Re: Is there a better algorithm?
On Jan 3, 4:38 am, mr mario.rugg...@gmail.com wrote: As has been noted, the best is to fix the input to be regular-3- tuples. For the fun of it, here's another variation of a solution: snip Yet another solution: for i in l: k, u, v = i[0], None if len(i) == 2 else i[1], i[-1] -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
On Jan 3, 11:55 am, Kottiyath n.kottiy...@gmail.com wrote: On Jan 3, 2:38 am, mr mario.rugg...@gmail.com wrote: snip It is a code to post some data to HTML server. Even though usually the POST values are of type(name, value), if file transfer is involved, then POST values change to (name, filename, value). My view was that since filename is a rare occurance and doesnt make sense in a usual POST, I had not kept it as a full 3 tuple. Since so many programmers (that too much more capable than me) are suggesting that it is code smell, I am reviewing my decision. Is it possible to change (name, filename, value) into (name, value, filename) instead? In most cases optional arguments should be last. There are some very exceptional case, where early optional argument might be better, like python's range, which type signature is: [[start,] stop[, step]] (i.e. with one argument, it is assigned to stop, instead of start) -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
En Sat, 03 Jan 2009 02:55:12 -0200, Kottiyath n.kottiy...@gmail.com
escribió:
tuples = [(1, 2), (3, 4, 5), (6, 7)]
It is a code to post some data to HTML server.
Even though usually the POST values are of type(name, value), if file
transfer is involved, then POST values change to (name, filename,
value).
My view was that since filename is a rare occurance and doesnt make
sense in a usual POST, I had not kept it as a full 3 tuple.
Since so many programmers (that too much more capable than me) are
suggesting that it is code smell, I am reviewing my decision.
What about using another data structure instead - like this:
class entry:
filename = None
def __init__(self, name, value, filename=None):
self.name = name
self.value = value
if filename is not None:
self.filename = filename
[entry('name', 'Gabriel'), entry('age', 18), entry('pic', picture_data,
'path/to/file')]
--
Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
class entry:
filename = None
def __init__(self, name, value, filename=None):
self.name = name
self.value = value
if filename is not None:
self.filename = filename
[entry('name', 'Gabriel'), entry('age', 18), entry('pic', picture_data,
'path/to/file')]
Gabriel thank you for this. I just now realized I've being doing things the
hard way. :)
-Alex Goretoy
http://www.alexgoretoy.com
somebodywhoca...@gmail.com
On Mon, Jan 5, 2009 at 8:58 PM, Gabriel Genellina gagsl-...@yahoo.com.arwrote:
En Sat, 03 Jan 2009 02:55:12 -0200, Kottiyath n.kottiy...@gmail.com
escribió:
tuples = [(1, 2), (3, 4, 5), (6, 7)]
It is a code to post some data to HTML server.
Even though usually the POST values are of type(name, value), if file
transfer is involved, then POST values change to (name, filename,
value).
My view was that since filename is a rare occurance and doesnt make
sense in a usual POST, I had not kept it as a full 3 tuple.
Since so many programmers (that too much more capable than me) are
suggesting that it is code smell, I am reviewing my decision.
What about using another data structure instead - like this:
class entry:
filename = None
def __init__(self, name, value, filename=None):
self.name = name
self.value = value
if filename is not None:
self.filename = filename
[entry('name', 'Gabriel'), entry('age', 18), entry('pic', picture_data,
'path/to/file')]
--
Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
--
http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
bearophile wrote: Fuzzyman: for i in l: u = None if len(i) == 2: k, v = i else: k, u, v = i That's the best solution I have seen in this thread so far (but I suggest to improve indents and use better variable names). In programming it's generally better to follow the KISS principle. Strange that nobody has suggested: for Tup in Lst: for item in Tup: print item, print Because if I recall correctly, all the OP was doing was printing the stuff. Gets a bit hairier if you want to return the values though. - if you replace the first print statement above with a yield and delete the second, bare print, you lose knowledge of which tuple the item belongs to. It is not trivial to do if you don't have a priori knowledge of the maximum and minimum tuple length. Seems to me it is one of those gems that is a PITA to do completely generally - basically because you don't know when you call it how many items to unpack - so you have to return a count or something, or just fall back on a two step approach - get the tuple from the list, then do something with the items in the tuple. - Hendrik -- http://mail.python.org/mailman/listinfo/python-list
Is there a better algorithm?
I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
On Jan 2, 6:11 pm, Kottiyath n.kottiy...@gmail.com wrote: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? for i in l: u = None if len(i) == 2: k, v = i else: k, u, v = i Best I could come up with. Alternatively: def mangle(i): if len(i) == 3: return i k, v = i return k, None, v for i in l: k, u, v = mangle(i) I'm sure there is a clever one liner using the Python 2.5 ternary expression syntax. On the other hand I'm not sure it would be very readable, so a straightforward (if less clever) solution is probably better. Michael -- http://www.ironpythoninaction.com/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath wrote: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? L = [(1, 2), (3, 4, 5), (6, 7)] for i in L: if len(i) == 2: k, v, u = i + (None, ) else: k, u, v = i print k, u, v regards Steve -- Steve Holden+1 571 484 6266 +1 800 494 3119 Holden Web LLC http://www.holdenweb.com/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Hi, Fuzzyman wrote: I'm sure there is a clever one liner using the Python 2.5 ternary expression syntax. On the other hand I'm not sure it would be very readable, so a straightforward (if less clever) solution is probably better. that would be something like this (using a generator) L = [(1, 2), (3, 4, 5), (6, 7)] g = ( ((e[0],None,e[1]) if len(e)==2 else (e[0],e[1],e[2])) for e in L) for elem in g: print elem So long, Markus P.S: No, I don't consider that more readable than your example ;-) -- Hacking is like having sex: you get in, you get out and you hope that you did not leave anything which can be traced back to you - Unbekannt -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath a écrit : I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? complement = lambda t: (t[0], None, t[1]) if len(t) == 2 else t for k, u, v in map(complement, L): print k, u, v -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath ha scritto: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? One way to avoid explicit checks on tuple size (but making the code a bit less clear and probably slower): for i in l: k, v, u = (i[:3]+(None,))[:3] ... Ciao - FB -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
On Fri, 02 Jan 2009 19:55:43 +0100 Markus Brueckner n...@slash-me.net wrote: g = ( ((e[0],None,e[1]) if len(e)==2 else (e[0],e[1],e[2])) for e in L) If this isn't proof of Python's versatility, I don't know what is. In one line it can mimic both Lisp and Perl. Sweet. :) /W -- My real email address is constructed by swapping the domain with the recipient (local part). -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath n.kottiy...@gmail.com writes: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? Just a note: this isn't really an algorithm problem. ;) It's more of a grammar obstruction. To make your code more simple, it would be nice if the assignment operator would return, None, in the case where there are too few values to unpack from the right-operand of the assignment operator. Aside from the typical workarounds that first came to mind, I started wondering whether it was possible to expose the problem and solve it directly. Sadly, it appears to be non-trivial (or at least, well hidden from the unwashed masses). I'd be really curious if the unpacking machinery were exposed to the developer. I started poking around the operator and types modules, but the implementation isn't obvious. What methods are being called on the operands by the assignment operator in the following statement: a, b, c = some_tuple I'm sorry if this line of inquiry is not very pythonic; but one is curious if this is some python magic happening here. After using the idiom for years I hadn't really thought about it much until recently. -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath n.kottiy...@gmail.com writes: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. Others have suggested messy ways to code what you're asking. At another level, that list format seems like a code smell. You may be better off organizing the program so that L = [(1, None, 2), (3, 4, 5), (6, None, 7)] after which unpacking becomes trivial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Fuzzyman: for i in l: u = None if len(i) == 2: k, v = i else: k, u, v = i That's the best solution I have seen in this thread so far (but I suggest to improve indents and use better variable names). In programming it's generally better to follow the KISS principle. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Paul Rubin http://phr...@nospam.invalid writes: Kottiyath n.kottiy...@gmail.com writes: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. Others have suggested messy ways to code what you're asking. At another level, that list format seems like a code smell. You may be better off organizing the program so that L = [(1, None, 2), (3, 4, 5), (6, None, 7)] after which unpacking becomes trivial. Very true. +1 -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
In article 85aba9h1e5@dozer.localdomain, J Kenneth King ja...@agentultra.com wrote: Kottiyath n.kottiy...@gmail.com writes: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? Just a note: this isn't really an algorithm problem. ;) It's more of a grammar obstruction. To make your code more simple, it would be nice if the assignment operator would return, None, in the case where there are too few values to unpack from the right-operand of the assignment operator. Looks like the extended iterable unpacking feature of Python 3.0, described in PEP 3132, comes to the rescue here: L = [(1, 2), (3, 4, 5), (6, 7)] for i in L: ... k, *u, v = i ... print(k, u, v) ... 1 [] 2 3 [4] 5 6 [] 7 -- Ned Deily, n...@acm.org -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
Kottiyath wrote: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: [...] If this is part of a real program, instead of an exercise, you should fix the code that creates this list of tuples so that they have a uniform length of 3. And if the third element is missing, it should be None. This saves lots of trouble later on. -- Gerhard -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
On Jan 2, 12:11 pm, Kottiyath n.kottiy...@gmail.com wrote: I have the following list of tuples: L = [(1, 2), (3, 4, 5), (6, 7)] I want to loop through the list and extract the values. The only algorithm I could think of is: for i in l: ... u = None ... try: ... (k, v) = i ... except ValueError: ... (k, u, v) = i ... print k, u, v - 1 None 2 3 4 5 6 None 7 - But, this algorithm doesnt look very beautiful - like say - for k, u, v in L: Can anyone suggest a better algorithm to get the values? for i in L: k, u, v= i[ 0 ], i[ 1 ], i[ -1 ] if len( i )== 2: u= None -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
As has been noted, the best is to fix the input to be regular-3- tuples. For the fun of it, here's another variation of a solution: tuples = [(1, 2), (3, 4, 5), (6, 7)] def triple_or_pair(seq): u = None try: k, u, v = seq except ValueError: k, v = seq return k, u, v for k, u, v in [ triple_or_pair(seq) for seq in tuples ]: print k, u, v -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
On Jan 3, 2:38 am, mr mario.rugg...@gmail.com wrote: As has been noted, the best is to fix the input to be regular-3- tuples. For the fun of it, here's another variation of a solution: tuples = [(1, 2), (3, 4, 5), (6, 7)] def triple_or_pair(seq): u = None try: k, u, v = seq except ValueError: k, v = seq return k, u, v for k, u, v in [ triple_or_pair(seq) for seq in tuples ]: print k, u, v It is a code to post some data to HTML server. Even though usually the POST values are of type(name, value), if file transfer is involved, then POST values change to (name, filename, value). My view was that since filename is a rare occurance and doesnt make sense in a usual POST, I had not kept it as a full 3 tuple. Since so many programmers (that too much more capable than me) are suggesting that it is code smell, I am reviewing my decision. -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a better algorithm?
I personally like this solution the best. Thanks mr. tuples = [(1, 2), (3, 4, 5), (6, 7)] def triple_or_pair(seq): u = None try: k, u, v = seq except ValueError: k, v = seq return k, u, v for k, u, v in [ triple_or_pair(seq) for seq in tuples ]: print k, u, v On Sat, Jan 3, 2009 at 4:55 AM, Kottiyath n.kottiy...@gmail.com wrote: On Jan 3, 2:38 am, mr mario.rugg...@gmail.com wrote: As has been noted, the best is to fix the input to be regular-3- tuples. For the fun of it, here's another variation of a solution: tuples = [(1, 2), (3, 4, 5), (6, 7)] def triple_or_pair(seq): u = None try: k, u, v = seq except ValueError: k, v = seq return k, u, v for k, u, v in [ triple_or_pair(seq) for seq in tuples ]: print k, u, v It is a code to post some data to HTML server. Even though usually the POST values are of type(name, value), if file transfer is involved, then POST values change to (name, filename, value). My view was that since filename is a rare occurance and doesnt make sense in a usual POST, I had not kept it as a full 3 tuple. Since so many programmers (that too much more capable than me) are suggesting that it is code smell, I am reviewing my decision. -- http://mail.python.org/mailman/listinfo/python-list -- А-Б-В-Г-Д-Е-Ё-Ж-З-И-Й-К-Л-М-Н-О-П-Р-С-Т-У-Ф-Х-Ц-Ч-Ш-Щ-Ъ-Ы-Ь-Э-Ю-Я а-б-в-г-д-е-ё-ж-з-и-й-к-л-м-н-о-п-р-с-т-у-ф-х-ц-ч-ш-щ-ъ-ы-ь-э-ю-я -- http://mail.python.org/mailman/listinfo/python-list
