blocks|key|63720|text|当然可以，也可以为此创建单个字符串键-只需合并键的字符串结果，如'，'.join(k1，k2，k3)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|63721|Read+more+about+dictionaries.|offset|length|63722|63723|字典通过键进行索引，键可以是任何不可变的类型；字符串和数字始终可以是键。如果只包含字符串、数字或元组，则可以将元组用作键；如果元组直接或间接包含任何可变对象，则不能将其用作键。您不能将列表用作键，因为可以使用索引分配、片分配或append()和extend()等方法在适当的位置修改列表。|blockquote|63724|63725|63726|因此，您尝试使用元组作为键，这是可以的。|63727|请注意，python中的字典是不排序的。您可以使用collections.OrderedDict来实现这一点。此外，要构建正确的排序，请使用指定了参数键的sort/sorted+functions来按您想要的方式进行排序。|63728|编辑后的示例：|63729|from+itertools+import+product
myDict+=+{}
for+x,y,z+in+product(range(10),+range(10,20),+range(20,30)):
++++myDict[(x,y,z)]+=+sum([x,y,z])|code-block|syntax|javascript|63730|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/tutorial/datastructures.html#dictionaries|1|http://docs.python.org/2/library/functions.html#sorted^0|0|0|T|0|0|0|0|0|0|0|25|L|1|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@$D|18|E|19|1|1A]]|A|$]]|$1|F|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|1C|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1F|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1G|8|@]|9|@$D|1H|E|1I|1|1J]]|A|$]]|$1|P|3|Q|5|6|7|1K|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|1L|8|@]|9|@]|A|$U|V]]|$1|W|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]]]

Sure you can, as well as to create single string key for this - just merge string results for your keys like ','.join([k1,k2,k3]) 

<a href="http://docs.python.org/2/tutorial/datastructures.html#dictionaries">Read more about dictionaries.</a>

<blockquote>
 dictionaries are indexed by keys, which can be any immutable type;
 strings and numbers can always be keys. Tuples can be used as keys if
 they contain only strings, numbers, or tuples; if a tuple contains any
 mutable object either directly or indirectly, it cannot be used as a
 key. You can’t use lists as keys, since lists can be modified in place
 using index assignments, slice assignments, or methods like append()
 and extend().
</blockquote>

So you trying to use tuples as a keys and this is OK.

Note that dictionaries in python are not sorted. You can use collections.OrderedDict for this. Also to build correct sorting use <a href="http://docs.python.org/2/library/functions.html#sorted">sort/sorted functions</a> with parameter key specified to sort the way you want.

EDITED sample:

<pre><code>from itertools import product
myDict = {}
for x,y,z in product(range(10), range(10,20), range(20,30)):
 myDict[(x,y,z)] = sum([x,y,z])
</code></pre>

blocks|key|63732|text|希望你会发现这一点很有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|63733|>>>+from+math+import+sqrt
>>>+def+dist(p1,+p2):
...+++++x1,+y1,+z1+=+p1
...+++++x2,+y2,+z2+=+p2
...+++++xd+=+x1+-+x2
...+++++yd+=+y1+-+y2
...+++++zd+=+z1+-+z2
...+++++return+sqrt((xd+**+2+%2B+yd+**+2+%2B+zd+**+2))
>>>+myPoint+=+(0,0,0)
>>>+class+MyObject:+pass
>>>+myDict+=+{(1,2,3):MyObject(),+(4,5,6):MyObject()}
>>>+sorted([dist(myPoint,+point)+for+point+in+myDict])
10:+[3.7416573867739413,+8.774964387392123]|code-block|syntax|javascript|63734|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|$]]

Hopefully you will find this useful.

<pre><code>&gt;&gt;&gt; from math import sqrt
&gt;&gt;&gt; def dist(p1, p2):
... x1, y1, z1 = p1
... x2, y2, z2 = p2
... xd = x1 - x2
... yd = y1 - y2
... zd = z1 - z2
... return sqrt((xd ** 2 + yd ** 2 + zd ** 2))
&gt;&gt;&gt; myPoint = (0,0,0)
&gt;&gt;&gt; class MyObject: pass
&gt;&gt;&gt; myDict = {(1,2,3):MyObject(), (4,5,6):MyObject()}
&gt;&gt;&gt; sorted([dist(myPoint, point) for point in myDict])
10: [3.7416573867739413, 8.774964387392123]
</code></pre>

blocks|key|4820394|text|在我看来，您可能需要一个从键(有序的三元组)到值(对象)的映射的无序集合。如果是这样的话，很容易从一个有序的三元组中生成一个组合键：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4820395|假设obj0在x，y，z坐标(10,20,30)处，obj1在x，y，z坐标(11,21,31)处。然后：|offset|length|style|CODE|4820396|myObjects+=+{(10,20,30):+obj0,
+++++++++++++(11,21,31):+obj1
++++++++++++}|code-block|syntax|javascript|4820397|这之所以可行，是因为tuple和int都是不可变类型。|4820398|希望这能有所帮助|4820399|entityMap^0|0|2|4|E|A|Q|4|12|A|0|0|A|5|G|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

It seems to me that you want a possibly unordered collection of mappings from keys (ordered triples) to values (objects). If that's the case, it's very easy to make a composite key out of an ordered triple:

Suppose <code>obj0</code> is at x,y,z coordinates <code>(10,20,30)</code> and <code>obj1</code> is at x,y,z coordinates <code>(11,21,31)</code>. Then:

<pre><code>myObjects = {(10,20,30): obj0,
 (11,21,31): obj1
 }
</code></pre>

This works because both <code>tuple</code>s and <code>int</code>s are immutable types

Hope that helps

blocks|key|63784|text|我解决了这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|63785|list+=+[
+++++++++[1,'a','school'],
+++++++++[2,'b','church'],
+++++++++[3,'c','library'],
+++++++++[4,'d','station']
++++]
dic_key+=+{}
for+i+in+list:
+++if+(i[0],i[1])+not+in+dic_key:
++++++dic_key[i[0],i[1]]+=+[]
+++dic_key[i[0],i[1]].append(i[2])

print(dic_key)|code-block|syntax|javascript|63786|输出：{(1，'a')：‘学校’，(2，'b')：‘教堂’，(3，'c')：'library'，(4，'d')：'station'}|63787|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|$]]

I solved doing this:
<pre><code>list = [
 [1,'a','school'],
 [2,'b','church'],
 [3,'c','library'],
 [4,'d','station']
 ]
dic_key = {}
for i in list:
 if (i[0],i[1]) not in dic_key:
 dic_key[i[0],i[1]] = []
 dic_key[i[0],i[1]].append(i[2])

print(dic_key)
</code></pre>
output: {(1, 'a'): ['school'], (2, 'b'): ['church'], (3, 'c'): ['library'], (4, 'd'): ['station']}

My question pertains to dictionary keys. I want to set up a dictionary that has 3 keys for any single object. The keys must be in order and can have a wide range of values. For instance,

<pre><code>dictionary = {(key1,key2,key3) : object}
</code></pre>

key1 can be any value between 1 and 10
key2 can be any value between 11 and 20
key3 can be any value between 21 and 30

The order in which the keys are placed does matter.

More specifically, my keys correspond to a range of x,y,z cartesian coordinates in which many objects are floating around in. I want to be able to sort the relative position of the objects based off their x,y,z positions.

Is there any way to set this up or will I have to take a different approach?
Thanks for any help!

Python complex dictionary keys

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我的问题与字典中的关键字有关。我想建立一个字典，有3个关键字的任何单一的对象。键必须是有序的，并且可以有很大范围的值。例如,dictionary = {(key1,key2,key3) : object}key1可以是1到10之间的任何值key2可以是11到20之间的任何值key3可以是21到30之间的任何值键的放置顺...

问Python复杂字典键
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Python复杂字典键EN