Creating formatted output using picture strings
in [Python]
|
Prev: come and join www.pakdub.com where u can find friends, classifieds, games, music albums, events, blogs, chatrooms, video songs and lot more.... for free
Next: Generic spreadsheet interface
From: Peter Otten on 10 Feb 2010 04:45 python(a)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
From: Olof Bjarnason on 10 Feb 2010 05:50 2010/2/10 Peter Otten <__peter__(a)web.de>: > python(a)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 > -- > http://mail.python.org/mailman/listinfo/python-list > 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]' >>> -- http://olofb.wordpress.com
From: Alf P. Steinbach on 10 Feb 2010 06:45 * Olof Bjarnason: > 2010/2/10 Peter Otten <__peter__(a)web.de>: >> python(a)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 >> -- >> http://mail.python.org/mailman/listinfo/python-list >> > > 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
From: Olof Bjarnason on 10 Feb 2010 07:46 2010/2/10 Alf P. Steinbach <alfps(a)start.no>: > * Olof Bjarnason: >> >> 2010/2/10 Peter Otten <__peter__(a)web.de>: >>> >>> python(a)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 >>> -- >>> http://mail.python.org/mailman/listinfo/python-list >>> >> >> 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. Thanks! Here's another version (maybe a little more readable?): def first(s): return s[0] def rest(s): return s[1:] def picture(s, pic): if not s: return pic if first(pic)=='#': return first(s)+picture(rest(s), rest(pic)) return first(pic)+picture(s, rest(pic)) > > 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 > -- > http://mail.python.org/mailman/listinfo/python-list > -- http://olofb.wordpress.com
From: MRAB on 10 Feb 2010 11:02
Olof Bjarnason wrote: > 2010/2/10 Peter Otten <__peter__(a)web.de>: >> python(a)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 >> -- >> http://mail.python.org/mailman/listinfo/python-list >> > > 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]' > Here's a version without recursion: >>> def picture(s, pic): .... pic = pic.replace("%", "%%").replace("#", "%s") .... return pic % tuple(s) .... >>> picture("123456789", "(###)-##-(###)[#]") '(123)-45-(678)[9]' |