Re: Last iteration?

[Gabriel Genellina]

En Fri, 19 Oct 2007 19:12:49 -0300, Michael J. Fromberger  
<[EMAIL PROTECTED]> escribió:

> Before I affront you with implementation details, here's an example:
>
> | from __future__ import with_statement
>
> | with last_of(enumerate(file('/etc/passwd', 'rU'))) as fp:
> | for pos, line in fp:
> | if fp.marked():
> | print "Last line, #%d = %s" % (pos + 1, line.strip())
>
> In short, last_of comprises a trivial context manager that knows how to
> iterate over its input, and can also indicate that certain elements are
> "marked".  In this case, only the last element is marked.

The name is very unfortunate. I'd expect that last_of(something) would  
return its last element, not an iterator.
Even with a different name, I don't like that marked() (a method of the  
iterator) should be related to the current element being iterated.

> We could also make the truth value of the context manager indicate the
> marking, as illustrated here:
>
> | with last_of("alphabet soup") as final:
> | for c in final:
> | if final:
> | print "Last character: %s" % c
>
> This is bit artificial, perhaps, but effective enough.  Of course, there

Again, why should the trueness of final be related to the current element  
being iterated?

> is really no reason you have to use "with," since we don't really care
> what happens when the object goes out of scope:  You could just as
> easily write:
>
> | end = last_of(xrange(25))
> | for x in end:
> |   if end:
> | print "Last element: %s" % x

Again, why the truth value of "end" is related to the current "x" element?

> But you could also handle nested context, using "with".  Happily, the
> machinery to do all this is both simple and easily generalized to other
> sorts of "marking" tasks.  For example, we could just as well do
> something special with all the elements that are accepted by a predicate
> function, e.g.,
>
> | def isinteger(obj):
> | return isinstance(obj, (int, long))
>
> | with matching(["a", 1, "b", 2, "c"], isinteger) as m:
> | for elt in m:
> | if m.marked():
> | print '#%s' % elt,
> | else:
> | print '(%s)' % elt,
> |
> | print

I think you are abusing context managers *a*lot*!
Even accepting such evil thing as matching(...), the above code could be  
equally written as:

m = matching(...)
for elt in m:
   ...

Anyway, a simple generator that yields (elt, function(elt)) would be  
enough...

-- 
Gabriel Genellina

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Re: Last iteration?

[Michael J. Fromberger]

In article <[EMAIL PROTECTED]>,
 Raymond Hettinger <[EMAIL PROTECTED]> wrote:

> [Diez B. Roggisch]
> > > out:) But I wanted a general purpose based solution to be available that
> > > doesn't count on len() working on an arbitrary iterable.
> 
> [Peter Otten]
> > You show signs of a severe case of morbus itertools.
> > I, too, am affected and have not yet fully recovered...
> 
> Maybe you guys were secretly yearning for a magical last element
> detector used like this: [...]


Although there have already been some nice solutions to this problem, 
but I'd like to propose another, which mildly abuses some of the newer 
features of Python  It is perhaps not as concise as the previous 
solutions, nor do I claim it's better; but I thought I'd share it as an 
alternative approach.

Before I affront you with implementation details, here's an example:

| from __future__ import with_statement

| with last_of(enumerate(file('/etc/passwd', 'rU'))) as fp:
| for pos, line in fp:
| if fp.marked():
| print "Last line, #%d = %s" % (pos + 1, line.strip())

In short, last_of comprises a trivial context manager that knows how to 
iterate over its input, and can also indicate that certain elements are 
"marked".  In this case, only the last element is marked.

We could also make the truth value of the context manager indicate the 
marking, as illustrated here:

| with last_of("alphabet soup") as final:
| for c in final:
| if final:
| print "Last character: %s" % c

This is bit artificial, perhaps, but effective enough.  Of course, there 
is really no reason you have to use "with," since we don't really care 
what happens when the object goes out of scope:  You could just as 
easily write:

| end = last_of(xrange(25))
| for x in end:
|   if end:
| print "Last element: %s" % x

But you could also handle nested context, using "with".  Happily, the 
machinery to do all this is both simple and easily generalized to other 
sorts of "marking" tasks.  For example, we could just as well do 
something special with all the elements that are accepted by a predicate 
function, e.g.,

| def isinteger(obj):
| return isinstance(obj, (int, long))

| with matching(["a", 1, "b", 2, "c"], isinteger) as m:
| for elt in m:
| if m.marked():
| print '#%s' % elt,
| else:
| print '(%s)' % elt,
|
| print

Now that you've seen the examples, here is an implementation.  The 
"marker" class is an abstract base that does most of the work, with the 
"last_of" and "matching" examples implemented as subclasses:

| class marker (object):
| """Generate a trivial context manager that flags certain elements
| in a sequence or iterable.
| 
| Usage sample:
|   with marker(ITERABLE) as foo:
| for elt in foo:
|   if foo.marked():
|  print 'this is a marked element'
|   else:
|  print 'this is an unmarked element'
| 
| Subclass overrides:
|  .next()   -- return the next unconsumed element from the input.
|  .marked() -- return True iff the last element returned is marked.
| 
| By default, no elements are marked.
| """
| def __init__(self, seq):
| self._seq = iter(seq)
| try:
| self._fst = self._seq.next()
| except StopIteration:
| self.next = self._empty
| 
| def _empty(self):
| raise StopIteration
| 
| def _iter(self):
| while True:
| yield self.next()
| 
| def next(self):
| out = self._fst
| try:
| self._fst = self._seq.next()
| except StopIteration:
| self.next = self._empty
| 
| return out
| 
| def marked(self):
| return False
| 
| def __iter__(self):
| return iter(self._iter())
| 
| def __nonzero__(self):
| return self.marked()
| 
| def __enter__(self):
| return self
| 
| def __exit__(self, *args):
| pass

A bit verbose, but uncomplicated apart from the subtlety in handling the 
end case.  Here's last_of, implemented as a subclass:

| class last_of (marker):
| def __init__(self, seq):
| super(last_of, self).__init__(seq)
| self._end = False
| 
| def next(self):
| out = super(last_of, self).next()
| if self.next == self._empty:
| self._end = True
| 
| return out
| 
| def marked(self):
| return self._end

And finally, matching:

| class matching (marker):
| def __init__(self, seq, func):
| super(matching, self).__init__(seq)
| self._func = func
| self._mark = False
| 
| def next(self):
| out = super(matching, self).next()
| self._mark = self._func(out)
| return out
| 
| def marked(self):
| return self._mark

Generally speaking, you should only have to override .next() and 
.marked() to make a useful subclass of mark

Re: Last iteration?

[Paul Hankin]

On Oct 17, 11:45 pm, Raymond Hettinger <[EMAIL PROTECTED]> wrote:
> [Paul Hankin]
>
> > def lastdetector(iterable):
> > t, u = tee(iterable)
> > return izip(chain(imap(lambda x: False, islice(u, 1, None)),
> > [True]), t)
>
> Sweet!  Nice, clean piece of iterator algebra.
>
> We need a C-speed verion of the lambda function, something like a K
> combinator that consumes arguments and emits constants.

Perhaps, but I think you'll need a better use-case than this :)

Actually, would a c-version be much faster?

--
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Re: Last iteration?

[Hendrik van Rooyen]

"Raymond Hettinger"  wrote:

> More straight-forward version:
> 
> def lastdetecter(iterable):
> t, lookahead = tee(iterable)
> lookahead.next()
> return izip(chain(imap(itemgetter(0), izip(repeat(False),
> lookahead)), repeat(True)), t)

If this is what you call straightforward - heaven forfend
that I ever clap my orbs on something you call convoluted!

:-)

Faced with this problem, I would probably have used
enumerate with a look ahead and the IndexError would
have told me I am at the end...

- Hendrik

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Re: Last iteration?

[Paul Rubin]

Raymond Hettinger <[EMAIL PROTECTED]> writes:
> We need a C-speed verion of the lambda function, something like a K
> combinator that consumes arguments and emits constants.

Some discussion of this is at .
I had suggested implementing K through an optional second arg for a
proposed identity function but everyone hated that.  I'm amused.  I
had suggested it because I thought that it would be easier than
getting two separate functions accepted.
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Re: Last iteration?

[Raymond Hettinger]

[Paul Hankin]
> def lastdetector(iterable):
> t, u = tee(iterable)
> return izip(chain(imap(lambda x: False, islice(u, 1, None)),
> [True]), t)

Sweet!  Nice, clean piece of iterator algebra.

We need a C-speed verion of the lambda function, something like a K
combinator that consumes arguments and emits constants.


Raymond



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Re: Last iteration?

[Paul Hankin]

On Oct 17, 8:16 am, Raymond Hettinger <[EMAIL PROTECTED]> wrote:
> > def lastdetecter(iterable):
> > "fast iterator algebra"
> > lookahead, t = tee(iterable)
> > lookahead.next()
> > t = iter(t)
> > return chain(izip(repeat(False), imap(itemgetter(1),
> > izip(lookahead, t))), izip(repeat(True),t))
>
> More straight-forward version:
>
> def lastdetecter(iterable):
> t, lookahead = tee(iterable)
> lookahead.next()
> return izip(chain(imap(itemgetter(0), izip(repeat(False),
> lookahead)), repeat(True)), t)

def lastdetector(iterable):
t, u = tee(iterable)
return izip(chain(imap(lambda x: False, islice(u, 1, None)),
[True]), t)

--
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Re: Last iteration?

[Peter Otten]

Raymond Hettinger wrote:

> [Diez B. Roggisch]
>> > out:) But I wanted a general purpose based solution to be available that
>> > doesn't count on len() working on an arbitrary iterable.
> 
> [Peter Otten]
>> You show signs of a severe case of morbus itertools.
>> I, too, am affected and have not yet fully recovered...
> 
> Maybe you guys were secretly yearning for a magical last element
> detector used like this:

Not secretly...

> def lastdetecter(iterable):
> it = iter(iterable)
> value = it.next()
> for nextvalue in it:
> yield (False, value)
> value = nextvalue
> yield (True, value)

as that's what I posted above...

> def lastdetecter(iterable):
> "fast iterator algebra"
> lookahead, t = tee(iterable)

  # make it cope with zero-length iterables 
  lookahead = islice(lookahead, 1, None)

> return chain(izip(repeat(False), imap(itemgetter(1),
> izip(lookahead, t))), izip(repeat(True),t))

and that's the "somebody call the doctor -- now!" version ;)

Peter
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Re: Last iteration?

[Raymond Hettinger]

> def lastdetecter(iterable):
> "fast iterator algebra"
> lookahead, t = tee(iterable)
> lookahead.next()
> t = iter(t)
> return chain(izip(repeat(False), imap(itemgetter(1),
> izip(lookahead, t))), izip(repeat(True),t))

More straight-forward version:

def lastdetecter(iterable):
t, lookahead = tee(iterable)
lookahead.next()
return izip(chain(imap(itemgetter(0), izip(repeat(False),
lookahead)), repeat(True)), t)


Raymond

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Re: Last iteration?

[Raymond Hettinger]

[Diez B. Roggisch]
> > out:) But I wanted a general purpose based solution to be available that
> > doesn't count on len() working on an arbitrary iterable.

[Peter Otten]
> You show signs of a severe case of morbus itertools.
> I, too, am affected and have not yet fully recovered...

Maybe you guys were secretly yearning for a magical last element
detector used like this:

for islast, value in lastdetecter([1,2,3]):
if islast:
print 'Last', value
else:
print value

Perhaps it could be written plainly using generators:

def lastdetecter(iterable):
it = iter(iterable)
value = it.next()
for nextvalue in it:
yield (False, value)
value = nextvalue
yield (True, value)

Or for those affected by "morbus itertools", a more arcane incantation
would be preferred:

from itertools import tee, chain, izip, imap
from operator import itemgetter

def lastdetecter(iterable):
"fast iterator algebra"
lookahead, t = tee(iterable)
lookahead.next()
t = iter(t)
return chain(izip(repeat(False), imap(itemgetter(1),
izip(lookahead, t))), izip(repeat(True),t))


Raymond

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Re: Last iteration?

[Bruno Desthuilliers]

Paul Hankin a écrit :
> On Oct 12, 11:58 am, Florian Lindner <[EMAIL PROTECTED]> wrote:
> 
>>Hello,
>>can I determine somehow if the iteration on a list of values is the last
>>iteration?
>>
>>Example:
>>
>>for i in [1, 2, 3]:
>>   if last_iteration:
>>  print i*i
>>   else:
>>  print i
> 
> 
> Yes, either use enumerate or just stop the loop early and deal with
> the last element outside the loop.
> 
> xs = [1, 2, 3]
> for x in xs[:-1]:
> print x
> print xs[-1] * xs[-1]

At least something sensible !-)
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Re: Last iteration?

[Peter Otten]

Diez B. Roggisch wrote:

> out:) But I wanted a general purpose based solution to be available that 
> doesn't count on len() working on an arbitrary iterable.

You show signs of a severe case of morbus itertools.
I, too, am affected and have not yet fully recovered...

Peter
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Re: Last iteration?

[Paul McGuire]

On Oct 14, 5:58 am, Paul Hankin <[EMAIL PROTECTED]> wrote:
> On Oct 14, 8:00 am, Paul McGuire <[EMAIL PROTECTED]> wrote:
> > def signal_last(lst):
> > last2 = None
> > it = iter(lst)
> > try:
> > last = it.next()
> > except StopIteration:
> > last = None
> > for last2 in it:
> > yield False, last
> > last = last2
> > yield True, last
>
> This yields a value when the iterator is empty, which Diez's solution
> didn't. Logically, there is no 'last' element in an empty sequence,
> and it's obscure to add one. Peter Otten's improvement to Diez's code
> looks the best to me: simple, readable and correct.
>

Of course!  For some reason I thought I was improving Peter Otten's
version, but when I modified my submission to behave as you stated, I
ended right back with what Peter had submitted.  Agreed - nice and
neat!

-- Paul


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Re: Last iteration?

[Diez B. Roggisch]

Peter Otten schrieb:
> Diez B. Roggisch wrote:
> 
>> Florian Lindner wrote:
> 
>>> can I determine somehow if the iteration on a list of values is the
>>> last iteration?
> 
>>  def last_iter(iterable):
>> it = iter(iterable)
>> buffer = [it.next()]
>> for i in it:
>> buffer.append(i)
>> old, buffer = buffer[0], buffer[1:]
>> yield False, old
>> yield True, buffer[0]
> 
> This can be simplified a bit since you never have to remember more than on
> item:
> 
>>>> def mark_last(items):
> ... items = iter(items)
> ... last = items.next()
> ... for item in items:
> ... yield False, last
> ... last = item
> ... yield True, last
> ...
>>>> list(mark_last([]))
> []
>>>> list(mark_last([1]))
> [(True, 1)]
>>>> list(mark_last([1,2]))
> [(False, 1), (True, 2)]

Nice.

I tried to come up with that solution in the first place - but most 
probably due to an java-coding induced brain overload it didn't work 
out:) But I wanted a general purpose based solution to be available that 
doesn't count on len() working on an arbitrary iterable.

Diez
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Re: Last iteration?

[Paul Hankin]

On Oct 14, 8:00 am, Paul McGuire <[EMAIL PROTECTED]> wrote:
> On Oct 12, 5:58 am, Florian Lindner <[EMAIL PROTECTED]> wrote:
>
>
>
> > Hello,
> > can I determine somehow if the iteration on a list of values is the last
> > iteration?
>
> > Example:
>
> > for i in [1, 2, 3]:
> >if last_iteration:
> >   print i*i
> >else:
> >   print i
>
> > that would print
>
> > 1
> > 2
> > 9
>
> > Can this be acomplished somehow?
>
> > Thanks,
>
> > Florian
>
> Maybe it's a leftover from my C++ days, but I find the iteration-based
> solutions the most appealing.  This is a refinement of the previous
> post by Diez Roggisch.  The test method seems to match your desired
> idiom pretty closely:
>
> def signal_last(lst):
> last2 = None
> it = iter(lst)
> try:
> last = it.next()
> except StopIteration:
> last = None
> for last2 in it:
> yield False, last
> last = last2
> yield True, last

This yields a value when the iterator is empty, which Diez's solution
didn't. Logically, there is no 'last' element in an empty sequence,
and it's obscure to add one. Peter Otten's improvement to Diez's code
looks the best to me: simple, readable and correct.

--
Paul Hankin

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Re: Last iteration?

[Paul McGuire]

On Oct 12, 5:58 am, Florian Lindner <[EMAIL PROTECTED]> wrote:
> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
>
> Example:
>
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
>
> that would print
>
> 1
> 2
> 9
>
> Can this be acomplished somehow?
>
> Thanks,
>
> Florian

Maybe it's a leftover from my C++ days, but I find the iteration-based
solutions the most appealing.  This is a refinement of the previous
post by Diez Roggisch.  The test method seems to match your desired
idiom pretty closely:

def signal_last(lst):
last2 = None
it = iter(lst)
try:
last = it.next()
except StopIteration:
last = None
for last2 in it:
yield False, last
last = last2
yield True, last

def test(t):
for isLast, item in signal_last(t):
if isLast:
print "...and the last item is", item
else:
print item

test("ABC")
test([])
test([1,2,3])

Prints:

A
B
...and the last item is C
...and the last item is None
1
2
...and the last item is 3

-- Paul

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Re: Last iteration?

[Robin Kåveland]

On Oct 12, 12:58 pm, Florian Lindner <[EMAIL PROTECTED]> wrote:
> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
>
> Example:
>
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
>
> that would print
>
> 1
> 2
> 9
>
> Can this be acomplished somehow?
>
> Thanks,
>
> Florian


If you want to do this over a list or a string, I'd just do:

for element in iterable[:-1]: print iterable
print iterable[-1]  * iterable[-1]

No need for it to get more advanced than that :)

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Re: Last iteration?

[Peter Otten]

Diez B. Roggisch wrote:

> Florian Lindner wrote:

>> can I determine somehow if the iteration on a list of values is the
>> last iteration?

>  def last_iter(iterable):
> it = iter(iterable)
> buffer = [it.next()]
> for i in it:
> buffer.append(i)
> old, buffer = buffer[0], buffer[1:]
> yield False, old
> yield True, buffer[0]

This can be simplified a bit since you never have to remember more than on
item:

>>> def mark_last(items):
... items = iter(items)
... last = items.next()
... for item in items:
... yield False, last
... last = item
... yield True, last
...
>>> list(mark_last([]))
[]
>>> list(mark_last([1]))
[(True, 1)]
>>> list(mark_last([1,2]))
[(False, 1), (True, 2)]

Peter
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Re: Last iteration?

[Paul Hankin]

On Oct 12, 2:18 pm, Carsten Haese <[EMAIL PROTECTED]> wrote:
> On Fri, 2007-10-12 at 12:58 +0200, Florian Lindner wrote:
> > Hello,
> > can I determine somehow if the iteration on a list of values is the last
> > iteration?
>
> > Example:
>
> > for i in [1, 2, 3]:
> >if last_iteration:
> >   print i*i
> >else:
> >   print i
>
> > that would print
>
> > 1
> > 2
> > 9
>
> Here's another solution:
>
> mylist = [1,2,3]
> for j,i in reversed(list(enumerate(reversed(mylist:
>   if j==0:
>  print i*i
>   else:
>  print i

Nice! A more 'readable' version is:

mylist = [1,2,3]
for not_last, i in reversed(list(enumerate(reversed(mylist:
  if not_last:
 print i
  else:
 print i * i

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Re: Last iteration?

[Carsten Haese]

On Fri, 2007-10-12 at 12:58 +0200, Florian Lindner wrote:
> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
> 
> Example:
> 
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
> 
> that would print
> 
> 1
> 2
> 9

Here's another solution:

mylist = [1,2,3]
for j,i in reversed(list(enumerate(reversed(mylist:
  if j==0:
 print i*i
  else:
 print i

;)

-- 
Carsten Haese
http://informixdb.sourceforge.net


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Re: Last iteration?

[tasjaevan]

On Oct 12, 11:58 am, Florian Lindner <[EMAIL PROTECTED]> wrote:
> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
>
> Example:
>
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
>
> that would print
>
> 1
> 2
> 9
>
> Can this be acomplished somehow?
>

Another suggestion:

  l = [1, 2, 3]
  for i in l[:-1]: print i
  i = l[-1]
  print i*i


James



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Re: Last iteration?

[Paul Hankin]

On Oct 12, 11:58 am, Florian Lindner <[EMAIL PROTECTED]> wrote:
> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
>
> Example:
>
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i

Yes, either use enumerate or just stop the loop early and deal with
the last element outside the loop.

xs = [1, 2, 3]
for x in xs[:-1]:
print x
print xs[-1] * xs[-1]

--
Paul Hankin

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Re: Last iteration?

[Diez B. Roggisch]

Florian Lindner wrote:

> Hello,
> can I determine somehow if the iteration on a list of values is the last
> iteration?
> 
> Example:
> 
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
> 
> that would print
> 
> 1
> 2
> 9
> 
> 
> Can this be acomplished somehow?

 def last_iter(iterable):
it = iter(iterable)
buffer = [it.next()]
for i in it:
buffer.append(i)
old, buffer = buffer[0], buffer[1:]
yield False, old
yield True, buffer[0]

  

for last, i in last_iter(xrange(4)):
   if last:
   print i*i
   else:
   print i



Diez
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Re: Last iteration?

[Stefan Behnel]

Florian Lindner wrote:
> can I determine somehow if the iteration on a list of values is the last
> iteration?
> 
> Example:
> 
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
> 
> that would print
> 
> 1
> 2
> 9
> 
> 
> Can this be acomplished somehow?

You could do this:

  l = [1,2,3]
  s = len(l) - 1
  for i, item in enumerate(l): # Py 2.4
  if i == s:
  print item*item
  else:
  print item

Or, you could look one step ahead:

   l = [1,2,3]
   next = l[0]
   for item in l[1:]:
   print next
   next = item
   print next * next

Stefan
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RE: Last iteration?

[Andreas Tawn]

> Hello,
> can I determine somehow if the iteration on a list of values 
> is the last
> iteration?
> 
> Example:
> 
> for i in [1, 2, 3]:
>if last_iteration:
>   print i*i
>else:
>   print i
> 
> that would print
> 
> 1
> 2
> 9

Something like:

myList = [1, 2, 3]
for i, j in enumerate(myList):
if i == len(myList)-1:
print j*j
else:
print j

Cheers,

Andreas Tawn
Lead Technical Artist
Ubisoft Reflections
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Last iteration?

[Florian Lindner]

Hello,
can I determine somehow if the iteration on a list of values is the last
iteration?

Example:

for i in [1, 2, 3]:
   if last_iteration:
  print i*i
   else:
  print i

that would print

1
2
9


Can this be acomplished somehow?

Thanks,

Florian
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