It seems that I rather frequently need a list or iterator of the form [x for x in <> while <>] And there is no one like this. May be there is another short way to write it (not as a loop). Is there? Thanks
> It seems that I rather frequently need a list or iterator of the form > [x for x in <> while <>] > And there is no one like this. > May be there is another short way to write it (not as a loop). Is > there?
The answer is very probably in the itertools module.
urikaluzhny <ukaluz...@nds.com> writes: > It seems that I rather frequently need a list or iterator of the form > [x for x in <> while <>] > And there is no one like this. > May be there is another short way to write it (not as a loop). Is there?
The following proposed solution is not intended to be a solution, it goes completely against the zen of python. [Type import this into the python command interpreter]
I brought it down to two lines:-
l = range(6) [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
itertools would still be a better approach in my opinion.
Just because I'm curious to know, can anyone bring it shorter[even if its cryptic] than this without invoking any Python Library.
P.S. Once again I would not recommend using this as Explicit is better than Implicit P.P.S. It is strongly undesirable for us humans to use anything starting with __ :)
> > I can think of two ways to interpret that. > >> I mean like [x for x in <A> if <B>], only that it breaks the loop when > >> the expression <B> is false.
> def gen(a): > for x in a: > if B: break > yield x
> The following proposed solution is not intended to be a solution, it > goes completely against the zen of python. [Type import this into the > python command interpreter]
> I brought it down to two lines:-
> l = range(6) > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> itertools would still be a better approach in my opinion.
> Just because I'm curious to know, can anyone bring it shorter[even if > its cryptic] than this without invoking any Python Library.
> P.S. Once again I would not recommend using this as Explicit is better > than Implicit > P.P.S. It is strongly undesirable for us humans to use anything > starting with __ :)
> On May 15, 5:10 pm, "Geoffrey Clements"
> <geoffrey.clement...@SPAMbaesystems.com> wrote: > > "urikaluzhny" <ukaluz...@nds.com> wrote in message
> > > I can think of two ways to interpret that. > > >> I mean like [x for x in <A> if <B>], only that it breaks the loop when > > >> the expression <B> is false.
> > def gen(a): > > for x in a: > > if B: break > > yield x
> > a_gen = gen(A)
> > # now iterate over a_gen
> > -- > > Geoff- Hide quoted text -
> - Show quoted text -
In your original, you have:
> l = range(6) > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
You may be hyperextending the use of '..if..else..', which is one of my fears regarding 'with x as y'. "l.__delslice__(0,len(l))" is not an expression.
"An expression in a programming language is a combination of values, variables, operators, and functions that are interpreted (evaluated) according to the particular rules of precedence and of association for a particular programming language, which computes and then produces (returns, in a stateful environment) another value."
l.__delslice__(0,len(l)) is an expression because it returns None [which also happens to be a value] in this case.
On May 16, 4:23 am, castironpi <castiro...@gmail.com> wrote:
> On May 15, 6:07 pm, afrobeard <afrobe...@gmail.com> wrote:
> > The following proposed solution is not intended to be a solution, it > > goes completely against the zen of python. [Type import this into the > > python command interpreter]
> > I brought it down to two lines:-
> > l = range(6) > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> > itertools would still be a better approach in my opinion.
> > Just because I'm curious to know, can anyone bring it shorter[even if > > its cryptic] than this without invoking any Python Library.
> > P.S. Once again I would not recommend using this as Explicit is better > > than Implicit > > P.P.S. It is strongly undesirable for us humans to use anything > > starting with __ :)
> > On May 15, 5:10 pm, "Geoffrey Clements"
> > <geoffrey.clement...@SPAMbaesystems.com> wrote: > > > "urikaluzhny" <ukaluz...@nds.com> wrote in message
> > > > I can think of two ways to interpret that. > > > >> I mean like [x for x in <A> if <B>], only that it breaks the loop when > > > >> the expression <B> is false.
> > > def gen(a): > > > for x in a: > > > if B: break > > > yield x
> > > a_gen = gen(A)
> > > # now iterate over a_gen
> > > -- > > > Geoff- Hide quoted text -
> > - Show quoted text -
> In your original, you have:
> > l = range(6) > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> You may be hyperextending the use of '..if..else..', which is one of > my fears regarding 'with x as y'. "l.__delslice__(0,len(l))" is not > an expression.
> On May 15, 6:07 pm, afrobeard <afrobe...@gmail.com> wrote:
> > The following proposed solution is not intended to be a solution, it > > goes completely against the zen of python. [Type import this into the > > python command interpreter]
> > I brought it down to two lines:-
> > l = range(6) > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> > itertools would still be a better approach in my opinion.
> > Just because I'm curious to know, can anyone bring it shorter[even if > > its cryptic] than this without invoking any Python Library.
> > P.S. Once again I would not recommend using this as Explicit is better > > than Implicit > > P.P.S. It is strongly undesirable for us humans to use anything > > starting with __ :)
> > On May 15, 5:10 pm, "Geoffrey Clements"
> > <geoffrey.clement...@SPAMbaesystems.com> wrote: > > > "urikaluzhny" <ukaluz...@nds.com> wrote in message
> > > > I can think of two ways to interpret that. > > > >> I mean like [x for x in <A> if <B>], only that it breaks the loop when > > > >> the expression <B> is false.
> > > def gen(a): > > > for x in a: > > > if B: break > > > yield x
> > > a_gen = gen(A)
> > > # now iterate over a_gen
> > > -- > > > Geoff- Hide quoted text -
> > - Show quoted text -
> In your original, you have:
> > l = range(6) > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> You may be hyperextending the use of '..if..else..', which is one of > my fears regarding 'with x as y'. "l.__delslice__(0,len(l))" is not > an expression.
> l.__delslice__(0,len(l)) is an expression as it returns None which is > a value
> On May 16, 4:23 am, castironpi <castiro...@gmail.com> wrote:
> > On May 15, 6:07 pm, afrobeard <afrobe...@gmail.com> wrote:
> > > The following proposed solution is not intended to be a solution, it > > > goes completely against the zen of python. [Type import this into the > > > python command interpreter]
> > > I brought it down to two lines:-
> > > l = range(6) > > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> > > itertools would still be a better approach in my opinion.
> > > Just because I'm curious to know, can anyone bring it shorter[even if > > > its cryptic] than this without invoking any Python Library.
> > > P.S. Once again I would not recommend using this as Explicit is better > > > than Implicit > > > P.P.S. It is strongly undesirable for us humans to use anything > > > starting with __ :)
> > > On May 15, 5:10 pm, "Geoffrey Clements"
> > > <geoffrey.clement...@SPAMbaesystems.com> wrote: > > > > "urikaluzhny" <ukaluz...@nds.com> wrote in message
> > > > > I can think of two ways to interpret that. > > > > >> I mean like [x for x in <A> if <B>], only that it breaks the loop when > > > > >> the expression <B> is false.
> > > > def gen(a): > > > > for x in a: > > > > if B: break > > > > yield x
> > > > a_gen = gen(A)
> > > > # now iterate over a_gen
> > > > -- > > > > Geoff- Hide quoted text -
> > > - Show quoted text -
> > In your original, you have:
> > > l = range(6) > > > [1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
> > You may be hyperextending the use of '..if..else..', which is one of > > my fears regarding 'with x as y'. "l.__delslice__(0,len(l))" is not > > an expression.- Hide quoted text -
> - Show quoted text -
Functional programming is really important to a former professor of me. I like to say that None returns as a value. I think you can call functional "evaluational".
urikaluzhny wrote: >> | It seems that I rather frequently need a list or iterator of the form >> | [x for x in <> while <>] >> I can think of two ways to interpret that. > I mean like [x for x in <A> if <B>], only that it breaks the loop when > the expression <B> is false.
How do you plan to modify B during iteration? May be [x for x in itertools.takewhile(<B>, <A>)] when <B> function accept element <A> and return True or False
urikaluzhny wrote: > It seems that I rather frequently need a list or iterator of the form > [x for x in <> while <>] > And there is no one like this. > May be there is another short way to write it (not as a loop). Is > there? > Thanks
I usually have the same problem and i came up with an solution like that:
from operator import ne def test(iterable, value, op=ne): _n = iter(iterable).next while True: _x = _n() if op(_x, value): yield _x else: raise StopIteration
> urikaluzhny wrote: > > It seems that I rather frequently need a list or iterator of the form > > [x for x in <> while <>] > > And there is no one like this. > > May be there is another short way to write it (not as a loop). Is > > there? > > Thanks
> I usually have the same problem and i came up with an solution like that:
> from operator import ne > def test(iterable, value, op=ne): > _n = iter(iterable).next > while True: > _x = _n() > if op(_x, value): > yield _x > else: > raise StopIteration
This is better written using takewhile... itertools.takewhile(lambda x: x != value, iterable)
But if you really need to reinvent the wheel, perhaps this is simpler?
def test(iterable, value, op=operator.ne): for x in iterable: if not op(x, value): return yield x
Paul Hankin wrote: > This is better written using takewhile... > itertools.takewhile(lambda x: x != value, iterable)
> But if you really need to reinvent the wheel, perhaps this is simpler?
> def test(iterable, value, op=operator.ne): > for x in iterable: > if not op(x, value): > return > yield x
yes you are right it is. However as i mentioned in my post i came up with an solution 'like' that. In fact my original code was to complex to post. While simplifying it, i've overseen the obvious solution.
For special cases where you need to do more complex tests, the best solution is IMHO to hide it in an generator function like above.
