在Google工作表中使用ifs，isblank和AND / OR的arrayformula

Question

我想要实现的是一个检查顶行为＃34的公式;所有都是空白的＆＃34;或者＆＃34;所有都不是空白＆＃34;在指定范围内。根据条件，具有公式的单元格会返回3个单词中的1个单词或将其留空。

我用以下图像中的颜色进一步说明了这一点

到目前为止的公式：

=ArrayFormula(ifs((not(isblank(A1:B1*C1:E1*G1:I1))*(isblank(J1:L1*N1:P1))),"SEND", not(isblank(A1:B1*J1:L1*N1:P1))*isblank(C1:E1*G1:I1),"RECEIVE", not(isblank(A1:B1*C1:E1*G1:I1*J1:L1*N1:P1)),"TRANSFER", ISBLANK(A1:B1+C1:E1+G1:I1+J1:L1+N1:P1),""))

我不明白配方有什么问题。例如，在填充J1:L1*N1:P1的情况下，isblank(J1:L1*N1:P1)会在我希望false时返回true。

Answer 1

以下是各个IF声明：

=IF((COUNTIF(A1:E1,"?*")=5)*(COUNTIF(G1:L1,"?*")=6)*(COUNTIF(N1:P1,"?*")=3)=TRUE,"TRANSFER","")

=IF((COUNTIF(A1:E1,"<>")=5)*(COUNTIF(G1:I1,"<>")=3)*(COUNTIF(J1:L1,"")=3)*COUNTIF(N1:P1,"")=3,"SEND","")

=IF((COUNTIF(C1:E1,"")=3)*(COUNTIF(G1:I1,"")=3)*(COUNTIF(J1:L1,"<>")=3)*COUNTIF(N1:P1,"<>")=3,"RECEIVE","")

以下是嵌套的IF语句。

=IF((COUNTIF(A1:E1,"?*")=5)*(COUNTIF(G1:L1,"?*")=6)*(COUNTIF(N1:P1,"?*")=3)=TRUE,"TRANSFER",IF((COUNTIF(A1:E1,"<>")=5)*(COUNTIF(G1:I1,"<>")=3)*(COUNTIF(J1:L1,"")=3)*COUNTIF(N1:P1,"")=3,"SEND",IF((COUNTIF(C1:E1,"")=3)*(COUNTIF(G1:I1,"")=3)*(COUNTIF(J1:L1,"<>")=3)*COUNTIF(N1:P1,"<>")=3,"RECEIVE","")))

Answer 2

所以在Nate的帮助下，我提出了以下公式：

=IFS(ISODD((COUNTBLANK(A1:E1)=0)*(COUNTBLANK(G1:L1)=0)*(COUNTBLANK(N1:P1)=0)),"TRANSFER", (ISODD((COUNTIF(A1:E1,"<>")=5)*(COUNTIF(G1:I1,"<>")=3)*(COUNTIF(J1:L1,"")=3)*COUNTIF(N1:P1,"")=3)),"SEND",ISODD((COUNTIF(C1:E1,"")=3)*(COUNTIF(G1:I1,"")=3)*(COUNTIF(J1:L1,"<>")=3)*(COUNTIF(N1:P1,"<>")=3)),"RECEIVE",true,"")