在KDB中如何对列表进行分块？

huangapple
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英文:

How do I chunk a list in KDB?

问题

我想将一个列表重新排列成大小为n的“块”：

0 1 2 3 4 5 6 7 8 9

0 1 2
1 2 3
2 3 4

在这种情况下，n=3。

在KDB中以有效的方式完成这个任务的方法是什么？

英文:

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

  1. 0 1 2 3 4 5 6 7 8 9
  3. 0 1 2
  4. 1 2 3
  5. 2 3 4

where n=3 in this case.

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

答案1

得分: 2

0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

英文:

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.

  1. l:til 10;
  2. m:{[n;l] @[l;(til n)+/: til neg[n-1]+count l]}
  4. q)m[3;l]
  5. 0 1 2
  6. 1 2 3
  7. 2 3 4
  8. 3 4 5
  9. 4 5 6
  10. 5 6 7
  11. 6 7 8
  12. 7 8 9
  14. q)m[4;l]
  15. 0 1 2 3
  16. 1 2 3 4
  17. 2 3 4 5
  18. 3 4 5 6
  19. 4 5 6 7
  20. 5 6 7 8
  21. 6 7 8 9
  23. m2:{[n;l] @[l;(til n)+/: til count l]}
  24. q)m2[4;l]
  25. 0 1 2 3
  26. 1 2 3 4
  27. 2 3 4 5
  28. 3 4 5 6
  29. 4 5 6 7
  30. 5 6 7 8
  31. 6 7 8 9
  32. 7 8 9
  33. 8 9
  34. 9
  36. </details>
  40. # 答案2
  41. **得分**: 1
  43. ## 滑动窗口
  45. 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
...

  2. ## 切割列表
  4. 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

  2. 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

  2. 如文档中所述：
  3. &gt; x 中的空值将导致该维度变为最大值。
  5. <details>
  6. <summary>英文:</summary>
  8. ## Sliding window
  10. 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
...

  2. ## Chunking a list
  4. 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

  2. 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

  2. As stated in the docs:
  3. &gt; A null in x will cause that dimension to be maximal.
  5. </details>
  9. # 答案3
  10. **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

  2. <details>
  3. <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

  2. </details>
  6. # 答案4
  7. **得分**: 0
  9. 以下是您要翻译的内容：
  11. 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）：
  1. \ts:100 {x[] each -2_(til count x)+(count x)#enlist til y}[til 1000000;3]
  2. 8513 153166544
  2. 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

  2. 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

  2. 然后，这用作我们要用来索引列表的值的列表，以生成切分。
  4. <details>
  5. <summary>英文:</summary>
  7. 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):
  9.     \ts:100 {x[] each -2_(til count x)+(count x)#enlist til y}[til 1000000;3]
  10.     8513 153166544
  12. Here, working for a list of syms:
  14.     q){x[] each -2_(til count x)+(count x)#enlist til y}[`a`b`c`d`e`f`g`h`i`j`k;3]
  15.     a b c
  16.     b c d
  17.     c d e
  18.     d e f
  19.     e f g
  20.     f g h
  21.     g h i
  22.     h i j
  23.     i j k
  25. 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.
  27.     0 1 2
  28.     1 2 3
  29.     2 3 4
  31. This then serves as the list of values we want to index our list by to producde the cut 
  33. </details>
  37. # 答案5
  38. **得分**: 0
  40. 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)
  41. ```q
  42. q)flip{(til[x]-x-1)_&#39;(x-1)_[1;]\y}[3] til 10
  43. 0 1 2
  44. 1 2 3
  45. 2 3 4
  46. 3 4 5
  47. 4 5 6
  48. 5 6 7
  49. 6 7 8
  50. 7 8 9

或者使用Drop Each Left和Drop Each

  1. q)flip{neg[reverse tx]_&#39;(tx:til x)_\:y}[3] til 10
  2. 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
英文:

Each column of the result is a 1_ of the preceding column. Use the Scan form of Do

  1. q)flip{(til[x]-x-1)_&#39;(x-1)_[1;]\y}[3] til 10
  2. 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

or Drop Each Left and Drop Each

  1. q)flip{neg[reverse tx]_&#39;(tx:til x)_\:y}[3] til 10
  2. 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

huangapple
  • 本文由 发表于 2023年3月7日 01:32:29
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