随心所愿 发表于 2021-12-26 13:18:21

见附件~~~~~~

吖柱 发表于 2021-12-29 14:27:26

谢谢提供解法,整理操作方法如下:
Power Query是我遇到的第2道习题,看来,还需要多学习下Power Query

泰陽 发表于 2022-1-7 22:05:17

=OFFSET($A$2,INT((ROW(A1)-1)/2001),)&OFFSET($B$2,INT(MOD(ROW(A1)-1,2001)/667),)&OFFSET($C$2,INT(MOD(ROW(A1)-1,667)/29),)&OFFSET($D$2,MOD(ROW(A1)-1,29),)这样行不行

哇咔哇咔 发表于 2022-1-11 16:11:13

谢谢提供算法,数组公式如下:
{=INDEX(A:A,MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)*(COUNTA(B:B)-1)),)-1,COUNTA(A:A)-1)+2)&INDEX(B:B,MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)),)-1,COUNTA(B:B)-1)+2)&INDEX(C:C,MOD(ROUNDUP(ROW(1:1)/(COUNTA(D:D)-1),)-1,COUNTA(C:C)-1)+2)&INDEX(D:D,MOD(ROW(1:1)-1,COUNTA(D:D)-1)+2)}

郑明耀 发表于 2022-1-17 05:53:40

谢谢,经过运算,这个公式也比较高效

小桂子 发表于 2022-1-18 13:06:01

这个速度能快点
=INDIRECT("A"&MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)*(COUNTA(B:B)-1)),)-1,COUNTA(A:A)-1)+2)&INDIRECT("B"&MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)),)-1,COUNTA(B:B)-1)+2)&INDIRECT("C"&MOD(ROUNDUP(ROW(1:1)/(COUNTA(D:D)-1),)-1,COUNTA(C:C)-1)+2)&INDIRECT("D"&MOD(ROW(1:1)-1,COUNTA(D:D)-1)+2)

电脑计算机监控 发表于 2022-1-26 07:16:33

谢谢。公式比较长,我分解了一下,便于公式的理解,方便大家套用:
INDIRECT("A"&MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)*(COUNTA(B:B)-1)),)-1,COUNTA(A:A)-1)+2)&
INDIRECT("B"&MOD(ROUNDUP(ROW(1:1)/((COUNTA(D:D)-1)*(COUNTA(C:C)-1)),)-1,COUNTA(B:B)-1)+2)&
INDIRECT("C"&MOD(ROUNDUP(ROW(1:1)(COUNTA(D:D)-1),)-1,COUNTA(C:C)-1)+2)&
INDIRECT("D"&MOD(ROW(1:1)-1,COUNTA(D:D)-1)+2)
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