Quick help needed: how to format an integer ?

[[EMAIL PROTECTED]]

Hi !

I need to convert some integer values.

"1622" ->"1 622"

or

"10001234" -> ""10.001.234""

So I need thousand separators.

Can anyone helps me with a simply solution (like %xxx) ?

Thanx for it: dd

Ps:
Now I use this proc:

def toths(i):
s=str(i)
l=[]
ls=len(s)
for i in range(ls):
c=s[ls-i-1]
if i%3==0 and i<>0:
   c=c+"."
l.append(c)
l.reverse()
return "".join(l)



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Re: Quick help needed: how to format an integer ?

[Satchidanand Haridas]

One possible solution. Don't know how efficient it is though. :-)

 >>> def put_decimal(s):
... return ''.join( [ [s[i], '.%s' % s[i]][(len(s)-i)%3 == 0] for i 
in range(0, len(s))])
...
 >>> put_decimal("10001234")
'10.001.234'
 >>> put_decimal("12622")
'12.622'

thanks,
Satchit


[EMAIL PROTECTED] wrote:

>Hi !
>
>I need to convert some integer values.
>
>"1622" ->"1 622"
>
>or
>
>"10001234" -> ""10.001.234""
>
>So I need thousand separators.
>
>Can anyone helps me with a simply solution (like %xxx) ?
>
>Thanx for it: dd
>
>Ps:
>Now I use this proc:
>
>def toths(i):
>s=str(i)
>l=[]
>ls=len(s)
>for i in range(ls):
>c=s[ls-i-1]
>if i%3==0 and i<>0:
>   c=c+"."
>l.append(c)
>l.reverse()
>return "".join(l)
>
>
>
>  
>
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Re: Quick help needed: how to format an integer ?

[Diez B. Roggisch]

[EMAIL PROTECTED] wrote:
> Hi !
> 
> I need to convert some integer values.
> 
> "1622" ->"1 622"
> 
> or
> 
> "10001234" -> ""10.001.234""
> 
> So I need thousand separators.
> 
> Can anyone helps me with a simply solution (like %xxx) ?

The module locale does what you need, look at ist docs, especially


locale.str
locale.format


Regards,

Diez
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Re: Quick help needed: how to format an integer ?

[Paul Rubin]

"[EMAIL PROTECTED]" <[EMAIL PROTECTED]> writes:
> "10001234" -> ""10.001.234""
> So I need thousand separators.
> Can anyone helps me with a simply solution (like %xxx) ?

I think you're supposed to do a locale-specific conversion (I've never
understood that stuff).  You could also do something like this:

>>> def f(n):
if n < 0: return '-' + f(-n)
if n < 1000: return '%d' % n
return f(n//1000) + '.' + '%03d' % (n%1000)

>>> f(3900900090090909009)
'39.009.000.900.909.090.000.009'
>>> f(39802183)
'39.802.183'
>>> f(3008)
'3.008'
>>> f(0)
'0'
>>> f(-9898239839)
'-9.898.239.839'
>>> f(12345)
'12.345'
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