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#1
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Consecutive occurences
Hi all
have data displayed on column C as follows: 3 5 4 5 3 4 4 2 4 3 4 4 3 5 6 4 4 3 3 4 2 3 2 4 4 4 4 4 Is there a way to count the large number of occurences of the same number, which in this case should be number 4 five times?? Tks in advance |
#2
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Consecutive occurences
This will work if there's only one number with the most number of occurrances
C1: =MODE(A1:A10) D1: =COUNTIF(A1:A10,C1) -- HTH, Barb Reinhardt "Antonio" wrote: Hi all have data displayed on column C as follows: 3 5 4 5 3 4 4 2 4 3 4 4 3 5 6 4 4 3 3 4 2 3 2 4 4 4 4 4 Is there a way to count the large number of occurences of the same number, which in this case should be number 4 five times?? Tks in advance |
#3
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Consecutive occurences
Tks 4 the help. but the result I get is the same as if I only use the countif
function "Barb Reinhardt" wrote: This will work if there's only one number with the most number of occurrances C1: =MODE(A1:A10) D1: =COUNTIF(A1:A10,C1) -- HTH, Barb Reinhardt "Antonio" wrote: Hi all have data displayed on column C as follows: 3 5 4 5 3 4 4 2 4 3 4 4 3 5 6 4 4 3 3 4 2 3 2 4 4 4 4 4 Is there a way to count the large number of occurences of the same number, which in this case should be number 4 five times?? Tks in advance |
#4
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Consecutive occurences
How about adapting the expression that Ron C. gave you in an earlier posting?
To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#5
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Consecutive occurences
then combine them together?
=MODE(A3:A22)&" = "&COUNTIF(A3:A22,MODE(A3:A22)) "Antonio" wrote: Tks 4 the help. but the result I get is the same as if I only use the countif function "Barb Reinhardt" wrote: This will work if there's only one number with the most number of occurrances C1: =MODE(A1:A10) D1: =COUNTIF(A1:A10,C1) -- HTH, Barb Reinhardt "Antonio" wrote: Hi all have data displayed on column C as follows: 3 5 4 5 3 4 4 2 4 3 4 4 3 5 6 4 4 3 3 4 2 3 2 4 4 4 4 4 Is there a way to count the large number of occurences of the same number, which in this case should be number 4 five times?? Tks in advance |
#6
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Consecutive occurences
Tks Max.
Tried it, but, got the result n/a..., will copy past your to see the result Tks "Max" wrote: How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#7
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Consecutive occurences
Tried it, but, got the result n/a...,
Try this quick sample which shows it working the way it should: http://www.freefilehosting.net/download/3g6h9 Max Consecutives in Col.xls -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#8
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Consecutive occurences
What does "the large number of occurences of the same number" mean?
Tyro "Antonio" wrote in message ... Hi all have data displayed on column C as follows: 3 5 4 5 3 4 4 2 4 3 4 4 3 5 6 4 4 3 3 4 2 3 2 4 4 4 4 4 Is there a way to count the large number of occurences of the same number, which in this case should be number 4 five times?? Tks in advance |
#9
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Consecutive occurences
That can be reduced to:
=MAX(FREQUENCY(IF(C1:C27=C2:C28,ROW(C1:C27)),IF(C1 :C27<C2:C28,ROW(C1:C27))))+1 Note that it will count empty cells as consecutives *and* empty cells and numeric 0 will evaluate as the same. -- Biff Microsoft Excel MVP "Max" wrote in message ... How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#10
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Consecutive occurences
Tks Max + Valko
Both formulas work, in giving the largest number of occurences, but unable to specify if it is from number one, two, three.... Any ideas?? Tks "Max" wrote: Tried it, but, got the result n/a..., Try this quick sample which shows it working the way it should: http://www.freefilehosting.net/download/3g6h9 Max Consecutives in Col.xls -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#11
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Consecutive occurences
Hi
the formula presently used as follows: =MAX(FREQUENCY(IF(B9:OFFSET(INDEX(B:B;MATCH(99^99; B:B));-1;0)=B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0);R OW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)));IF(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)<B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0); ROW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)))))+1 but still unable to get the desired result, that for the given set of numbers should be: 1 = 0 (consecutive) 2 = 0 (consecutive) 3 = 2 (Consecutive) 4 = 5 (Consecutive) 5 = 0 6 = 0 Tks for the assistance n bst rgds "T. Valko" wrote: That can be reduced to: =MAX(FREQUENCY(IF(C1:C27=C2:C28,ROW(C1:C27)),IF(C1 :C27<C2:C28,ROW(C1:C27))))+1 Note that it will count empty cells as consecutives *and* empty cells and numeric 0 will evaluate as the same. -- Biff Microsoft Excel MVP "Max" wrote in message ... How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#12
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Consecutive occurences
Assume your range of numbers are in B9:B30.
List the unique numbers starting in D9 on down. Enter this array formula** in E9 and copy down as needed: =MAX(FREQUENCY(IF(B$9:B$30=D9,ROW(B$9:B$30)),IF(B$ 9:B$30<D9,ROW(B$9:B$30)))) ** array formulas need to be entered using the key combination of CTRL,SHIFT,ENTER (not just ENTER) -- Biff Microsoft Excel MVP "Antonio" wrote in message ... Hi the formula presently used as follows: =MAX(FREQUENCY(IF(B9:OFFSET(INDEX(B:B;MATCH(99^99; B:B));-1;0)=B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0);R OW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)));IF(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)<B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0); ROW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)))))+1 but still unable to get the desired result, that for the given set of numbers should be: 1 = 0 (consecutive) 2 = 0 (consecutive) 3 = 2 (Consecutive) 4 = 5 (Consecutive) 5 = 0 6 = 0 Tks for the assistance n bst rgds "T. Valko" wrote: That can be reduced to: =MAX(FREQUENCY(IF(C1:C27=C2:C28,ROW(C1:C27)),IF(C1 :C27<C2:C28,ROW(C1:C27))))+1 Note that it will count empty cells as consecutives *and* empty cells and numeric 0 will evaluate as the same. -- Biff Microsoft Excel MVP "Max" wrote in message ... How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#13
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Consecutive occurences
Tks all for the precious help...
with this one obtained the desired result "T. Valko" wrote: Assume your range of numbers are in B9:B30. List the unique numbers starting in D9 on down. Enter this array formula** in E9 and copy down as needed: =MAX(FREQUENCY(IF(B$9:B$30=D9,ROW(B$9:B$30)),IF(B$ 9:B$30<D9,ROW(B$9:B$30)))) ** array formulas need to be entered using the key combination of CTRL,SHIFT,ENTER (not just ENTER) -- Biff Microsoft Excel MVP "Antonio" wrote in message ... Hi the formula presently used as follows: =MAX(FREQUENCY(IF(B9:OFFSET(INDEX(B:B;MATCH(99^99; B:B));-1;0)=B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0);R OW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)));IF(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)<B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0); ROW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)))))+1 but still unable to get the desired result, that for the given set of numbers should be: 1 = 0 (consecutive) 2 = 0 (consecutive) 3 = 2 (Consecutive) 4 = 5 (Consecutive) 5 = 0 6 = 0 Tks for the assistance n bst rgds "T. Valko" wrote: That can be reduced to: =MAX(FREQUENCY(IF(C1:C27=C2:C28,ROW(C1:C27)),IF(C1 :C27<C2:C28,ROW(C1:C27))))+1 Note that it will count empty cells as consecutives *and* empty cells and numeric 0 will evaluate as the same. -- Biff Microsoft Excel MVP "Max" wrote in message ... How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
#14
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Consecutive occurences
You're welcome. Thanks for the feedback!
-- Biff Microsoft Excel MVP "Antonio" wrote in message ... Tks all for the precious help... with this one obtained the desired result "T. Valko" wrote: Assume your range of numbers are in B9:B30. List the unique numbers starting in D9 on down. Enter this array formula** in E9 and copy down as needed: =MAX(FREQUENCY(IF(B$9:B$30=D9,ROW(B$9:B$30)),IF(B$ 9:B$30<D9,ROW(B$9:B$30)))) ** array formulas need to be entered using the key combination of CTRL,SHIFT,ENTER (not just ENTER) -- Biff Microsoft Excel MVP "Antonio" wrote in message ... Hi the formula presently used as follows: =MAX(FREQUENCY(IF(B9:OFFSET(INDEX(B:B;MATCH(99^99; B:B));-1;0)=B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0);R OW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)));IF(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)<B10:OFFSET(INDEX(B:B;MATCH(99^99;B:B));0;0); ROW(B9:OFFSET(INDEX(B:B;MATCH(99^99;B:B));-1;0)))))+1 but still unable to get the desired result, that for the given set of numbers should be: 1 = 0 (consecutive) 2 = 0 (consecutive) 3 = 2 (Consecutive) 4 = 5 (Consecutive) 5 = 0 6 = 0 Tks for the assistance n bst rgds "T. Valko" wrote: That can be reduced to: =MAX(FREQUENCY(IF(C1:C27=C2:C28,ROW(C1:C27)),IF(C1 :C27<C2:C28,ROW(C1:C27))))+1 Note that it will count empty cells as consecutives *and* empty cells and numeric 0 will evaluate as the same. -- Biff Microsoft Excel MVP "Max" wrote in message ... How about adapting the expression that Ron C. gave you in an earlier posting? To suit it here for your col C data in C1:C28, try array-entered: =MAX(FREQUENCY(IF(C1:C27=(C2:C28),ROW(C1:C27)),IF( C1:C27<(C2:C28),ROW(C1:C27))))+(COUNT(1/(C1:C27=(C2:C28)))0) -- Max Singapore http://savefile.com/projects/236895 xdemechanik --- |
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