Question

我需要分析过去1年期间的每周订单频率，以找出每种产品的订单的最小/最大/平均频率。无论是新的还是旧的，系统都应该计算当年第一次出现的订单作为订单的起始周。最小订单频率是连续订购周之间的差异。如果第一个订单在wk 3中并且第二个订单在wk6中，则暗示订单频率是3周（=> 6-3）。订单可以在过去52周内的任何一周。平均订单频率=（52 - 第一个订单周）/没有订单的周数。

附加excel以更好地理解该问题。

Original image

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| Product | wk1 | wk2 | wk3 | wk4 | wk5 | wk6 | wk7 | wk8 | wk9 | wk10 | wk11 | wk12 | wk13 | wk14 | wk15 | wk16 | wk17 | wk18 | wk19 | wk20 | wk21 | wk22 | wk23 | wk24 | wk25 | wk26 | wk27 | wk28 | wk29 | wk30 | wk31 | wk32 | wk33 | wk34 | wk35 | wk36 | wk37 | wk38 | wk39 | wk40 | wk41 | wk42 | wk43 | wk44 | wk45 | wk46 | wk47 | wk48 | wk49 | wk50 | wk51 | wk52 | Order start wk | Order frequency (Weeks) |     |                                   |  |
+---------+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+----------------+-------------------------+-----+-----------------------------------+--+
|         |     |     |     |     |     |     |     |     |     |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |                | Min                     | Max | Average                           |  |
|         |     |     |     |     |     |     |     |     |     |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      |                |                         |     | (End wk - Start week)/No of times |  |
| SKU 1   |     |     |     |     |     |     |     |     | y   |      | y    |      | y    |      | y    |      | y    |      | y    |      | y    |      | y    | y    |      |      | y    |      | y    |      | y    |      | y    |      |      |      |      |      | y    |      | y    |      | y    |      | y    |      | y    |      | y    |      | y    |      |              9 | 1                       | 6   | 2.15                              |  |
| SKU 2   |     |     |     |     |     |     | y   |     |     |      |      |      | y    |      |      |      |      |      | y    |      |      |      |      |      | y    |      |      |      |      |      | y    |      |      |      |      |      | y    |      |      |      |      |      | y    |      |      |      |      |      | y    |      |      |      |              1 | 0                       | 0   | 7.29                              |  |
| SKU 3   |     |     |     |     |     |     |     |     |     |      |      |      |      |      | y    |      |      |      |      |      |      |      |      |      |      |      |      |      |      |      | y    |      |      |      |      |      |      |      | y    |      |      |      |      |      |      |      | y    |      |      |      |      |      |             15 | 8                       | 15  | 9.25                              |  |
+---------+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+----------------+-------------------------+-----+-----------------------------------+--+

Answer 1

正如前面提到的那样，@ Barry Houdini解决了找到由here优雅分隔的最长零序列的问题<{3}}

您只需稍微更改它以检查由'y'分隔的重复空白单元格。唯一的事情是你不想在第一个'y'之前包含单元格，并且（虽然这不清楚）可能不希望在最后一个'y'之后包含空白单元格。

MIN的公式变为

=MIN(IF((ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))>1)*(ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))<COUNTA(B4:BA4)+1),FREQUENCY(IF(B4:BA4="",COLUMN(B4:BA4)),IF(B4:BA4="y",COLUMN(B4:BA4)))))+1

并且MAX的公式变为（相同）

=MAX(IF((ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))>1)*(ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))<COUNTA(B4:BA4)+1),FREQUENCY(IF(B4:BA4="",COLUMN(B4:BA4)),IF(B4:BA4="y",COLUMN(B4:BA4)))))+1

你需要加1以使结果与问题一致，因为@Barry的公式计算空白数，但OP需要两个连续y之间的间隔。生成一个ny + 1个元素的数组，其中ny是y的个数。这是因为FREQUENCY函数返回一个包含n + 1个元素的数组，其中n是切割点的数量（documentation中的bins_array，因为包含y的单元格的列数用作切割点，因此有ny个

这些都是数组公式，需要输入 Ctrl Shift Enter

平均值的公式只是

=(COLUMNS(B4:BA4)-MATCH("y",B4:BA4,0))/COUNTA(B4:BA4)

计算Excel

1 个答案: