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Well, if I didn;t see it myself I'd never beleive it, but closing and
reopening Excel did the trick. I cannot explain why I had trouble earlier, but at least it's nice to know that my solution was tenable. "Wazooli" wrote: You know - I have been around Windows long enough to know that sometimes a good swift restart is all things take. I am going to try that. Maybe some other workbook I have open is "confusing" Excel. And I have checked 3 times to make sure everything is in numeric formatting. No conditional formatting exists, so I've no idea. Like I said earlier, I like these problems because they chellenge my Excel skills. Ultimately, however, I will never have the need to do a manipulation like the OP. "Bob Phillips" wrote: You're lucky I didn't call you Wazzer :-) My A part resolves to {1;1;0;0;0;0;0;0;1;1;1;1;1;1;1;1;1;1;1;1} I have no idea what you have done, are you sure they are not text imn column A? -- HTH RP (remove nothere from the email address if mailing direct) "Wazooli" wrote in message ... Lol - haven't been called Waz since college. Anyway, why do the first resolve to 0;0;0;...? This seems to me to be a mistake. If the query is A=1, and B=0, there are 3 in the original data posted. Shouldn't the first expression resolve to 0;0;1;1;1;0;0;0;0;...? "Bob Phillips" wrote: Waz, I have replicated what you did and it still works for me. I get lots of 0 0 entries, because A10-A20 all resolve to 0 0 (that is why my formula tested for them), but it works. The B part resolves to {TRUE;FALSE;TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;TRUE ;TRUE;TRUE;TRUE;TRUE;TRUE; TRUE;TRUE;TRUE;TRUE;TRUE;TRUE} for me, or {1;0;1;1;0;0;0;0;1;1;1;1;1;1;1;1;1;1;1;1} if I use the double unary. Thnis is in cell E2. -- HTH RP (remove nothere from the email address if mailing direct) "Wazooli" wrote in message ... I used A1:B20 because JulieD suggested those. I merely filled in to complete the ranges. If my table is D1:I6, with cell D1 being empty (the numbering starts beneath and to the right), and I enter =SUMPRODUCT(--($A$1:$A$20=$D2),--($B$1:$B$20=E$1), filldown and then to the right, I get all zeroes. When I look at how Excel evaluates the formula, I can verify that the '--' is working, but the values for the first part of the expression resolve to '0;0;0;0;...' The second part is correct. "Bob Phillips" wrote: It does! I just followed Domenic's instructions and it worked perfectly for me. Where does A20 and B20 com e into it in your reply? -- HTH RP (remove nothere from the email address if mailing direct) "Wazooli" wrote in message ... Doesn't. When I look at how Excel evaluates the formula, it seems to have a problem with $A$20 representing the first range. The sumproduct function gives all zeroes. The range $B$20, on the other hand, presents no problem. wazooli "Domenic" wrote: Assumptions: 1) Sheet1 contains your source table 2) the first row in Sheet2 contains your numbers (0 through 4) starting at B1 3) the first column in Sheet2 contains your other set of numbers (0 through 4) starting at A2 Formula: Sheet2!B2, copied across and down: =SUMPRODUCT(--(Sheet1!$A$1:$A$9=$A2),--(Sheet1!$B$1:$B$9=B$1)) Hope this helps! In article 34, Marc Fleury wrote: I'm pretty sure I need an array function for this, but nothing I do works. My data is basically two colums: A B 0 0 0 2 1 0 1 0 1 4 2 1 3 3 3 3 4 0 etc Now, I need to create a table that counts how many instance of each pair there are. Like so: 0 1 2 3 4 ___________ 0 |1 0 1 0 0 1 |2 0 0 0 1 2 |0 1 0 0 0 3 |0 0 0 2 0 4 |1 0 0 0 0 So the upper left cell counts how many times there is a 0 in column A AND a 0 in colum B (once). The next cell to the right counts how many times there is a 0 in column A and a 1 in colum B (zero times). Etc. The closest that I have come is {=SUM(IF(($A$1:$A$10=0)+($B$1:$B$10=0),1,0))} I actually have reference cells for the "0" that's being compared to, so that I can use the same formula for every cell in the table, but I know how to do that part. The problem with this formula is that it counts how many times there is a zero in column A OR a zero in column B. HELP! -- Marc. |
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