What is the pythonic way to do this ?
For my part, i reach to this rather complicated code:
# ----------------------
def comaSep(z,k=3, sep=','):
z=z[::-1]
x=[z[k*i:k*(i+1)][::-1] for i in range(1+(len(z)-1)/k)][::-1]
return sep.join(x)
# Test
for z in ["75096042068045", "509", "12024", "7", "2009"]:
print z+" --> ", comaSep(z)
Just if you are interested, a recursive solution:
>>> def comaSep(z,k=3,sep=","):
return comaSep(z[:-3],k,sep)+sep+z[-3:] if len(z)>3 else z
>>> comaSep("7")
'7'
>>> comaSep("2007")
'2,007'
>>> comaSep("12024")
'12,024'
>>> comaSep("509")
'509'
>>> comaSep("75096042068045")
'75,096,042,068,045'
>>>
Gregor
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