3 number and dot..

[Abandoned]

Hi..
I want to do this:
for examle:
12332321 ==12.332.321

How can i do?

[Paul McNett]

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

[Abandoned]

On Oct 31, 10:18 pm, Paul McNett <[e-mail]:

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

[Paul McGuire]

On Oct 31, 2:58 pm, Abandoned <[e-mail]:

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

[Abandoned]

On Oct 31, 10:38 pm, Paul McGuire <[e-mail]:

On Oct 31, 2:58 pm, Abandoned <[e-mail]:

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!

[Chris Mellon]

On Oct 31, 2007 3:24 PM, Abandoned <be*****@gmail.comwrote:

On Oct 31, 10:18 pm, Paul McNett <[e-mail]:

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', 10000, 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.

[Roberto Bonvallet]

On 31 oct, 16:58, Abandoned <[e-mail]:

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

[Abandoned]

On Oct 31, 10:38 pm, Paul McGuire <[e-mail]:

On Oct 31, 2:58 pm, Abandoned <[e-mail]:

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 ?

[elf]

On Oct 31, 9:58 pm, Abandoned <[e-mail]:

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('1234567890000000')
print conv('1')
print conv('123')
print conv('1234')

>>>

1.234.567.890.000.000
1
123
1.234

[Abandoned]

On Oct 31, 10:50 pm, Roberto Bonvallet <[e-mail]:

On 31 oct, 16:58, Abandoned <[e-mail]:

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

[elf]

On Oct 31, 9:58 pm, Abandoned <[e-mail]:

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('1234567890000000')
print conv('1')
print conv('123')
print conv('1234')

>>>

1.234.567.890.000.000
1
123
1.234

[Abandoned]

On Oct 31, 10:50 pm, Roberto Bonvallet <[e-mail]:

On 31 oct, 16:58, Abandoned <[e-mail]:

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

[Paul McNett]

Abandoned wrote:

On Oct 31, 10:18 pm, Paul McNett <[e-mail]:

>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

[Hrvoje Niksic]

Abandoned <be*****@gmail.comwrites:

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'\g<0>.', str(1234567))

'1.234.567'

[Steven D'Aprano]

On Wed, 31 Oct 2007 21:39:05 +0000, Abandoned wrote:

On Oct 31, 10:50 pm, Roberto Bonvallet <[e-mail]:

>On 31 oct, 16:58, Abandoned <[e-mail]:

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.

[Hrvoje Niksic]

Paul McNett <[e-mail]:

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.

[bearophileHUGS]

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'\g<0>.', 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

[Paul Hankin]

On Oct 31, 7:58 pm, Abandoned <[e-mail]:

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

[George Sakkis]

On Oct 31, 7:32 pm, [e-mail] 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'\g<0>.', 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 '-'*(num<0) + (localize(d)+sep+'%03d'%m if d else str(m))
George

[Paul Rubin]

Abandoned <be*****@gmail.comwrites:

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)

[Roberto Bonvallet]

On 31 oct, 22:21, Paul Rubin <http://[e-mail]:

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

[Steven D'Aprano]

On Fri, 02 Nov 2007 16:39:12 +0000, Roberto Bonvallet wrote:

On 31 oct, 22:21, Paul Rubin <http://[e-mail]:

>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.

[Roberto Bonvallet]

On 2 nov, 14:54, Steven D'Aprano <st...@REMOVE-THIS-
cybersource.com.auwrote:

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