blocks|key|2622795|text|您可以使用enumerate()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2622796|for+i,+obj+in+enumerate(object_list):
++++do_something
++++if+i+==+0:
++++++++do_something_else
++++if+i+==+len(object_list)+-+1:
++++++++do_another_thing|code-block|syntax|javascript|2622797|但是，与其在每次迭代中检查您正在处理的对象，也许这样做会更好：|2622798|def+do_with_list(object_list):
++++for+obj+in+object_list:
++++++++do_something(obj)
++++do_something_else(object_list[0])
++++do_another_thing(object_list[-1])|2622799|假设您有一个包含100个对象的列表，然后您进行了198个不必要的比较，因为当前元素不能是列表中的第一个或最后一个元素。|2622800|但这取决于语句是否必须以特定的顺序执行，以及它们正在做什么。|2622801|2622802|顺便说一句。不要隐藏object，它已经是Python中的一个标识符;)|2622803|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/functions.html#enumerate^0|5|B|5|B|0|0|0|0|0|0|0|0|A|6|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

You can use <a href="http://docs.python.org/library/functions.html#enumerate" rel="noreferrer"><code>enumerate()</code></a>:

<pre><code>for i, obj in enumerate(object_list):
 do_something
 if i == 0:
 do_something_else
 if i == len(object_list) - 1:
 do_another_thing
</code></pre>

But instead of checking in every iteration which object you are dealing with, maybe something like this is better:

<pre><code>def do_with_list(object_list):
 for obj in object_list:
 do_something(obj)
 do_something_else(object_list[0])
 do_another_thing(object_list[-1])
</code></pre>

Imagine you have a list of 100 objects, then you make 198 unnecessary comparisons, because the current element cannot be the first or last element in the list.

But it depends on whether the statements have to be executed in a certain order and what they are doing.

<hr>

Btw. don't shadow <code>object</code>, it is already an identifier in Python ;)

blocks|key|1576702|text|for+index,+obj+in+enumerate(object_list):
++++do_something
++++if+index+==+0:
++++++++do_something_else
++++if+index+==+len(object_list)-1:
++++++++do_another_thing|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1576703|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>for index, obj in enumerate(object_list):
 do_something
 if index == 0:
 do_something_else
 if index == len(object_list)-1:
 do_another_thing
</code></pre>

blocks|key|4030932|text|你只需要使用一个计数器，不需要range：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4030933|假设您的计数器是cpt。|offset|length|style|CODE|4030934|if+cpt+==+0:
++print+"first"

if+cpt+==+len(object_list)+-+1:
++print+"last"|code-block|syntax|javascript|4030935|编辑：枚举解决方案可能更优雅。|BOLD|4030936|entityMap^0|0|8|3|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|V|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|W|8|@$D|X|E|Y|F|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

You just have to use a counter, no need for range :

Let's say your counter is <code>cpt</code>.

<pre><code>if cpt == 0:
 print "first"

if cpt == len(object_list) - 1:
 print "last"
</code></pre>

edit : the enumerate solution may be more elegant.

blocks|key|4030987|text|在迭代中，你可以使i=0，然后i%2B%2B。所以如果是第一次迭代，如果I+==+list.size()-1，那么就是最后一次迭代。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4030988|对象必须有像indexOf(|4030989|)这样的smth，它返回元素在列表中的位置。if+0-比第一次迭代，if+size()-1比最后一次迭代。但它很贵。|ordered-list-item|4030990|4030991|4030992|使用枚举。请参阅琥珀或菲利克斯的帖子。|4030993|4030994|4030995|将当前元素与列表和列表-1进行比较。在第一种情况下-第一次迭代，在最后一种情况-最后一次迭代。它也很贵。|4030996|4030997|所以我会选择1或3。|4030998|PS:+list.size()当然是len(列表)。|4030999|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|10|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|13|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|15|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|16|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|U|$]]

<ol>
<li>You can make i = 0 and then i++ in iterations. So if i==0 it is first iteration, if i == list.size()-1 than it is last iteration. </li>
<li>There must be smth like indexOf(object), which returns position of element in list. if 0 - than first iteration, if size()-1 than last. But it is expensive. </li>
<li>use enumerate. See Amber or Felix posts. </li>
<li>compare current element to list[0] and to list[-1]. in first case - first iteration, in last case - last iteration. It is expensive too. </li>
</ol>

So i'd choose 1 or 3. 
PS: list.size() is len(list) of course.

blocks|key|2622909|text|li+=+iter(object_list)

obj+=+next(li)

do_first_thing_with(obj)

while+True:
++++try:
++++++++do_something_with(obj)
++++++++obj+=+next(li)
++++except+StopIteration:
++++++++do_final_thing_with(obj)
++++++++break|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2622910|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>li = iter(object_list)

obj = next(li)

do_first_thing_with(obj)

while True:
 try:
 do_something_with(obj)
 obj = next(li)
 except StopIteration:
 do_final_thing_with(obj)
 break
</code></pre>

blocks|key|827805|text|first+=+True
for+o+in+object_list:
++++do_something(o)
++++if+first:
++++++++first+=+False
++++++++do_something_with_first(o)
if+'o'+in+locals():
++++do_something_with_last(o)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|827806|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>first = True
for o in object_list:
 do_something(o)
 if first:
 first = False
 do_something_with_first(o)
if 'o' in locals():
 do_something_with_last(o)
</code></pre>

Assuming I have object_list list which contains objects.

I want to check if my current iteration is is at the first or the last.

<pre><code>for object in object_list:
 do_something
 if first_indexed_element:
 do_something_else
 if last_indexed_element:
 do_another_thing
</code></pre>

How can this be achieved? I know that I can use range and count index but if feels clumsy.

Regards

Python check first and last index of a list

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我有包含对象的object_list列表。我想检查我当前的迭代是在第一次还是最后一次。for object in object_list:    do_something    if first_indexed_element:        do_something_else    if last_indexed...

问Python检查列表的第一个和最后一个索引
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