Sometimes I want to compare a "pointer" to more then one others. The "in" operator would be handy, but it uses the "==" operator instead of the "is" operator. So a "is in" operator would be nice. Though I don't know how easy it is for a newbie to see what does what. # This: if x is in (a, b, c): ... # would be equivalent to this: if x is a or x is b or x is c: ... # And of course there should be a "is not in" operator, too: if x is not in (a, b, c): ... # this would be equivalent to tis: if x is not a and x is not b and x is not c: ... Hmmm, maybe a way to apply some kind of comparison between a value and more other values would be better. But that already exists, so screw this msg: if any(x is y for y in (a, b, c)): ... if all(x is not y for y in (a, b, c)): ...
On 9/26/07, Mathias Panzenböck <grosser.meister.morti@gmx.net> wrote:
Or in a more obfuscated way: import operator as op from itertools import imap from functools import partial if any(imap(partial(op.is_,x), (a, b, c))): ... if all(imap(partial(op.is_not,x), (a, b, c))): ... George
George Sakkis wrote:
Or in haskell (assuming haskell would have "is" and "is not"): if any (x is) [a, b, c] then ... else ... if all (x is not) [a, b c] then ... else ... I'm not sure if "any" and "all" are the ones with 2 parameters (function and list) or if that would be "or" and "and". -panzi
On 9/26/07, Mathias Panzenböck <grosser.meister.morti@gmx.net> wrote:
There's many different ways you might want to do a comparison. That's why sorted() has a cmp=func argument. A new API won't work though, as dicts or sets need to know the hash in advance, and lists are O(n) anyway (so there's little appropriate use.) To solve your problem you should be using a decorate/undecorate pattern, possibly encapsulated into a custom container type. There doesn't appear to be any in the python cookbook (so it may be a very rare need), but assuming you did use a container type your code might be rewritten as such: if x in idset([a, b, c]): But decorating is almost as simple: if id(x) in [id(a), id(b), id(c)]: (Caveat: id(obj) assumes you have another reference to the obj, to prevent the identity from being reused.)
-- Adam Olsen, aka Rhamphoryncus
On 9/26/07, Mathias Panzenböck <grosser.meister.morti@gmx.net> wrote:
Or in a more obfuscated way: import operator as op from itertools import imap from functools import partial if any(imap(partial(op.is_,x), (a, b, c))): ... if all(imap(partial(op.is_not,x), (a, b, c))): ... George
George Sakkis wrote:
Or in haskell (assuming haskell would have "is" and "is not"): if any (x is) [a, b, c] then ... else ... if all (x is not) [a, b c] then ... else ... I'm not sure if "any" and "all" are the ones with 2 parameters (function and list) or if that would be "or" and "and". -panzi
On 9/26/07, Mathias Panzenböck <grosser.meister.morti@gmx.net> wrote:
There's many different ways you might want to do a comparison. That's why sorted() has a cmp=func argument. A new API won't work though, as dicts or sets need to know the hash in advance, and lists are O(n) anyway (so there's little appropriate use.) To solve your problem you should be using a decorate/undecorate pattern, possibly encapsulated into a custom container type. There doesn't appear to be any in the python cookbook (so it may be a very rare need), but assuming you did use a container type your code might be rewritten as such: if x in idset([a, b, c]): But decorating is almost as simple: if id(x) in [id(a), id(b), id(c)]: (Caveat: id(obj) assumes you have another reference to the obj, to prevent the identity from being reused.)
-- Adam Olsen, aka Rhamphoryncus
participants (3)
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Adam Olsen -
George Sakkis -
Mathias Panzenböck