excel - Differentiating between cells that have the same information for match function -
lets have these 3 tables show interest rates various different things (e.g. auto loan, mortage, credit cards). "######"s showing there values in cells used calculate numbers @ bottom (0.01, 0.03, etc.). lets range in excel these 3 data tables a1:i6.
| | datatable 1 | | | datatable 2 | | | datatable 3 | | |:------------:|:---------------:|:-------------:|:------------:|:---------------:|:-------------:|:------------:|:---------------:|:-------------:| | low,interest | medium,interest | high,interest | low,interest | medium,interest | high,interest | low,interest | medium,interest | high,interest | |--------------|-----------------|---------------|--------------|-----------------|---------------|:-------------|-----------------|---------------| | ####### | ####### | ####### | ####### | ####### | ####### | ####### | ####### | ####### | | ####### | ####### | ####### | ####### | ####### | ####### | ####### | ####### | ####### | | 0.01 | 0.03 | 0.05 | 0.02 | 0.04 | 0.06 | 0.10 | 0.20 | 0.30 |
i have drop down list in a8 contains values data table 1, data table 2, , data table 3.
lets have table (range k1:m14).
| month | balance | medium interest | |:---------:|:-------:|:---------------:| | january | $100.00 | 3% | | february | $103.00 | 3% | | march | $106.09 | 3% | | april | $109.27 | 3% | | may | $112.55 | 3% | | june | $115.93 | 3% | | july | $119.41 | 3% | | august | $122.99 | 3% | | september | $126.68 | 3% | | october | $130.48 | 3% | | november | $134.39 | 3% | | december | $138.42 | 3% |
i wrote formula determine how 3% gets medium interest column.
=index($a$6:$i$6,match($a$8,$a$1:$i$1,0),match($m$2,$a$2:$i$2,0))
it works when choose data table 1 in drop down list. correctly places 3% medium interest rate data table 1 when choose either of other 2 data tables, invalid cell reference error. appear me problem formula cannot differentiate between low, medium, , high columns different tables.
this need in real spreadsheet.
please try:
=index($a$6:$i$6,match($a$8,$a$1:$i$1,0)+match($m$2,$a$2:$c$2,0)-1)
but note matches must exact, in particular label columnm , corresponding indicators in columnsa:i.
the first match provides index number position of value selected drop-down because data in blocks identifies of 3 blocks. in addition necessary identify column block, achieved matching columnm label within first block - return 1 (low), 2 (medium) or 3 (high). since each block starts low, adding 1 (for low) , subtracting 1 provides offset required other match find anyway (if first column of whichever block) + 0 still first column (low) of whichever block. 2 (medium) -1 1 index function takes 1 more index number returned first match.
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