Last iteration?
Florian Lindner
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
Andreas Tawn
> 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
Stefan Behnel
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
Diez B. Roggisch
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
Paul Hankin
> 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
tasj...@gmail.com
Another suggestion:
l = [1, 2, 3]
for i in l[:-1]: print i
i = l[-1]
print i*i
James
Carsten Haese
> 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
Paul Hankin
> 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
--
Paul Hankin
Peter Otten
> 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
Robin Kåveland
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 :)
Paul McGuire
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
Paul Hankin
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
Diez B. Roggisch
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
Paul McGuire
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
Peter Otten
> 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
Bruno Desthuilliers
At least something sensible !-)
Raymond Hettinger
> > 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
Raymond Hettinger
> "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
Peter Otten
> [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
Paul Hankin
def lastdetector(iterable):
t, u = tee(iterable)
return izip(chain(imap(lambda x: False, islice(u, 1, None)),
[True]), t)
--
Paul Hankin
Raymond Hettinger
> 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
Paul Rubin
> 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 <http://bugs.python.org/issue1673203>.
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.
Hendrik van Rooyen
> 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
Paul Hankin
Perhaps, but I think you'll need a better use-case than this :)
Actually, would a c-version be much faster?
--
Paul Hankin
Michael J. Fromberger
Raymond Hettinger <pyt...@rcn.com> 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 marker -- and possibly also
__init__ if you need some additional setup.
Cheers,
-M
--
Michael J. Fromberger | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA
Gabriel Genellina
<Michael.J....@Clothing.Dartmouth.EDU> 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