* Olof Bjarnason:
2010/2/10 Peter Otten <__pete...@web.de>:
pyt...@bdurham.com wrote:
Does Python provide a way to format a string according to a
'picture' format?
For example, if I have a string '123456789' and want it formatted
like '(123)-45-(678)[9]', is there a module or function that will
allow me to do this or do I need to code this type of
transformation myself?
A basic implementation without regular expressions:
def picture(s, pic, placeholder="@"):
... parts = pic.split(placeholder)
... result = [None]*(len(parts)+len(s))
... result[::2] = parts
... result[1::2] = s
... return "".join(result)
...
picture("123456789", "(@@@)-@@-(@@@)[...@]")
'(123)-45-(678)[9]'
Peter
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Inspired by your answer here's another version:
def picture(s, pic):
... if len(s)==0: return pic
... if pic[0]=='#': return s[0]+picture(s[1:], pic[1:])
... return pic[0]+picture(s, pic[1:])
...
picture("123456789", "(###)-##-(###)[#]")
'(123)-45-(678)[9]'
I learned a bit by Peter Otten's example; I would have gotten to that notation
sooner or later, but that example made it 'sooner' :-).
I think your version is cute.
I'd probably write it in a non-recursive way, though, like
def picture( s, pic, placeholder = "@" ):
result = ""
char_iter = iter( s )
for c in pic:
result += c if c != placeholder else next( char_iter )
return result
Of course this is mostly personal preference, but there is also a functional
difference.
With your version an IndexError will be raised if there are too /many/
characters in s, while too few characters in s will yield "#" in the result.
With my version a StopIteration will be raised if there are to /few/ characters
in s, while too many characters will just have the extraneous chars ignored.
Cheers,
- Alf
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