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Last iteration?

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
Oct 12 '07 #1
23 16805
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
Oct 12 '07 #2
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
Oct 12 '07 #3
On Oct 12, 11:58 am, Florian Lindner <Florian.Lind...@xgm.dewrote:
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

Oct 12 '07 #4
On Oct 12, 11:58 am, Florian Lindner <Florian.Lind...@xgm.dewrote:
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

Oct 12 '07 #5
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
Oct 12 '07 #6
On Oct 12, 2:18 pm, Carsten Haese <cars...@uniqsys.comwrote:
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

Oct 12 '07 #7
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
Oct 12 '07 #8
On Oct 12, 12:58 pm, Florian Lindner <Florian.Lind...@xgm.dewrote:
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 :)

Oct 13 '07 #9
On Oct 12, 5:58 am, Florian Lindner <Florian.Lind...@xgm.dewrote:
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

Oct 14 '07 #10
On Oct 14, 8:00 am, Paul McGuire <pt...@austin.rr.comwrote:
On Oct 12, 5:58 am, Florian Lindner <Florian.Lind...@xgm.dewrote:
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

Oct 14 '07 #11
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
Oct 14 '07 #12
On Oct 14, 5:58 am, Paul Hankin <paul.han...@gmail.comwrote:
On Oct 14, 8:00 am, Paul McGuire <pt...@austin.rr.comwrote:
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
Oct 14 '07 #13
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
Oct 15 '07 #14
Paul Hankin a écrit :
On Oct 12, 11:58 am, Florian Lindner <Florian.Lind...@xgm.dewrote:
>>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 !-)
Oct 16 '07 #15
[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

Oct 17 '07 #16
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

Oct 17 '07 #17
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
Oct 17 '07 #18
On Oct 17, 8:16 am, Raymond Hettinger <pyt...@rcn.comwrote:
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)

--
Paul Hankin

Oct 17 '07 #19
[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

Oct 17 '07 #20
"Raymond Hettinger" <pyt...cn.comwrote:
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

Oct 18 '07 #21
On Oct 17, 11:45 pm, Raymond Hettinger <pyt...@rcn.comwrote:
[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?

--
Paul Hankin

Oct 18 '07 #22
In article <11*********************@e34g2000pro.googlegroups. com>,
Raymond Hettinger <py****@rcn.comwrote:
[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
Oct 19 '07 #23
En Fri, 19 Oct 2007 19:12:49 -0300, Michael J. Fromberger
<Mi******************@Clothing.Dartmouth.EDUescrib ió:
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

Oct 20 '07 #24

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