blocks|key|1081881|text|l[index]+if+index+<+len(l)+else+default|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1081882|要支持负指数，我们可以使用：|unstyled|1081883|l[index]+if+-len(l)+<=+index+<+len(l)+else+default|1081884|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>l[index] if index &lt; len(l) else default
</code></pre>

To support negative indices we can use:

<pre><code>l[index] if -len(l) &lt;= index &lt; len(l) else default
</code></pre>

blocks|key|1081912|text|try:
+++a+=+b[n]
except+IndexError:
+++a+=+default|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1081913|编辑:我删除了对TypeError的检查--也许让调用者来处理这件事更好。|unstyled|1081914|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>try:
 a = b[n]
except IndexError:
 a = default
</code></pre>

Edit: I removed the check for TypeError - probably better to let the caller handle this.

blocks|key|1081948|text|(a[n:]%2B[default])[0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1081949|随着a变得越来越大，这可能会更好|unstyled|offset|length|style|CODE|1081950|(a[n:n%2B1]%2B[default])[0]|1081951|这之所以有效，是因为如果a[n:]是一个空列表，如果n+=>+len(a)|1081952|下面是一个使用range(5)的示例|1081953|>>>+range(5)[3:4]
[3]
>>>+range(5)[4:5]
[4]
>>>+range(5)[5:6]
[]
>>>+range(5)[6:7]
[]|1081954|和完整的表达式|1081955|>>>+(range(5)[3:4]%2B[999])[0]
3
>>>+(range(5)[4:5]%2B[999])[0]
4
>>>+(range(5)[5:6]%2B[999])[0]
999
>>>+(range(5)[6:7]%2B[999])[0]
999|1081956|entityMap^0|0|2|1|0|0|C|5|Q|B|0|7|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Z|8|@$G|10|H|11|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|13|8|@$G|14|H|15|I|J]|$G|16|H|17|I|J]]|9|@]|A|$]]|$1|O|3|P|5|F|7|18|8|@$G|19|H|1A|I|J]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|S|3|T|5|F|7|1C|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$B|C]]|$1|W|3|-4|5|F|7|1E|8|@]|9|@]|A|$]]]|X|$]]

<pre><code>(a[n:]+[default])[0]
</code></pre>

This is probably better as <code>a</code> gets larger 

<pre><code>(a[n:n+1]+[default])[0]
</code></pre>

This works because if <code>a[n:]</code> is an empty list if <code>n =&gt; len(a)</code>

Here is an example of how this works with <code>range(5)</code>

<pre><code>&gt;&gt;&gt; range(5)[3:4]
[3]
&gt;&gt;&gt; range(5)[4:5]
[4]
&gt;&gt;&gt; range(5)[5:6]
[]
&gt;&gt;&gt; range(5)[6:7]
[]
</code></pre>

And the full expression

<pre><code>&gt;&gt;&gt; (range(5)[3:4]+[999])[0]
3
&gt;&gt;&gt; (range(5)[4:5]+[999])[0]
4
&gt;&gt;&gt; (range(5)[5:6]+[999])[0]
999
&gt;&gt;&gt; (range(5)[6:7]+[999])[0]
999
</code></pre>

blocks|key|1081999|text|我刚刚发现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082000|next(iter(myList),+5)|code-block|syntax|javascript|1082001|iter(l)在myList上返回迭代器，next()使用迭代器的第一个元素，并引发StopIteration错误，除非使用默认值调用，此处是第二个参数5|offset|length|style|CODE|1082002|只有当你想要第一个元素时才有效，这是你的例子中的情况，但不是在你的问题文本中，所以...|1082003|此外，它不需要在内存中创建临时列表，并且它适用于任何类型的迭代器，即使它没有名称(请参阅熊奇米奥夫对gruszczy回答的评论)|1082004|entityMap^0|0|0|0|7|8|6|L|6|16|D|24|1|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

Just discovered that : 

<pre><code>next(iter(myList), 5)
</code></pre>

<code>iter(l)</code> returns an iterator on <code>myList</code>, <code>next()</code> consumes the first element of the iterator, and raises a <code>StopIteration</code> error except if called with a default value, which is the case here, the second argument, <code>5</code>

This only works when you want the 1st element, which is the case in your example, but not in the text of you question, so...

Additionally, it does not need to create temporary lists in memory and it works for any kind of iterable, even if it does not have a name (see Xiong Chiamiov's comment on gruszczy's answer)

blocks|key|1296874|text|(L[n:n%2B1]+or+[somedefault])[0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1296875|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>(L[n:n+1] or [somedefault])[0]
</code></pre>

blocks|key|1297005|text|将@Joachim与上面的代码结合起来，您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1297006|next(iter(my_list[index:index%2B1]),+default)|code-block|syntax|javascript|1297007|示例：|1297008|next(iter(range(10)[8:9]),+11)
8
>>>+next(iter(range(10)[12:13]),+11)
11|1297009|1297010|或者，也许更清楚，但没有len|offset|length|style|CODE|1297011|my_list[index]+if+my_list[index:index+%2B+1]+else+default|1297012|entityMap^0|0|0|0|0|0|C|3|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|10|8|@$N|11|O|12|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

Combining @Joachim's with the above, you could use
<pre><code>next(iter(my_list[index:index+1]), default)
</code></pre>
Examples:
<pre><code>next(iter(range(10)[8:9]), 11)
8
&gt;&gt;&gt; next(iter(range(10)[12:13]), 11)
11
</code></pre>
<hr />
Or, maybe more clear, but without the <code>len</code>
<pre><code>my_list[index] if my_list[index:index + 1] else default
</code></pre>

blocks|key|1297124|text|一个简单的解决方案是使用enumerate创建一个字典，并照常使用.get()，如下所示|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1297125|+dict(enumerate(l)).get(7,+my_default)|code-block|syntax|javascript|1297126|entityMap^0|X|6|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|$]]

A cheap solution is to really make a dict with enumerate and use <code>.get()</code> as usual, like

<pre><code> dict(enumerate(l)).get(7, my_default)
</code></pre>

blocks|key|1082081|text|使用Python3.4的contextlib.suppress(exceptions)构建一个类似于getattr()的getitem()方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1082082|import+contextlib

def+getitem(iterable,+index,+default=None):
++++"""Return+iterable[index]+or+default+if+IndexError+is+raised."""
++++with+contextlib.suppress(IndexError):
++++++++return+iterable[index]
++++return+default|code-block|syntax|javascript|1082083|entityMap^0|C|V|1E|9|1O|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Using Python 3.4's <code>contextlib.suppress(exceptions)</code> to build a <code>getitem()</code> method similar to <code>getattr()</code>.

<pre><code>import contextlib

def getitem(iterable, index, default=None):
 """Return iterable[index] or default if IndexError is raised."""
 with contextlib.suppress(IndexError):
 return iterable[index]
 return default
</code></pre>

blocks|key|1082162|text|尽管这不是一行解决方案，但您可以使用默认值定义函数，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082163|def+get_val(myList,+idx,+default=5):
++++try:
++++++++return+myList[idx]
++++except+IndexError:
++++++++return+default|code-block|syntax|javascript|1082164|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|$]]

Althought this is not a one-liner solution, you can define a function with a default value like so:
<pre class="lang-py prettyprint-override"><code>def get_val(myList, idx, default=5):
 try:
 return myList[idx]
 except IndexError:
 return default
</code></pre>

blocks|key|1082118|text|在通读答案之后，我将使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082119|(L[n:]+or+[somedefault])[0]|code-block|syntax|javascript|1082120|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|$]]

After reading through the answers, I'm going to use:
<pre class="lang-py prettyprint-override"><code>(L[n:] or [somedefault])[0]
</code></pre>

I'm looking for an equivalent in python of <code>dictionary.get(key, default)</code> for lists. Is there any one liner idiom to get the nth element of a list or a default value if not available?

For example, given a list myList I would like to get <code>myList[0]</code>, or 5 if<code>myList</code> is an empty list.

Thanks.

How to get the nth element of a python list or a default if not available

我正在寻找在dictionary.get(key, default)的python中用于列表的等价物。有没有任何一种线性习惯用法来获取列表的第n个元素，或者如果不可用，则获取默认值？例如，给定一个列表myList，我希望获取myList[0]，如果myList为空列表，则为5。谢谢。

问如何获取python列表的第n个元素，如果不可用，则获取默认值
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回答 10

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问如何获取python列表的第n个元素，如果不可用，则获取默认值EN

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