计算数组中前面重复项的数量

如何解决计算数组中前面重复项的数量

我有以下查询，它遍历数组中的每个项目，然后回过头看看有多少重复的 f 包括它自己。

它可以工作，但是在大量行中会很慢 - 有没有更简洁的方法来处理数组中的序列？

SELECT
['p','p','f','f'] AS sequence,arrayMap( (x,y) -> (x,if (x='f',(arrayFirstIndex( k -> k=0,arrayCumSumNonNegative((n,index) -> n = 'f' ? 1 : -index,arrayReverse(arraySlice(sequence,1,y)) as arr,arrayEnumerate(arr)))
   )-1,0)),sequence,arrayEnumerate(sequence))

result:

[('p',0),('p',('f',1),2),3),4),3)]

提前致谢

解决方法

试试这个查询：

WITH 'f' AS ch
SELECT　
  arraySplit((x,i) -> x = ch and sequence[i - 1] != ch or x != ch and sequence[i - 1] = ch,sequence,arrayEnumerate(sequence)) parts,arrayMap(part -> arrayMap((x,index) -> (x,x = ch ? index : 0),part,arrayEnumerate(part)),parts) parts_and_number,arrayFlatten(parts_and_number) result
FROM (
  SELECT arrayJoin([
    ['p','p','f','f'],['p','w',['f','w'],['f']
  ]) as sequence)

/*
Row 1:
──────
parts:            [['p','p'],['p'],'f']]
parts_and_number: [[('p',0),('p',0)],[('f',1),('f',2),3),4)],[('p',3)]]
result:           [('p',4),3)]

Row 2:
──────
parts:            [['p',('w',3)]

Row 3:
──────
parts:            [['f','f']]
parts_and_number: [[('f',3)]]
result:           [('f',3)]

Row 4:
──────
parts:            [['p','w']]
parts_and_number: [[('p',0)]]
result:           [('p',0)]

Row 5:
──────
parts:            [['f',2)]]
result:           [('f',2)]

Row 6:
──────
parts:            [['f']]
parts_and_number: [[('f',1)]]
result:           [('f',1)]
*/