blocks|key|599865|text|要跳过Python中的第一个元素，您可以简单地编写|type|unstyled|depth|inlineStyleRanges|entityRanges|data|599866|for+car+in+cars[1:]:
++++#+Do+What+Ever+you+want|code-block|syntax|javascript|599867|或者跳过最后一个元素|599868|for+car+in+cars[:-1]:
++++#+Do+What+Ever+you+want|599869|您可以将此概念用于任何sequence+(但不适用于任何iterable+)。|offset|length|style|CODE|599870|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/glossary.html#term-sequence|1|https://docs.python.org/3/glossary.html#term-iterable^0|0|0|0|0|B|8|S|8|B|8|0|S|8|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@$M|19|N|1A|1|1B]|$M|1C|N|1D|1|1E]]|A|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

To skip the first element in Python you can simply write
<pre><code>for car in cars[1:]:
 # Do What Ever you want
</code></pre>
or to skip the last elem
<pre><code>for car in cars[:-1]:
 # Do What Ever you want
</code></pre>
You can use this concept for any <a href="https://docs.python.org/3/glossary.html#term-sequence" rel="noreferrer"><code>sequence</code></a> (not for any <a href="https://docs.python.org/3/glossary.html#term-iterable" rel="noreferrer"><code>iterable</code></a> though).

blocks|key|384073|text|其他答案仅适用于序列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|384074|对于任何可迭代对象，跳过第一项：|384075|itercars+=+iter(cars)
next(itercars)
for+car+in+itercars:
++++#+do+work|code-block|syntax|javascript|384076|如果你想跳过最后一个，你可以这样做：|384077|itercars+=+iter(cars)
#+add+'next(itercars)'+here+if+you+also+want+to+skip+the+first
prev+=+next(itercars)
for+car+in+itercars:
++++#+do+work+on+'prev'+not+'car'
++++#+at+end+of+loop:
++++prev+=+car
#+now+you+can+do+whatever+you+want+to+do+to+the+last+one+on+'prev'|384078|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The other answers only work for a sequence.

For any iterable, to skip the first item:

<pre><code>itercars = iter(cars)
next(itercars)
for car in itercars:
 # do work
</code></pre>

If you want to skip the last, you could do:

<pre><code>itercars = iter(cars)
# add 'next(itercars)' here if you also want to skip the first
prev = next(itercars)
for car in itercars:
 # do work on 'prev' not 'car'
 # at end of loop:
 prev = car
# now you can do whatever you want to do to the last one on 'prev'
</code></pre>

blocks|key|600154|text|跳过第一个项目的最佳方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|600155|from+itertools+import+islice
for+car+in+islice(cars,+1,+None):
++++pass
++++#+do+something|code-block|syntax|javascript|600156|在本例中，调用islice的起点为1，终点为None，表示iterable的结束。|offset|length|style|CODE|600157|为了能够跳过iterable末尾的项目，您需要知道它的长度(对于list总是可能的，但不一定对于您可以迭代的所有内容)。例如，islice(cars,+1,+len(cars)-1)将跳过cars中的第一项和最后一项。|600158|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/itertools.html#itertools.islice|1|https://docs.python.org/3/glossary.html#term-iterable^0|0|0|7|6|H|1|M|4|T|8|7|6|0|T|8|1|0|6|8|1R|S|2M|4|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@$I|19|J|1A|1|1B]|$I|1C|J|1D|1|1E]]|A|$]]|$1|M|3|N|5|6|7|1F|8|@$I|1G|J|1H|K|L]|$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

The best way to skip the first item(s) is:
<pre><code>from itertools import islice
for car in islice(cars, 1, None):
 pass
 # do something
</code></pre>
<a href="https://docs.python.org/3/library/itertools.html#itertools.islice" rel="nofollow noreferrer"><code>islice</code></a> in this case is invoked with a start-point of <code>1</code>, and an end point of <code>None</code>, signifying the end of the <a href="https://docs.python.org/3/glossary.html#term-iterable" rel="nofollow noreferrer"><code>iterable</code></a>.
To be able to skip items from the end of an <code>iterable</code>, you need to know its length (always possible for a list, but not necessarily for everything you can iterate on). for example, <code>islice(cars, 1, len(cars)-1)</code> will skip the first and last items in <code>cars</code>.

blocks|key|384232|text|下面是一个更通用的生成器函数，它可以跳过从迭代体的开头和结尾开始的任意数量的项：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|384233|def+skip(iterable,+at_start=0,+at_end=0):
++++it+=+iter(iterable)
++++for+x+in+itertools.islice(it,+at_start):
++++++++pass
++++queue+=+collections.deque(itertools.islice(it,+at_end))
++++for+x+in+it:
++++++++queue.append(x)
++++++++yield+queue.popleft()|code-block|syntax|javascript|384234|示例用法：|384235|>>>+list(skip(range(10),+at_start=2,+at_end=2))
[2,+3,+4,+5,+6,+7]|384236|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is a more general generator function that skips any number of items from the beginning and end of an iterable:

<pre><code>def skip(iterable, at_start=0, at_end=0):
 it = iter(iterable)
 for x in itertools.islice(it, at_start):
 pass
 queue = collections.deque(itertools.islice(it, at_end))
 for x in it:
 queue.append(x)
 yield queue.popleft()
</code></pre>

Example usage:

<pre><code>&gt;&gt;&gt; list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
</code></pre>

blocks|key|383977|text|for+item+in+do_not_use_list_as_a_name[1:-1]:
++++#...do+whatever|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|383978|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 item in do_not_use_list_as_a_name[1:-1]:
 #...do whatever
</code></pre>

blocks|key|600449|text|示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|600450|mylist=['one','two','three','four','five']
for+i+in+mylist[1:]:
+++print(i)|code-block|syntax|javascript|600451|在python中，索引从0开始，我们可以使用切片操作符在迭代中进行操作。|600452|for+i+in+range(1,-1):|600453|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Example:
<pre><code>mylist=['one','two','three','four','five']
for i in mylist[1:]:
 print(i)
</code></pre>
In python index start from 0, We can use slicing operator to make manipulations in iteration.
<pre><code>for i in range(1,-1):
</code></pre>

blocks|key|600420|text|这是我的首选。它不需要在循环中添加太多内容，并且只使用内置工具。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|600421|始发地：|600422|for+item+in+my_items:
++do_something(item)|code-block|syntax|javascript|600423|至：|600424|for+i,+item+in+enumerate(my_items):
++if+i+==+0:
++++continue
++do_something(item)|600425|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's my preferred choice. It doesn't require adding on much to the loop, and uses nothing but built in tools.

Go from:

<pre class="lang-py prettyprint-override"><code>for item in my_items:
 do_something(item)
</code></pre>

to:

<pre class="lang-py prettyprint-override"><code>for i, item in enumerate(my_items):
 if i == 0:
 continue
 do_something(item)
</code></pre>

blocks|key|384150|text|嗯，你的语法从一开始就不是真正的Python。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|384151|Python中的迭代是在容器的内容上进行的(从技术上讲，它是在迭代器上进行的)，其语法是for+item+in+container。在本例中，容器是cars列表，但是您希望跳过第一个和最后一个元素，因此这意味着cars[1:-1]+(python列表是从零开始的，负数从末尾开始计数，:是分片语法。|offset|length|style|CODE|384152|所以你想|384153|for+c+in+cars[1:-1]:
++++do+something+with+c|code-block|syntax|javascript|384154|entityMap^0|0|18|L|22|4|2X|A|3Y|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|11|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|P|$]]

Well, your syntax isn't really Python to begin with.

Iterations in Python are over he contents of containers (well, technically it's over iterators), with a syntax <code>for item in container</code>. In this case, the container is the <code>cars</code> list, but you want to skip the first and last elements, so that means <code>cars[1:-1]</code> (python lists are zero-based, negative numbers count from the end, and <code>:</code> is slicing syntax.

So you want 

<pre><code>for c in cars[1:-1]:
 do something with c
</code></pre>

blocks|key|600202|text|基于@SvenMarnach的答案，但有点简单，没有使用deque|type|unstyled|depth|inlineStyleRanges|entityRanges|data|600203|>>>+def+skip(iterable,+at_start=0,+at_end=0):
++++it+=+iter(iterable)
++++it+=+itertools.islice(it,+at_start,+None)
++++it,+it1+=+itertools.tee(it)
++++it1+=+itertools.islice(it1,+at_end,+None)
++++return+(next(it)+for+_+in+it1)

>>>+list(skip(range(10),+at_start=2,+at_end=2))
[2,+3,+4,+5,+6,+7]
>>>+list(skip(range(10),+at_start=2,+at_end=5))
[2,+3,+4]|code-block|syntax|javascript|600204|还要注意，根据我的timeit结果，这比deque解决方案稍微快一点|offset|length|style|CODE|600205|>>>+iterable=xrange(1000)
>>>+stmt1="""
def+skip(iterable,+at_start=0,+at_end=0):
++++it+=+iter(iterable)
++++it+=+itertools.islice(it,+at_start,+None)
++++it,+it1+=+itertools.tee(it)
++++it1+=+itertools.islice(it1,+at_end,+None)
++++return+(next(it)+for+_+in+it1)
list(skip(iterable,2,2))
++++"""
>>>+stmt2="""
def+skip(iterable,+at_start=0,+at_end=0):
++++it+=+iter(iterable)
++++for+x+in+itertools.islice(it,+at_start):
++++++++pass
++++queue+=+collections.deque(itertools.islice(it,+at_end))
++++for+x+in+it:
++++++++queue.append(x)
++++++++yield+queue.popleft()
list(skip(iterable,2,2))
++++++++"""
>>>+timeit.timeit(stmt+=+stmt1,+setup='from+__main__+import+iterable,+skip,+itertools',+number+=+10000)
2.0313770640908047
>>>+timeit.timeit(stmt+=+stmt2,+setup='from+__main__+import+iterable,+skip,+itertools,+collections',+number+=+10000)
2.9903135454296716|600206|entityMap^0|0|0|9|6|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Based on @SvenMarnach 's Answer, but bit simpler and without using deque

<pre><code>&gt;&gt;&gt; def skip(iterable, at_start=0, at_end=0):
 it = iter(iterable)
 it = itertools.islice(it, at_start, None)
 it, it1 = itertools.tee(it)
 it1 = itertools.islice(it1, at_end, None)
 return (next(it) for _ in it1)

&gt;&gt;&gt; list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
&gt;&gt;&gt; list(skip(range(10), at_start=2, at_end=5))
[2, 3, 4]
</code></pre>

Also Note, based on my <code>timeit</code> result, this is marginally faster than the deque solution

<pre><code>&gt;&gt;&gt; iterable=xrange(1000)
&gt;&gt;&gt; stmt1="""
def skip(iterable, at_start=0, at_end=0):
 it = iter(iterable)
 it = itertools.islice(it, at_start, None)
 it, it1 = itertools.tee(it)
 it1 = itertools.islice(it1, at_end, None)
 return (next(it) for _ in it1)
list(skip(iterable,2,2))
 """
&gt;&gt;&gt; stmt2="""
def skip(iterable, at_start=0, at_end=0):
 it = iter(iterable)
 for x in itertools.islice(it, at_start):
 pass
 queue = collections.deque(itertools.islice(it, at_end))
 for x in it:
 queue.append(x)
 yield queue.popleft()
list(skip(iterable,2,2))
 """
&gt;&gt;&gt; timeit.timeit(stmt = stmt1, setup='from __main__ import iterable, skip, itertools', number = 10000)
2.0313770640908047
&gt;&gt;&gt; timeit.timeit(stmt = stmt2, setup='from __main__ import iterable, skip, itertools, collections', number = 10000)
2.9903135454296716
</code></pre>

blocks|key|384027|text|如果cars是一个序列，你可以这样做|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|384028|for+car+in+cars[1:-1]:
++++pass|code-block|syntax|javascript|384029|entityMap^0|2|4|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|$]]

If <code>cars</code> is a sequence you can just do 

<pre><code>for car in cars[1:-1]:
 pass
</code></pre>

blocks|key|600219|text|另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|600220|for+idx,+car+in+enumerate(cars):
++++#+Skip+first+line.
++++if+not+idx:
++++++++continue
++++#+Skip+last+line.
++++if+idx+%2B+1+==+len(cars):
++++++++continue
++++#+Real+code+here.
++++print+car|code-block|syntax|javascript|600221|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|$]]

An alternative method:

<pre><code>for idx, car in enumerate(cars):
 # Skip first line.
 if not idx:
 continue
 # Skip last line.
 if idx + 1 == len(cars):
 continue
 # Real code here.
 print car
</code></pre>

blocks|key|600333|text|more_itertools项目扩展了itertools.islice以处理负索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|600334|示例|600335|import+more_itertools+as+mit

iterable+=+'ABCDEFGH'
list(mit.islice_extended(iterable,+1,+-1))
#+Out:+['B',+'C',+'D',+'E',+'F',+'G']|code-block|syntax|javascript|600336|因此，您可以优雅地在iterable的第一项和最后一项之间应用它的切片元素：|600337|for+car+in+mit.islice_extended(cars,+1,+-1):
++++#+do+something|600338|entityMap|0|LINK|mutability|MUTABLE|url|https://more-itertools.readthedocs.io/en/latest/^0|0|E|J|G|0|E|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

The <a href="https://more-itertools.readthedocs.io/en/latest/" rel="nofollow noreferrer"><code>more_itertools</code></a> project extends <code>itertools.islice</code> to handle negative indices.

Example

<pre><code>import more_itertools as mit

iterable = 'ABCDEFGH'
list(mit.islice_extended(iterable, 1, -1))
# Out: ['B', 'C', 'D', 'E', 'F', 'G']
</code></pre>

Therefore, you can elegantly apply it slice elements between the first and last items of an iterable:

<pre><code>for car in mit.islice_extended(cars, 1, -1):
 # do something
</code></pre>

blocks|key|384400|text|我是这样做的，尽管它看起来像一个黑客，但每次都能奏效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|384401|ls_of_things+=+['apple',+'car',+'truck',+'bike',+'banana']
first+=+0
last+=+len(ls_of_things)
for+items+in+ls_of_things:
++++if+first+==+0
++++++++first+=+first+%2B+1
++++++++pass
++++elif+first+==+last+-+1:
++++++++break
++++else:
++++++++do_stuff
++++++++first+=+first+%2B+1
++++++++pass|code-block|syntax|javascript|384402|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|$]]

I do it like this, even though it looks like a hack it works every time:

<pre><code>ls_of_things = ['apple', 'car', 'truck', 'bike', 'banana']
first = 0
last = len(ls_of_things)
for items in ls_of_things:
 if first == 0
 first = first + 1
 pass
 elif first == last - 1:
 break
 else:
 do_stuff
 first = first + 1
 pass
</code></pre>

In python, How do I do something like:

<pre><code>for car in cars:
 # Skip first and last, do work for rest
</code></pre>

Skip first entry in for loop in python?

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。

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在python中，我如何做一些类似的事情：for car in cars:   # Skip first and last, do work for rest

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