blocks|key|2751281|text|这看起来很不雅观，但它得到了正确的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2751282|(seq+[(1,+2,+3);+(1,+2,+3);+(1,+2,+3);])+
%7C>+Seq.fold+(fun+(sa,sb,sc)+(a,b,c)+->a::sa,b::sb,c::sc)+([],[],[])+
%7C>+fun+(a,b,c)+->+a::b::c::[]|code-block|syntax|javascript|2751283|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|$]]

This seems very inelegant but it gets the right answer:

<pre><code>(seq [(1, 2, 3); (1, 2, 3); (1, 2, 3);]) 
|&gt; Seq.fold (fun (sa,sb,sc) (a,b,c) -&gt;a::sa,b::sb,c::sc) ([],[],[]) 
|&gt; fun (a,b,c) -&gt; a::b::c::[]
</code></pre>

blocks|key|2751302|text|看起来像是矩阵转置。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2751303|let+data+=
++++seq+[
++++++++seq+[1;+2;+3]
++++++++seq+[1;+2;+3]
++++++++seq+[1;+2;+3]
++++]

let+rec+transpose+=+function
++++%7C+(_::_)::_+as+M+->+List.map+List.head+M+::+transpose+(List.map+List.tail+M)
++++%7C+_+->+[]

//+I+don't+claim+it+is+very+elegant,+but+no+doubt+it+is+readable
let+result+=
++++data
++++%7C>+List.ofSeq
++++%7C>+List.map+List.ofSeq
++++%7C>+transpose
++++%7C>+Seq.ofList
++++%7C>+Seq.map+Seq.ofList|code-block|syntax|javascript|2751304|或者，您可以对seq采用相同的方法，这要归功于this+answer提供的优雅的活动模式：|offset|length|style|CODE|2751305|let+(%7CSeqEmpty%7CSeqCons%7C)+(xs:+'a+seq)+=
++if+Seq.isEmpty+xs+then+SeqEmpty
++else+SeqCons(Seq.head+xs,+Seq.skip+1+xs)

let+rec+transposeSeq+=+function
++++%7C+SeqCons(SeqCons(_,_),_)+as+M+->
++++++++Seq.append
++++++++++++(Seq.singleton+(Seq.map+Seq.head+M))
++++++++++++(transposeSeq+(Seq.map+(Seq.skip+1)+M))
++++%7C+_+->+Seq.empty

let+resultSeq+=+data+%7C>+transposeSeq|2751306|另请参阅this+answer以获取技术细节和两个参考：PowerPack的Microsoft.FSharp.Math.Matrix和涉及可变数据的另一种方法。|BOLD|2751307|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/1757340/974789|1|https://stackoverflow.com/a/10849803/974789^0|0|0|7|3|N|B|0|0|0|S|9|1X|4|12|S|4|B|1|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|13|8|@$I|14|J|15|K|L]]|9|@$I|16|J|17|1|18]]|A|$]]|$1|M|3|N|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1A|8|@$I|1B|J|1C|K|Q]|$I|1D|J|1E|K|Q]|$I|1F|J|1G|K|L]]|9|@$I|1H|J|1I|1|1J]]|A|$]]|$1|R|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]|Z|$5|U|V|W|A|$X|10]]]]

It looks like matrix transposition.

<pre><code>let data =
 seq [
 seq [1; 2; 3]
 seq [1; 2; 3]
 seq [1; 2; 3]
 ]

let rec transpose = function
 | (_::_)::_ as M -&gt; List.map List.head M :: transpose (List.map List.tail M)
 | _ -&gt; []

// I don't claim it is very elegant, but no doubt it is readable
let result =
 data
 |&gt; List.ofSeq
 |&gt; List.map List.ofSeq
 |&gt; transpose
 |&gt; Seq.ofList
 |&gt; Seq.map Seq.ofList
</code></pre>

Alternatively, you may adopt the same method for <code>seq</code>, thanks to <a href="https://stackoverflow.com/a/1757340/974789">this answer</a> for an elegant Active pattern:

<pre><code>let (|SeqEmpty|SeqCons|) (xs: 'a seq) =
 if Seq.isEmpty xs then SeqEmpty
 else SeqCons(Seq.head xs, Seq.skip 1 xs)

let rec transposeSeq = function
 | SeqCons(SeqCons(_,_),_) as M -&gt;
 Seq.append
 (Seq.singleton (Seq.map Seq.head M))
 (transposeSeq (Seq.map (Seq.skip 1) M))
 | _ -&gt; Seq.empty

let resultSeq = data |&gt; transposeSeq
</code></pre>

See also <a href="https://stackoverflow.com/a/10849803/974789">this answer</a> for technical details and two references: to PowerPack's <code>Microsoft.FSharp.Math.Matrix</code> and yet another method involving mutable data.

blocks|key|4961865|text|如果你要寻找一个语义上是Seq的解决方案，你将不得不一直保持懒惰。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4961866|let+zip+seq+=+seq
++++++++++++%7C>+Seq.collect(fun+s+->+s+%7C>+Seq.mapi(fun+i+e+->+(i,+e)))+//wrap+with+index
++++++++++++%7C>+Seq.groupBy(fst)+//group+by+index
++++++++++++%7C>+Seq.map(fun+(i,+s)+->+s+%7C>+Seq.map+snd)+//unwrap|code-block|syntax|javascript|4961867|测试：|4961868|let+seq+=++Enumerable.Repeat((seq+[1;+2;+3]),+3)+//don't+want+to+while(true)+yield.+bleh.
printfn+"%25A"+(zip+seq)|4961869|输出：|4961870|seq+[seq+[1;+1;+1];+seq+[2;+2;+2];+seq+[3;+3;+3]]|4961871|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you're going for a solution which is semantically Seq, you're going to have to stay lazy all the time.

<pre><code>let zip seq = seq
 |&gt; Seq.collect(fun s -&gt; s |&gt; Seq.mapi(fun i e -&gt; (i, e))) //wrap with index
 |&gt; Seq.groupBy(fst) //group by index
 |&gt; Seq.map(fun (i, s) -&gt; s |&gt; Seq.map snd) //unwrap
</code></pre>

Test:

<pre><code>let seq = Enumerable.Repeat((seq [1; 2; 3]), 3) //don't want to while(true) yield. bleh.
printfn "%A" (zip seq)
</code></pre>

Output:

<pre><code>seq [seq [1; 1; 1]; seq [2; 2; 2]; seq [3; 3; 3]]
</code></pre>

blocks|key|3232718|text|这是与@Asti相同的答案，只是稍微清理了一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3232719|[[1;2;3];+[1;2;3];+[1;2;3]]+
++++%7C>+Seq.collect+Seq.indexed+
++++%7C>+Seq.groupBy+fst+
++++%7C>+Seq.map+(snd+>>+Seq.map+snd);;|code-block|syntax|javascript|3232720|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|$]]

This is the same answer as @Asti, just cleaned up a little:

<pre><code>[[1;2;3]; [1;2;3]; [1;2;3]] 
 |&gt; Seq.collect Seq.indexed 
 |&gt; Seq.groupBy fst 
 |&gt; Seq.map (snd &gt;&gt; Seq.map snd);;
</code></pre>

Let's say I have a sequence of sequences, e.g. 

<pre><code>{1, 2, 3}, {1, 2, 3}, {1, 2, 3}
</code></pre>

What is the best way to pivot or zip this sequence so I instead have,

<pre><code>{1, 1, 1}, {2, 2, 2}, {3, 3, 3}
</code></pre>

Is there a comprehensible way of doing so without resorting to manipulation of the underlying <code>IEnumerator&lt;_&gt;</code> type? 

To clarify, these are <code>seq&lt;seq&lt;int&gt;&gt;</code> objects. Each sequences (both internal and external) can have any number of items.

Pivot or zip a seq<seq<'a>> in F#

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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假设我有一系列的序列，例如{1, 2, 3}, {1, 2, 3}, {1, 2, 3}什么是最好的方法来旋转或压缩这个序列，所以我有，{1, 1, 1}, {2, 2, 2}, {3, 3, 3}有没有一种可以理解的方式，不用操作底层的IEnumerator<_>类型？为了清楚起见，这些是seq<seq<int>>对...

问在F#中旋转或压缩序列
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