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Default Interpretation for: =INDEX(MATCH(1 <--one

Interpretation for:
=INDEX(F1:F100,MATCH(1,(A1:A100="Custname")*(D1:D1 00="State"),0)) Enter with
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Dear All,

I came across this formula and was wondering how to interpret:

Match(1,(xxxxxx)*(xxxxxxx) = why 1 and not 2 or 3 or so forth (i tried 2 and
3 but doesn't work)


I tested the formula in my workbook

=INDEX(Dbase!B1:B22,MATCH(1,(Dbase!C1:C22=B1)*(Dba se!A1:A22=B2),0)) = works

=INDEX(Dbase!B1:B22,MATCH(2,(Dbase!C1:C22=B1)*(Dba se!A1:A22=B2),0)) =
doesn't work


Thanks.
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Default Interpretation for: =INDEX(MATCH(1 <--one

Match(1,(xxxxxx)*(xxxxxxx),0) = why 1 and not 2 or 3 ...

It's because the lookup array in the MATCH, this part:
(A1:A100="Custname")*(D1:D100="State")
will resolve to an array of ones/zeros, eg: {0;0;1;0;0;0;0;...}
depending on where the dual criteria is satisfied (1's) or not (0's)

Using the lookup value: 1 in MATCH would hence give us the (1st) matching
position within the array where the dual criteria is satisfied. It's presumed
of course, that there should be only a single matching position (ie a
single/unique instance of 1) to be returned within the array. That position
(a number) returned is then used by the INDEX part of the expression (eg:
INDEX(F1:F100, ...) to yield the required result.

Hope the above clarifies it.
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
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Default Interpretation for: =INDEX(MATCH(1 <--one

Thanks Max.

I get it, the 1 and 0. Looking back at the function box, the match criteria
is 1. That's how it was matched.

Appreciate it!!

"Max" wrote:

Match(1,(xxxxxx)*(xxxxxxx),0) = why 1 and not 2 or 3 ...


It's because the lookup array in the MATCH, this part:
(A1:A100="Custname")*(D1:D100="State")
will resolve to an array of ones/zeros, eg: {0;0;1;0;0;0;0;...}
depending on where the dual criteria is satisfied (1's) or not (0's)

Using the lookup value: 1 in MATCH would hence give us the (1st) matching
position within the array where the dual criteria is satisfied. It's presumed
of course, that there should be only a single matching position (ie a
single/unique instance of 1) to be returned within the array. That position
(a number) returned is then used by the INDEX part of the expression (eg:
INDEX(F1:F100, ...) to yield the required result.

Hope the above clarifies it.
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---

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Max Max is offline
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Default Interpretation for: =INDEX(MATCH(1 <--one

welcome, glad it clarifies.
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---
" wrote in
message ...
Thanks Max.

I get it, the 1 and 0. Looking back at the function box, the match
criteria
is 1. That's how it was matched.

Appreciate it!!



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