blocks|key|5371523|text|“一百万个元素”并不多。如果这足够快，请尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5371524|(defn+this-and-that
++[s]
++(map-indexed+(fn+[i+el]
+++++++++++++++++[el+(keep-indexed+#(if+(=+%251+i)
++++++++++++++++++++++++++++++++++++++++++++nil
++++++++++++++++++++++++++++++++++++++++++++%252)+s)])+s))|code-block|syntax|javascript|5371525|示例：|5371526|user>+(this-and-that+[1+2+3])
([1+(2+3)]+[2+(1+3)]+[3+(1+2)])|5371527|请注意，对于包含nil的seq，此代码不能正常工作。|offset|length|style|CODE|5371528|entityMap^0|0|0|0|0|8|3|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

"a million elements" is not a lot. Try if this is fast enough:

<pre><code>(defn this-and-that
 [s]
 (map-indexed (fn [i el]
 [el (keep-indexed #(if (= %1 i)
 nil
 %2) s)]) s))
</code></pre>

Example:

<pre><code>user&gt; (this-and-that [1 2 3])
([1 (2 3)] [2 (1 3)] [3 (1 2)])
</code></pre>

Note that this code doesn't work correctly for seqs containing <code>nil</code>.

blocks|key|90037|text|这将在不反复遍历整个向量的情况下工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|90038|(defn+this-and-that+[xs]
++(map-indexed+(fn+[i+x]
++++++++++++++++[x+(concat+(take+i+xs)+
+++++++++++++++++++++++++++(drop+(inc+i)+xs))])
++++++++++++++xs))|code-block|syntax|javascript|90039|也适用于nils：|90040|user=>+(this-and-that+[1+2+nil+3])
([1+(2+nil+3)]+[2+(1+nil+3)]+[nil+(1+2+3)]+[3+(1+2+nil)])|90041|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|$]]

This will work without traversing the entire vector over and over:

<pre><code>(defn this-and-that [xs]
 (map-indexed (fn [i x]
 [x (concat (take i xs) 
 (drop (inc i) xs))])
 xs))
</code></pre>

and also works for nils:

<pre><code>user=&gt; (this-and-that [1 2 nil 3])
([1 (2 nil 3)] [2 (1 nil 3)] [nil (1 2 3)] [3 (1 2 nil)])
</code></pre>

blocks|key|1109776|text|其他答案给出了实现这一点的好方法，但是您所描述的基本上是一个O(n%5E2)操作，假设您实际上实现了所有的懒惰序列等等。如果n+>=+1,000,000，这可能不是一个好主意|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1109777|您可能想要考虑扩大您正在考虑的代码的范围，看看是否可以找到一个整体上更有效的算法。|1109778|例如，您可能会发现，最好将整个列表转换为向量，并将索引访问方面的算法编写为向量。|1109779|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Other answers give good ways of doing this, however what you are describing is fundamentally an O(n^2) operation assuming that you actually realise all the lazy sequences etc. This probably isn't going to be a good idea if n >= 1,000,000

You might want to consider broadening the scope of the code you are considering to see if you can find a more efficient algorithm overall.

For example, you might find that is is better to convert the entire list into a vector, and writing your algorithm in terms of indexed access into a vector.

blocks|key|5371630|text|使用集合怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5371631|user>+(let+[xs+#{1+2+3}]
++++++++(for+[x+xs]+[x+(disj+xs+x)]))
([1+#{2+3}]+[2+#{1+3}]+[3+#{1+2}])|code-block|syntax|javascript|5371632|在一个有一百万个项目的集合上做这件事在时间上是不是太糟糕了：|5371633|(let+[xs+(set+(range+1000000))]
++(time+(count+(for+[x+xs]+[x+(disj+xs+x)]))))
"Elapsed+time:+841.668+msecs"
1000000|5371634|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|$]]

How about using sets?

<pre><code>user&gt; (let [xs #{1 2 3}]
 (for [x xs] [x (disj xs x)]))
([1 #{2 3}] [2 #{1 3}] [3 #{1 2}])
</code></pre>

Doing this on a set with a million items isn't too bad time-wise:

<pre><code>(let [xs (set (range 1000000))]
 (time (count (for [x xs] [x (disj xs x)]))))
"Elapsed time: 841.668 msecs"
1000000
</code></pre>

What is the fastest way to get the list containing pairs of the current element with the list containing every other element? This should be fast as the list could contain a million elements or more.

For example given the list <code>(1 2 3)</code>

I want to get the list <code>((1 (2 3)) (2 (1 3)) (3 (1 2)))</code>

Thanks for your help!

Get the current element and the list containing everything else in Clojure?

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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获取包含当前元素对的列表和包含所有其他元素的列表的最快方法是什么？这应该很快，因为列表可能包含一百万个或更多元素。例如，给定列表(1 2 3)我想要得到列表((1 (2 3)) (2 (1 3)) (3 (1 2)))谢谢你的帮忙!

问获取当前元素和包含Clojure中其他所有内容的列表？
EN

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问获取当前元素和包含Clojure中其他所有内容的列表？EN