I can think of two ways to interpret that.
| And there is no one like this.
| May be there is another short way to write it (not as a loop). Is
| there?
Using loops to write an example of what you mean would make the above
clearer.
tjr
The answer is very probably in the itertools module.
itertools.takewhile(condition, seq)
def gen(a):
for x in a:
if B: break
yield x
a_gen = gen(A)
# now iterate over a_gen
--
Geoff
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"
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:
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".
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
l = range(6)
print [x for x in test(l, 4)]
r@linux:~/tmp> python test18.py
[0, 1, 2, 3]
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
> 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.