blocks|key|764589|text|item+=+defaultvalue
for+item+in+my_iter:
++++pass|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|764590|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>item = defaultvalue
for item in my_iter:
 pass
</code></pre>

blocks|key|16568|text|如果您使用的是Python+3.x：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|16569|*_,+last+=+iterator+#+for+a+better+understanding+check+PEP+448
print(last)|code-block|syntax|javascript|16570|如果您使用的是python+2.7：|16571|last+=+next(iterator)
for+last+in+iterator:
++++continue
print+last|16572|旁注：|offset|length|style|BOLD|16573|通常，上面给出的解决方案是您在常规情况下所需要的，但是如果您要处理大量数据，使用大小为1的deque会更有效。(source)|CODE|16574|from+collections+import+deque

#aa+is+an+interator
aa+=+iter('apple')

dd+=+deque(aa,+maxlen=1)
last_element+=+dd.pop()|16575|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/3169701/5658350^0|0|0|0|0|0|3|0|19|5|1K|6|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|17|8|@$M|18|N|19|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@$M|1B|N|1C|O|S]]|9|@$M|1D|N|1E|1|1F]]|A|$]]|$1|T|3|U|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|V|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]]]

If you are using Python 3.x:
<pre><code>*_, last = iterator # for a better understanding check PEP 448
print(last)
</code></pre>
if you are using python 2.7:
<pre><code>last = next(iterator)
for last in iterator:
 continue
print last
</code></pre>
 Side Note: 
Usually, the solution presented above is what you need for regular cases, but if you are dealing with a big amount of data, it's more efficient to use a <code>deque</code> of size 1. (<a href="https://stackoverflow.com/a/3169701/5658350">source</a>)
<pre><code>from collections import deque

#aa is an interator
aa = iter('apple')

dd = deque(aa, maxlen=1)
last_element = dd.pop()
</code></pre>

blocks|key|996235|text|使用大小为1的deque。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|996236|from+collections+import+deque

#aa+is+an+interator
aa+=+iter('apple')

dd+=+deque(aa,+maxlen=1)
last_element+=+dd.pop()|code-block|syntax|javascript|996237|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/library/collections.html#collections.deque^0|7|5|7|5|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use a <a href="https://docs.python.org/library/collections.html#collections.deque" rel="noreferrer"><code>deque</code></a> of size 1.

<pre><code>from collections import deque

#aa is an interator
aa = iter('apple')

dd = deque(aa, maxlen=1)
last_element = dd.pop()
</code></pre>

blocks|key|2405432|text|如果__reversed__可用，可能值得使用它|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2405433|if+hasattr(my_iter,'__reversed__'):
++++last+=+next(reversed(my_iter))
else:
++++for+last+in+my_iter:
++++++++pass|code-block|syntax|javascript|2405434|entityMap^0|2|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Probably worth using <code>__reversed__</code> if it is available

<pre><code>if hasattr(my_iter,'__reversed__'):
 last = next(reversed(my_iter))
else:
 for last in my_iter:
 pass
</code></pre>

blocks|key|2405551|text|就像这样简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2405552|max(enumerate(the_iter))[1]|code-block|syntax|javascript|2405553|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

As simple as:



<pre class="lang-py prettyprint-override"><code>max(enumerate(the_iter))[1]
</code></pre>

blocks|key|16563|text|由于lambda的原因，这不太可能比empty+for循环快，但可能会给其他人一个启发。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|16564|reduce(lambda+x,y:y,my_iter)|code-block|syntax|javascript|16565|如果iter为空，则引发TypeError|16566|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is unlikely to be faster than the empty for loop due to the lambda, but maybe it will give someone else an idea

<pre><code>reduce(lambda x,y:y,my_iter)
</code></pre>

If the iter is empty, a TypeError is raised

blocks|key|16548|text|这就是这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|16549|list(+the_iter+)[-1]|code-block|syntax|javascript|16550|如果迭代的长度真的很长--具体化列表将耗尽内存--那么您真的需要重新考虑设计了。|16551|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

There's this

<pre><code>list( the_iter )[-1]
</code></pre>

If the length of the iteration is truly epic -- so long that materializing the list will exhaust memory -- then you really need to rethink the design.

blocks|key|764600|text|我会使用reversed，除非它只接受序列而不是迭代器，这似乎相当武断。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|764601|不管你怎么做，你都必须遍历整个迭代器。在效率最高的情况下，如果你不再需要迭代器，你可以丢弃所有的值：|764602|for+last+in+my_iter:
++++pass
#+last+is+now+the+last+item|code-block|syntax|javascript|764603|不过，我认为这是一个次优的解决方案。|764604|entityMap^0|4|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I would use <code>reversed</code>, except that it only takes sequences instead of iterators, which seems rather arbitrary. 

Any way you do it, you'll have to run through the entire iterator. At maximum efficiency, if you don't need the iterator ever again, you could just trash all the values:

<pre><code>for last in my_iter:
 pass
# last is now the last item
</code></pre>

I think this is a sub-optimal solution, though.

blocks|key|2405618|text|toolz库提供了一个很好的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2405619|from+toolz.itertoolz+import+last
last(values)|code-block|syntax|javascript|2405620|但添加非核心依赖项可能并不值得，因为只有在这种情况下才使用它。|2405621|entityMap|0|LINK|mutability|MUTABLE|url|https://toolz.readthedocs.io/^0|0|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

The <a href="https://toolz.readthedocs.io/" rel="nofollow noreferrer">toolz</a> library provides a nice solution:

<pre><code>from toolz.itertoolz import last
last(values)
</code></pre>

But adding a non-core dependency might not be worth it for using it only in this case.

blocks|key|2405539|text|类似的代码请看下面的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2405540|http://excamera.com/sphinx/article-islast.html|offset|length|2405541|您可以使用它来选取最后一项，如下所示：|2405542|[(last,+e)+for+(last,+e)+in+islast(the_iter)+if+last]|code-block|syntax|javascript|2405543|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|Z|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

See this code for something similar:

<a href="http://excamera.com/sphinx/article-islast.html" rel="nofollow">http://excamera.com/sphinx/article-islast.html</a>

you might use it to pick up the last item with:

<pre><code>[(last, e) for (last, e) in islast(the_iter) if last]
</code></pre>

blocks|key|996283|text|我只会使用next(reversed(myiter))|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|996284|entityMap^0|5|M|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

I would just use <code>next(reversed(myiter))</code>

blocks|key|16728|text|问题是如何获取迭代器的最后一个元素，但是如果您的迭代器是通过将条件应用于序列来创建的，那么reversed可以通过对序列本身应用reverse来查找反向序列的“第一个”，只查看所需的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|16729|一个人为的例子，|16730|>>>+seq+=+list(range(10))
>>>+last_even+=+next(_+for+_+in+reversed(seq)+if+_+%25+2+==+0)
>>>+last_even
8|code-block|syntax|javascript|16731|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The question is about getting the last element of an iterator, but if your iterator is created by applying conditions to a sequence, then reversed can be used to find the "first" of a reversed sequence, only looking at the needed elements, by applying reverse to the sequence itself.

A contrived example, 

<pre><code>&gt;&gt;&gt; seq = list(range(10))
&gt;&gt;&gt; last_even = next(_ for _ in reversed(seq) if _ % 2 == 0)
&gt;&gt;&gt; last_even
8
</code></pre>

blocks|key|16656|text|或者，对于无限迭代器，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|16657|from+itertools+import+islice+
last+=+list(islice(iterator(),+1000))[-1]+#+where+1000+is+number+of+samples+|code-block|syntax|javascript|16658|我认为它会比deque慢，但它同样快，而且实际上比for循环方法更快(不知何故)|offset|length|style|CODE|16659|entityMap^0|0|0|6|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Alternatively for infinite iterators you can use:

<pre><code>from itertools import islice 
last = list(islice(iterator(), 1000))[-1] # where 1000 is number of samples 
</code></pre>

I thought it would be slower then <code>deque</code> but it's as fast and it's actually faster then for loop method ( somehow )

blocks|key|996223|text|这个问题是错误的，只能导致一个复杂和低效的答案。要获得迭代器，当然要从可迭代的东西开始，这在大多数情况下会提供一种更直接的方式来访问最后一个元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|996224|一旦你从一个迭代器创建了一个迭代器，你就不得不遍历元素，因为这是一个迭代器唯一提供的东西。|996225|因此，最有效、最清晰的方法不是首先创建迭代器，而是使用迭代器的本机访问方法。|996226|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The question is wrong and can only lead to an answer that is complicated and inefficient. 
To get an iterator, you of course start out from something that is iterable, which will in most cases offer a more direct way of accessing the last element. 

Once you create an iterator from an iterable you are stuck in going through the elements, because that is the only thing an iterable provides.

So, the most efficient and clear way is not to create the iterator in the first place but to use the native access methods of the iterable.

What's the best way of getting the last item from an iterator in Python 2.6? For example, say

<pre><code>my_iter = iter(range(5))
</code></pre>

What is the shortest-code / cleanest way of getting <code>4</code> from <code>my_iter</code>?

I could do this, but it doesn't seem very efficient:

<pre><code>[x for x in my_iter][-1]
</code></pre>

Cleanest way to get last item from Python iterator

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在Python2.6中，从迭代器中获取最后一项的最佳方法是什么？例如，假设my_iter = iter(range(5))从my_iter获取4的最短代码/最简洁的方法是什么我可以这样做，但似乎效率不是很高：[x for x in my_iter][-1]

问从Python迭代器获取最后一项的最简洁方法
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