do?
x = 12332321
'.'.join(''.join(i
Hrvoje Niksic:
I'm surprised that no one has proposed a regex solution, such as:
I presume many Python programmers aren't much used in using REs.
import re
re.sub(r'\d{1,3}(?=(?:\d{3})+$)', r'\g0.', str(1234567))
'1.234.567'
It works with negative numbers too. It's a very nice solution
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
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
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)
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