每当位置和日期等于特定值时,我都会尝试计算不同的 ID 值。
有一些文章/问题概述了一些不同的方法来做到这一点(例如,如何在 Excel 中计算具有两个条件的唯一值),但由于某种原因,似乎没有一个对我有用(这可能是因为我'我做错了什么)。
我在excel中的表格是这样设置的
+------------------+--------------+------------+
| Col A - Location | Col B - Date | Col C - ID |
+------------------+--------------+------------+
| A | Mar 2018 | 1 |
| B | Mar 2018 | 2 |
| C | Mar 2018 | 3 |
| A | Mar 2018 | 4 |
| B | Mar 2018 | |
| C | Mar 2018 | 5 |
| A | Mar 2018 | 6 |
| B | Apr 2018 | 1 |
| C | Apr 2018 | 2 |
| C | Apr 2018 | 4 |
| C | Mar 2018 | 5 |
| A | Mar 2018 | 1 |
| B | Mar 2018 | 5 |
| B | Mar 2018 | 8 |
| B | Mar 2018 | |
| C | Mar 2018 | 1 |
| B | Mar 2018 | 3 |
+------------------+--------------+------------+
我尝试了 4 种我在网上找到的不同计算方法,并根据我的数据进行了修改:
Calc 1 (run by just hitting ENTER):
=COUNTIFS(A:A, "A",C:C,C:C,B:B, "Mar 2018")
Result = 2
Calc 2 (running by hitting CTRL + SHFT + ENTER):
=SUMPRODUCT( ( (C:C <> "") * (A:A = "A") * (B:B = "Mar 2018") ) / COUNTIFS(C:C,C:C & "",A:A, "A",B:B, "Mar 2018"))
Result = 0
Calc 3 (running by hitting CTRL + SHFT + ENTER):
=SUM(IF(FREQUENCY(IF(C:C<>"", MATCH(C:C,C:C,0)),ROW(C:C)-ROW(C2)+1),1))
Result = 0
Calc 4 (running by hitting CTRL + SHFT + ENTER):
=SUM(IF(FREQUENCY(IF((A:A="A")*(B:B="Mar 2018")*(C:C<>""),MATCH(C:C,C:C,0)),ROW(C:C)-MIN(ROW(C:C))+1),1))
Result = 0
(This one keeps giving an error message about running out of resources while trying to calculate)
我试图找到的实际结果是,对于值为 'A' 的 'Col A - Location' 和值为 'Mar 2018' 的 'Col B - Date' 的唯一 # of 'Col C - ID ' 应该是 3。