Oct 02 2018 07:32 AM
Hello.How can I select top numbers from my numbers list that contains less different digits ?for example there are 5 different numbers 3333333,3330333,3330633,3330623,3330628.How can i automatically select top 2 number (3333333 and 3330333) that contains fewer different digits?Thanks
Oct 02 2018 08:59 AM
Hi
To count unique digits for six digit numbers:
=MMULT({1,1,1,1,1,1,1},--(FREQUENCY(--MID(A1,{1;2;3;4;5;6},1),--MID(A1,{1;2;3;4;5;6},1))>0))
Oct 03 2018 12:59 AM
SolutionSee attached file.
Play with the hurdle and see how the conditional formatting changes.
Oct 03 2018 04:56 AM - edited Oct 03 2018 04:59 AM
Excellent!!! Thank you very very much.
Oct 10 2018 09:53 AM
Now i have got numbers list that contains only 2 and 3 unique digits.Can we add another parameter to this file to be able to select only some digit combinations from these numbers that contains 2-3 unique digits?For example when i add this combination 1133322 to the parameter,the numbers in this combinations (like 66888444,5577799,3366688)can be selected automatically.I have got cominations list ,i would like to add my combinations list to that paramater,not only one combination.Thank you
Oct 10 2018 10:03 AM
You want to match patterns?
I don't think that is possible with formulas. Maybe with VBA. But that's not my field of expertise.
Oct 10 2018 10:15 AM - edited Oct 10 2018 10:43 AM
Yes, from my numbers list i would like select numbers in patterns like these
1133322 |
1133323 |
1133332 |
1211113 |
1222213 |
1222223 |
1222231 |
1222232 |
1222233 |
1222333 |
1223222 |
I played with the formula that you gave and could filter and select some numbers with necessary patterns. for instance i type {1;3;5;6;7} instead of {1;2;3;4;5;6;7} and set hurdle to 1 to get numbers like 8687888.I hide two of the numbers each time and change pattern. but it take too much time in this way.
Oct 03 2018 12:59 AM
SolutionSee attached file.
Play with the hurdle and see how the conditional formatting changes.