blocks|key|771760|text|归根结底，它可能没有安全的.get方法，因为dict是一个关联集合(值与名称相关联)，在这种集合中，在不抛出异常的情况下检查键是否存在(并返回它的值)效率很低，而避免访问列表元素的异常非常简单(因为len方法非常快)。.get方法允许您查询与名称相关联的值，而不是直接访问字典中的第37个条目(这更像是您对列表的要求)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|771761|当然，您可以很容易地自己实现此功能：|771762|def+safe_list_get+(l,+idx,+default):
++try:
++++return+l[idx]
++except+IndexError:
++++return+default|code-block|syntax|javascript|771763|您甚至可以将其附加到__main__中的__builtins__.list构造函数中，但这将是一个不太普遍的更改，因为大多数代码都不使用它。如果您只是想将其与您自己的代码创建的列表一起使用，则可以简单地将list子类化并添加get方法。|771764|entityMap^0|D|4|M|4|2R|3|31|4|0|0|0|A|8|K|H|2U|4|34|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|P|$]]

Ultimately it probably doesn't have a safe <code>.get</code> method because a <code>dict</code> is an associative collection (values are associated with names) where it is inefficient to check if a key is present (and return its value) without throwing an exception, while it is super trivial to avoid exceptions accessing list elements (as the <code>len</code> method is very fast). The <code>.get</code> method allows you to query the value associated with a name, not directly access the 37th item in the dictionary (which would be more like what you're asking of your list).

Of course, you can easily implement this yourself:

<pre><code>def safe_list_get (l, idx, default):
 try:
 return l[idx]
 except IndexError:
 return default
</code></pre>

You could even monkeypatch it onto the <code>__builtins__.list</code> constructor in <code>__main__</code>, but that would be a less pervasive change since most code doesn't use it. If you just wanted to use this with lists created by your own code you could simply subclass <code>list</code> and add the <code>get</code> method.

blocks|key|771803|text|字典是用来查字典的。询问条目是否存在是有意义的。列表通常是迭代的。通常不会问L10是否存在，而是问L的长度是否为11。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771804|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Dictionaries are for look ups. It makes sense to ask if an entry exists or not. Lists are usually iterated. It isn't common to ask if L[10] exists but rather if the length of L is 11.

blocks|key|986659|text|可能是因为它对列表语义没有多大意义。但是，您可以通过子类化轻松地创建自己的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|986660|class+safelist(list):
++++def+get(self,+index,+default=None):
++++++++try:
++++++++++++return+self.__getitem__(index)
++++++++except+IndexError:
++++++++++++return+default

def+_test():
++++l+=+safelist(range(10))
++++print+l.get(20,+"oops")

if+__name__+==+"__main__":
++++_test()|code-block|syntax|javascript|986661|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|$]]

Probably because it just didn't make much sense for list semantics. However, you can easily create your own by subclassing.

<pre><code>class safelist(list):
 def get(self, index, default=None):
 try:
 return self.__getitem__(index)
 except IndexError:
 return default

def _test():
 l = safelist(range(10))
 print l.get(20, "oops")

if __name__ == "__main__":
 _test()
</code></pre>

blocks|key|771860|text|与使用.get相比，像这样使用lists应该是可以的。只是用法上的不同。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771861|>>>+l+=+[1]
>>>+l[10]+if+10+<+len(l)+else+'fail'
'fail'|code-block|syntax|javascript|771862|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|$]]

Instead of using .get, using like this should be ok for lists. Just a usage difference.

<pre><code>&gt;&gt;&gt; l = [1]
&gt;&gt;&gt; l[10] if 10 &lt; len(l) else 'fail'
'fail'
</code></pre>

blocks|key|986754|text|你的用例基本上只与处理固定长度的数组和矩阵有关，这样你就可以提前知道它们有多长。在这种情况下，您通常还会在手动将它们填充为None或0之前创建它们，因此实际上将使用的任何索引都已经存在。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|986755|您可以这样说:我经常需要字典中的.get()。作为一名十年的全职程序员，我想我从来没有在清单上需要过它。:)|986756|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Your usecase is basically only relevant for when doing arrays and matrixes of a fixed length, so that you know how long they are before hand. In that case you typically also create them before hand filling them up with None or 0, so that in fact any index you will use already exists.

You could say this: I need .get() on dictionaries quite often. After ten years as a full time programmer I don't think I have ever needed it on a list. :)

blocks|key|986809|text|如果您想要第一个元素，比如my_list.get(0)，这是可行的|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|986810|>>>+my_list+=+[1,2,3]
>>>+next(iter(my_list),+'fail')
1
>>>+my_list+=+[]
>>>+next(iter(my_list),+'fail')
'fail'|code-block|syntax|javascript|986811|我知道这不完全是你想要的，但它可能会对其他人有所帮助。|986812|entityMap^0|D|E|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

This works if you want the first element, like <code>my_list.get(0)</code>

<pre><code>&gt;&gt;&gt; my_list = [1,2,3]
&gt;&gt;&gt; next(iter(my_list), 'fail')
1
&gt;&gt;&gt; my_list = []
&gt;&gt;&gt; next(iter(my_list), 'fail')
'fail'
</code></pre>

I know it's not exactly what you asked for but it might help others.

blocks|key|986891|text|所以我对此做了更多的研究，结果发现没有任何具体的东西。当我发现list.index(+value+)时，我很兴奋，它返回指定项的索引，但没有任何东西可以获取特定索引处的值。因此，如果您不想使用safe_list_get解决方案，我认为这是非常好的。以下是一些可以根据场景为您完成工作的1linder+if语句：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|986892|>>>+x+=+[1,+2,+3]
>>>+el+=+x[4]+if+len(x)+>+4+else+'No'
>>>+el
'No'|code-block|syntax|javascript|986893|您也可以使用None而不是'No'，这更有意义：|986894|>>>+x+=+[1,+2,+3]
>>>+i+=+2
>>>+el_i+=+x[i]+if+len(x)+==+i%2B1+else+None|986895|此外，如果您只想获取列表中的第一项或最后一项，则可以使用此方法|986896|end_el+=+x[-1]+if+x+else+None|986897|您也可以将它们转换为函数，但我仍然喜欢IndexError异常解决方案。我尝试了safe_list_get解决方案的一个虚拟版本，并使其更简单(没有默认设置)：|offset|length|style|CODE|986898|def+list_get(l,+i):
++++try:
++++++++return+l[i]
++++except+IndexError:
++++++++return+None|986899|还没有进行基准测试，看看什么是最快的。|986900|entityMap^0|0|0|0|0|0|0|14|D|0|0|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]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@$Q|17|R|18|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

So I did some more research into this and it turns out there isn't anything specific for this. I got excited when I found list.index(value), it returns the index of a specified item, but there isn't anything for getting the value at a specific index. So if you don't want to use the safe_list_get solution which I think is pretty good. Here are some 1 liner if statements that can get the job done for you depending on the scenario:

<pre><code>&gt;&gt;&gt; x = [1, 2, 3]
&gt;&gt;&gt; el = x[4] if len(x) &gt; 4 else 'No'
&gt;&gt;&gt; el
'No'
</code></pre>

You can also use None instead of 'No', which makes more sense.:

<pre><code>&gt;&gt;&gt; x = [1, 2, 3]
&gt;&gt;&gt; i = 2
&gt;&gt;&gt; el_i = x[i] if len(x) == i+1 else None
</code></pre>

Also if you want to just get the first or last item in the list, this works

<pre><code>end_el = x[-1] if x else None
</code></pre>

You can also make these into functions but I still liked the IndexError exception solution. I experimented with a dummied down version of the <code>safe_list_get</code> solution and made it a bit simpler (no default):

<pre><code>def list_get(l, i):
 try:
 return l[i]
 except IndexError:
 return None
</code></pre>

Haven't benchmarked to see what is fastest.

blocks|key|772104|text|您可以做的合理的事情是将列表转换为字典，然后使用get方法访问它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|772105|>>>+my_list+=+['a',+'b',+'c',+'d',+'e']
>>>+my_dict+=+dict(enumerate(my_list))
>>>+print+my_dict
{0:+'a',+1:+'b',+2:+'c',+3:+'d',+4:+'e'}
>>>+my_dict.get(2)
'c'
>>>+my_dict.get(10,+'N/A')|code-block|syntax|javascript|772106|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|$]]

A reasonable thing you can do is to convert the list into a dict and then access it with the get method:
<pre><code>&gt;&gt;&gt; my_list = ['a', 'b', 'c', 'd', 'e']
&gt;&gt;&gt; my_dict = dict(enumerate(my_list))
&gt;&gt;&gt; print my_dict
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e'}
&gt;&gt;&gt; my_dict.get(2)
'c'
&gt;&gt;&gt; my_dict.get(10, 'N/A')
</code></pre>

blocks|key|986948|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|986949|>>>+i+=+3
>>>+a+=+[1,+2,+3,+4]
>>>+next(iter(a[i:]),+'fail')
4
>>>+next(iter(a[i+%2B+1:]),+'fail')
'fail'|code-block|syntax|javascript|986950|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|$]]

Try this:

<pre><code>&gt;&gt;&gt; i = 3
&gt;&gt;&gt; a = [1, 2, 3, 4]
&gt;&gt;&gt; next(iter(a[i:]), 'fail')
4
&gt;&gt;&gt; next(iter(a[i + 1:]), 'fail')
'fail'
</code></pre>

blocks|key|986982|text|感谢jose.angel.jimenez和Gus+Bus。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|986983|986984|对于"oneliner“的粉丝们，…|986985|986986|如果您想要一个列表的第一个元素，或者如果您想要一个默认值，如果列表是空的，请尝试：|986987|liste+=+['a',+'b',+'c']
value+=+(liste[0:1]+or+('default',))[0]
print(value)|code-block|syntax|javascript|986988|返回a|style|CODE|986989|和|986990|liste+=+[]
value+=+(liste[0:1]+or+('default',))[0]
print(value)|986991|返回default|986992|986993|其他元素…的示例|986994|liste+=+['a',+'b',+'c']
print(liste[0:1])++#+returns+['a']
print(liste[1:2])++#+returns+['b']
print(liste[2:3])++#+returns+['c']
print(liste[3:4])++#+returns+[]|986995|使用默认回退…|986996|liste+=+['a',+'b',+'c']
print((liste[0:1]+or+('default',))[0])++#+returns+a
print((liste[1:2]+or+('default',))[0])++#+returns+b
print((liste[2:3]+or+('default',))[0])++#+returns+c
print((liste[3:4]+or+('default',))[0])++#+returns+default|986997|986998|可能更短：|986999|liste+=+['a',+'b',+'c']
value,+=+liste[:1]+or+('default',)
print(value)++#+returns+a|987000|看起来您需要在等号、等号和后面的括号前加上逗号。|987001|987002|更一般的情况：|987003|liste+=+['a',+'b',+'c']
f+=+lambda+l,+x,+d:+l[x:x%2B1]+and+l[x]+or+d
print(f(liste,+0,+'default'))++#+returns+a
print(f(liste,+1,+'default'))++#+returns+b
print(f(liste,+2,+'default'))++#+returns+c
print(f(liste,+3,+'default'))++#+returns+default|987004|987005|使用Python+3.6.0+(v3.6.0:41df79263a11,+Dec+22+2016,+17:23:13)进行测试|987006|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/634355/jose-angel-jimenez|1|https://stackoverflow.com/users/9362404/gus-bus^0|2|I|0|L|7|1|0|0|0|0|0|0|2|1|0|0|0|2|7|0|0|0|0|0|0|0|0|0|0|0|0|0|0|2|1L|0^^$0|@$1|2|3|4|5|6|7|1W|8|@]|9|@$A|1X|B|1Y|1|1Z]|$A|20|B|21|1|22]]|C|$]]|$1|D|3|-4|5|6|7|23|8|@]|9|@]|C|$]]|$1|E|3|F|5|6|7|24|8|@]|9|@]|C|$]]|$1|G|3|-4|5|6|7|25|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|26|8|@]|9|@]|C|$]]|$1|J|3|K|5|L|7|27|8|@]|9|@]|C|$M|N]]|$1|O|3|P|5|6|7|28|8|@$A|29|B|2A|Q|R]]|9|@]|C|$]]|$1|S|3|T|5|6|7|2B|8|@]|9|@]|C|$]]|$1|U|3|V|5|L|7|2C|8|@]|9|@]|C|$M|N]]|$1|W|3|X|5|6|7|2D|8|@$A|2E|B|2F|Q|R]]|9|@]|C|$]]|$1|Y|3|-4|5|6|7|2G|8|@]|9|@]|C|$]]|$1|Z|3|10|5|6|7|2H|8|@]|9|@]|C|$]]|$1|11|3|12|5|L|7|2I|8|@]|9|@]|C|$M|N]]|$1|13|3|14|5|6|7|2J|8|@]|9|@]|C|$]]|$1|15|3|16|5|L|7|2K|8|@]|9|@]|C|$M|N]]|$1|17|3|-4|5|6|7|2L|8|@]|9|@]|C|$]]|$1|18|3|19|5|6|7|2M|8|@]|9|@]|C|$]]|$1|1A|3|1B|5|L|7|2N|8|@]|9|@]|C|$M|N]]|$1|1C|3|1D|5|6|7|2O|8|@]|9|@]|C|$]]|$1|1E|3|-4|5|6|7|2P|8|@]|9|@]|C|$]]|$1|1F|3|1G|5|6|7|2Q|8|@]|9|@]|C|$]]|$1|1H|3|1I|5|L|7|2R|8|@]|9|@]|C|$M|N]]|$1|1J|3|-4|5|6|7|2S|8|@]|9|@]|C|$]]|$1|1K|3|1L|5|6|7|2T|8|@$A|2U|B|2V|Q|R]]|9|@]|C|$]]|$1|1M|3|-4|5|6|7|2W|8|@]|9|@]|C|$]]]|1N|$1O|$5|1P|1Q|1R|C|$1S|1T]]|1U|$5|1P|1Q|1R|C|$1S|1V]]]]

Credits to <a href="https://stackoverflow.com/users/634355/jose-angel-jimenez">jose.angel.jimenez</a> and <a href="https://stackoverflow.com/users/9362404/gus-bus">Gus Bus</a>.
<hr />
For the &quot;oneliner&quot; fans…
<hr />
If you want the first element of a list or if you want a default value if the list is empty try:
<pre><code>liste = ['a', 'b', 'c']
value = (liste[0:1] or ('default',))[0]
print(value)
</code></pre>
returns <code>a</code>
and
<pre><code>liste = []
value = (liste[0:1] or ('default',))[0]
print(value)
</code></pre>
returns <code>default</code>
<hr />
Examples for other elements…
<pre><code>liste = ['a', 'b', 'c']
print(liste[0:1]) # returns ['a']
print(liste[1:2]) # returns ['b']
print(liste[2:3]) # returns ['c']
print(liste[3:4]) # returns []
</code></pre>
With default fallback…
<pre><code>liste = ['a', 'b', 'c']
print((liste[0:1] or ('default',))[0]) # returns a
print((liste[1:2] or ('default',))[0]) # returns b
print((liste[2:3] or ('default',))[0]) # returns c
print((liste[3:4] or ('default',))[0]) # returns default
</code></pre>
<hr />
Possibly shorter:
<pre><code>liste = ['a', 'b', 'c']
value, = liste[:1] or ('default',)
print(value) # returns a
</code></pre>
It looks like you need the comma before the equal sign, the equal sign and the latter parenthesis.
<hr />
More general:
<pre><code>liste = ['a', 'b', 'c']
f = lambda l, x, d: l[x:x+1] and l[x] or d
print(f(liste, 0, 'default')) # returns a
print(f(liste, 1, 'default')) # returns b
print(f(liste, 2, 'default')) # returns c
print(f(liste, 3, 'default')) # returns default
</code></pre>
<hr />
Tested with <code>Python 3.6.0 (v3.6.0:41df79263a11, Dec 22 2016, 17:23:13)</code>

blocks|key|987032|text|对于较小的索引值，您可以实现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|987033|my_list.get(index,+default)|offset|length|style|CODE|987034|作为|987035|(my_list+%2B+[default]+*+(index+%2B+1))[index]|987036|如果你事先知道索引是什么，那么这可以被简化，例如，如果你知道它是1，那么你可以这样做|987037|(my_list+%2B+[default,+default])[index]|987038|因为列表是前向打包的，所以我们需要担心的唯一失败情况是列表末尾的运行。这种方法用足够的默认值填充列表的末尾，以确保索引被覆盖。|987039|entityMap^0|0|0|R|0|0|0|16|0|0|0|11|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|15|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|S|$]]

For small index values you can implement
<code>my_list.get(index, default)</code>
as
<code>(my_list + [default] * (index + 1))[index]</code>
If you know in advance what index is then this can be simplified, for example if you knew it was 1 then you could do
<code>(my_list + [default, default])[index]</code>
Because lists are forward packed the only fail case we need to worry about is running off the end of the list. This approach pads the end of the list with enough defaults to guarantee that index is covered.

blocks|key|987112|text|这不是一个非常通用的解决方案，但是我遇到过这样的情况:我需要一个长度为3到5的列表(带有一个保护if)，并且我要将这些值分解为指定的变量。为此，我找到了一种简单而简洁的方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|987113|foo+=+(argv+%2B+[None,+None])[3]
bar+=+(argv+%2B+[None,+None])[4]|code-block|syntax|javascript|987114|现在，foo和bar是列表中的第4和第5个值，如果没有那么多值，则是None。|987115|entityMap^0|1C|2|0|0|3|3|7|3|Y|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

This isn't an extremely general-purpose solution, but I had a case where I expected a list of length 3 to 5 (with a guarding <code>if</code>), and I was breaking out the values to named variables. A simple and concise way I found for this involved:
<pre><code>foo = (argv + [None, None])[3]
bar = (argv + [None, None])[4]
</code></pre>
Now <code>foo</code> and <code>bar</code> are either the 4th and 5th values in the list, or <code>None</code> if there weren't that many values.

Why doesn't list have a safe "get" method like dictionary?

<pre><code>&gt;&gt;&gt; d = {'a':'b'}
&gt;&gt;&gt; d['a']
'b'
&gt;&gt;&gt; d['c']
KeyError: 'c'
&gt;&gt;&gt; d.get('c', 'fail')
'fail'

&gt;&gt;&gt; l = [1]
&gt;&gt;&gt; l[10]
IndexError: list index out of range
</code></pre>

Why doesn't list have safe "get" method like dictionary?

Python

为什么list没有像字典一样安全的"get“方法？>>> d = {'a':'b'}>>> d['a']'b'>>> d['c']KeyError: 'c'>>> d.get('c', 'fail')'fail'>>> l = [1]>>> l[10]IndexError: list index out of rang...

问为什么list没有像字典一样安全的get方法？
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