How do I chunk a list in KDB?

I want to rearrange a list into n-sized 'chunks':

0 1 2 3 4 5 6 7 8 9

0 1 2
1 2 3
2 3 4

where n=3 in this case.

What's an efficient way to do this in KDB that works for long lists?

Could build a list of indices and index into your list. Not really sure what you want to do when there are less than 3/n for the cut. m below stops before nulls are populated. m2 below includes the nulls.

l:til 10;
m:{[n;l] @[l;(til n)+/: til neg[n-1]+count l]}

q)m[3;l]
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

q)m[4;l]
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
6 7 8 9

m2:{[n;l] @[l;(til n)+/: til count l]}
q)m2[4;l]
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
6 7 8 9
7 8 9
8 9
9

</details>



# 答案2
**得分**: 1

## 滑动窗口

要实现你示例中所述的输出，可以使用修改过的版本的[滑动窗口](https://code.kx.com/q/kb/programming-idioms/#how-do-i-apply-a-function-to-a-sequence-sliding-window)函数：

q){(x-1)_{ 1_x,y }[x#0;y]}[3;til 10]
0 1 2
1 2 3
2 3 4
...

## 切割列表

要将列表切割成大小为n的块，请使用[take `#`](https://code.kx.com/q/ref/take/#atom-or-list)运算符。例如，当n=3时：

q)0N 3#til 10
0 1 2
3 4 5
6 7 8
,9

相同的符号也可用于将列表切割成n部分。例如，n=3：

q)3 0N#til 10
0 1 2
3 4 5
6 7 8 9

如文档中所述：
> x 中的空值将导致该维度变为最大值。

<details>
<summary>英文:</summary>

## Sliding window

To achieve the output set out by your example you could use a modified version of this [sliding window](https://code.kx.com/q/kb/programming-idioms/#how-do-i-apply-a-function-to-a-sequence-sliding-window) function:

q){(x-1)_{ 1_x,y }[x#0;y]}[3;til 10]
0 1 2
1 2 3
2 3 4
...

## Chunking a list

To cut a list into n sized chunks use the [take `#`](https://code.kx.com/q/ref/take/#atom-or-list) operator. For example, when n=3:

q)0N 3#til 10
0 1 2
3 4 5
6 7 8
,9

The same notation can also be used to cut a list into n parts. For example n=3:

q)3 0N#til 10
0 1 2
3 4 5
6 7 8 9

As stated in the docs:
> A null in x will cause that dimension to be maximal.

</details>



# 答案3
**得分**: 0

q)\ts:100 {flip til[x]+:til 1+y-x}[3;1000000]
2181 97555184

q)\ts:100 {(x-1)_{ 1_x,y }[x#0;y]}[3;til 1000000]
17120 89166928

q){flip til[x]+:til 1+y-x}[3;1000000]~{(x-1)_{ 1_x,y }[x#0;y]}[3;til 1000000]
1b

<details>
<summary>英文:</summary>

q)\ts:100 {flip til[x]+:til 1+y-x}[3;1000000]
2181 97555184

q)\ts:100 {(x-1)_{ 1_x,y }[x#0;y]}[3;til 1000000]
17120 89166928

q){flip til[x]+:til 1+y-x}[3;1000000]~{(x-1)_{ 1_x,y }[x#0;y]}[3;til 1000000]
1b

</details>



# 答案4
**得分**: 0

以下是您要翻译的内容：

这里有另一个更快的（尽管内存效率较低）解决方案，适用于所有类型的列表（我相信用户21344792的只适用于数字）：

\ts:100 {x[] each -2_(til count x)+(count x)#enlist til y}[til 1000000;3]
8513 153166544

在这里，适用于一组符号：

q){x[] each -2_(til count x)+(count x)#enlist til y}[abcdefghij`k;3]
a b c
b c d
c d e
d e f
e f g
f g h
g h i
h i j
i j k

该解决方案首先生成一个数字矩阵，其中每行元素的值比上面的行大一，即：

0 1 2
1 2 3
2 3 4

然后，这用作我们要用来索引列表的值的列表，以生成切分。

<details>
<summary>英文:</summary>

Here&#39;s another faster (though less memory efficient) solution that is generalisable to lists of all types (I believe user21344792&#39;s only works for numbers):

    \ts:100 {x[] each -2_(til count x)+(count x)#enlist til y}[til 1000000;3]
    8513 153166544

Here, working for a list of syms:

    q){x[] each -2_(til count x)+(count x)#enlist til y}[`a`b`c`d`e`f`g`h`i`j`k;3]
    a b c
    b c d
    c d e
    d e f
    e f g
    f g h
    g h i
    h i j
    i j k

The solution works by first generating a matrix of numbers, where the value of each row element is one bigger than the row &quot;above&quot; i.e.

    0 1 2
    1 2 3
    2 3 4

This then serves as the list of values we want to index our list by to producde the cut 

</details>



# 答案5
**得分**: 0

每一列的结果都是前一列的`1_`。使用[Do的扫描形式](https://code.kx.com/q/ref/accumulators/#do)
```q
q)flip{(til[x]-x-1)_&#39;(x-1)_[1;]\y}[3] til 10
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

或者使用Drop Each Left和Drop Each

q)flip{neg[reverse tx]_'(tx:til x)_\:y}[3] til 10
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

q)flip{(til[x]-x-1)_'(x-1)_[1;]\y}[3] til 10
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

q)flip{neg[reverse tx]_'(tx:til x)_\:y}[3] til 10
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9