I want to make the total of values every each member or names in every each their own group at the first match (or after blank space) or highest values positions of each them on column "D" according to column "B" with the result's row of an output like the exactly as an EXPECT OUTPUT as act of what I've just created on column "E". That's the replace a little bit down of just only one row from the column "B" positions or row must be the same as the column "C" and "D". Could we do this anyway ?
My achievements: I feel I've tried this before and got succeed to achieve this but I've forgot how to solve this when that happened. But it's look like kinda this code of formula:
=FILTER(IF(IFERROR(MATCH($B$3:$B;$B:$B;0);0)=ROW($B$3:$B);SUMIF($B$3:$B;$B$3:$B;$D$3:$D);"");$B$3:$B<>"0")
I don't know if I'm right or wrong but please see the table I'd created at the down below this and also see how I expected with that and feel free as well to edit to my doc file of google sheet I attached down below this.
THIS HERE YOU CAN EDIT TO MY SAMPLE G.SHEET TO SOLVE THIS MY QUIZ. THANKS IN ADVANCE!
A
B
C
D
E
1
2
N U M B
I D - M E M B E R
I D - C O D E
V A L U E S
E X P E C T O U T P U T
3
4
4
JYFI7
5
JYFI7
J3573
3
6
6
JYFI7
IYR
1
7
JYFI7
F498S
2
8
9
3
DFJ9F11
10
DFJ9F11
C684J
7
8
11
DFJ9F11
J58
1
12
13
2
H684K
14
H684K
JF585
2
2
15
16
1
FJSR
17
FJSR
4684
7
16
18
FJSR
834
1
19
FJSR
49
2
20
FJSR
9835
6
Here's a possible solution:
=ARRAYFORMULA(LAMBDA(cusum,IF(SCAN(,cusum,
LAMBDA(acc,cur,if(cur="",,acc+1)))=1,cusum,))
(SORT(SCAN(,SORT(D3:D,ROW(D3:D),0),
LAMBDA(acc,cur,if(cur="",,acc+cur))),ROW(D3:D),0)))
You can find it in tab 'z' cell F3.
Related
Using this table:
A
B
C
D
1
2
3
4
5
6
7
8
9
10
11
12
In Google Sheets if I do this here in column E:
={A1:B3;C1:D3}
Teremos:
E
F
1
2
5
6
9
10
3
4
7
8
11
12
But the result I want is this:
E
F
1
2
3
4
5
6
7
8
9
10
11
12
I tried multiple options with FLATTEN, but none of them returned what I wanted.
Well you can try:
=WRAPROWS(TOCOL(A1:D3),2)
You could try with MAKEARRAY
=MAKEARRAY(ROWS(A1:D3)*2,2,LAMBDA(r,c,INDEX(FLATTEN(A1:D3),c+(r-1)*2)))
GENERAL ANSWER
For you or anyone else: to do something similar but with a variable number of columns of origin or of destination, you can use this formula. Changing the range and amount of columns at the end of LAMBDA:
=LAMBDA(range,cols,MAKEARRAY(ROWS(range)*ROUNDUP(COLUMNS(range)/cols),cols,LAMBDA(r,c,IFERROR(INDEX(FLATTEN(range),c+(r-1)*cols)))))(A1:D3,2)
you can do:
={FLATTEN({A1:A3, C1:C3}), FLATTEN({B1:B3, D1:D3})}
for more columns, it could be automated with MOD
I would like to design an organization chart and I am building a supervisor tree like this.
9 8 7 6 5 4 3 2 1 0
A B C
D
E F B C
By using vlookup I got the table like above.
Now I would like the table to be like this:
9 8 7 6 5 4 3 2 1 0
A B C
D
E F B C
I tried
=IFS(ISBLANK(I2),H2,ISBLANK(H2),G2,ISBLANK(G2),F2,ISBLANK(F2),E2,ISBLANK(E2),D2,ISBLANK(D2),C2,ISBLANK(C2),B2,ISBLANK(B2),A2)
But it turned out to be #REF
I just would like to move the data to the right, to get rid of those empty cells.
Does anyone know how to do it? Or can we realize it in BigQuery?
try:
=ARRAYFORMULA(TRIM(SUBSTITUTE(IFERROR(SPLIT(REPT("♦♠",
IF(A2:A<>"", MMULT(IF(A2:J<>"", 0, 1), TRANSPOSE(COLUMN(A2:J)^0)), ))&
TRANSPOSE(QUERY(TRANSPOSE(IF(A2:J<>"", "♦"&A2:J, )),,999^99)), "♦")), "♠", )))
Example
3 2 5 5
a b c d
Joining first two
5 | 5 5
3 2 | c d
a b |
I have to put the new tree of five into the queue
Am I obligated to put it in the end like this:
5 5 5
c d / \
3 2
a b
Or can I put it in the beginning:
5 5 5
3 2 c d
a b
Or even in the middle of 'c' and 'd'
Is it my choice or is there a rule?
It's not your choice, the Queue needs to be sorted at all times (by it's number of occurrences and in case of equal number of occurrences by the depth of the tree). So it needs to be inserted where it belongs into the order.
This is needed to pick the sub-trees with the least amount of occurrences and if there is choice the most shallow one of them by simply pop-ing them.
If you simply resort after every insertion (this is inefficient and should not be done) the position obviously doesn't matter.
Yes, it's your choice. Whichever way you will get an optimal Huffman code, even though two resulting codes can be manifestly different.
You can get:
a - 00
b - 01
c - 10
d - 11
or you can get:
a - 111
b - 110
c - 10
d - 0
Now if I multiply the number of bits in each symbol times the number of occurrences, I get for the first code: 2*3 + 2*2 + 2*5 + 2*5 = 30 bits. For the second code: 3*3 + 3*2 + 2*5 + 1*5 = 30 bits. So both codes will code the original message to exactly 30 bits.
I wish to make a formula to sum up the value with 2 criteria, example show as below:-
A B C D E
1 1-Apr 2-Apr 3-Apr 4-Apr
2 aa 1 4 7 10
3 bb 2 5 8 11
4 cc 3 6 9 12
5
6 Criteria 1 bb
7 Range start 2-Apr-16
8 Range End 4-Apr-16
9 Total sum #VALUE!
tried formula
1 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(C7,B1:E1,0)))
* Only return 1 cell value
2 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(">="&C7,B1:E1,0)))
* Showed N/A error
3 SUMIFS(B2:E4,A2:A4,C6,B1:E1,">="&C7,B1:E1,"<="&C8)
* Showed #Value error
Hereby I attached a link of picture for better understanding :
Can anyone help me on the formula?
I figured out the solution with step evaluation:
=SUMIF(B1:F1,">="&C7,INDEX(B2:F4,MATCH(C6,A2:A4,0),0)) -
SUMIF(B1:F1,">"&C8,INDEX(B2:F4,MATCH(C6,A2:A4,0),0))
I have written this formula below. I do not know the correct part of this formula that will add the numbers I have in Column AB2:AB552. As it is, this formula is counting the number of cells in that range that has numbers in it, but I need it to total those numbers as my final result. Any help would be great.
=COUNTIFS(Cases!B2:B552,"1",Cases!G2:G552,"c*",Cases!X2:X552,"No",**Cases!AB2:AB552,">0"**)
Assuming you don't actually need the intermediate counts, the sumifs function should give you the final result:
=SUMIFS(Cases!AB2:AB552,Cases!B2:B552,1,Cases!G2:G552,"c",Cases!X2:X552,"No",Cases!AB2:AB552,">0")
Testing this with some limited data:
Row B G X AB
2 2 a No 10
3 1 c No 24
4 2 c No 4
5 1 c No 0
6 1 a Yes 9
7 2 c No 12
8 2 c No 6
9 2 b No 0
10 1 b No 0
11 1 a No 10
12 2 c No 6
13 1 c No 20
14 1 c No 4
15 1 b Yes 22
16 1 b Yes 22
the formula above returned 48, the sum of AB3, AB13, and AB14, which were the only rows matching all 4 criteria