I have a constant number X. I also have two numbers that add up to it. How can I make it so that if I change one number, the other number automatically changes so that it still adds up to X.
I have tried to take subtract the one number from X and add it to the other number, but instead I got two numbers in the thousands.
Assuming your constant value is 10, you can set this in a cell and make all your other calculations based on it.
For example, you can have cell C2 containing your constant, in this example, 10
Then in C4 you can have the number which you change, and the value of C5 will be equal to the value of the constant minus the value in C4.
You can then finally do your sum wherever you want, adding up the values of C4 and C5.
Here's an example Spreadsheet:
Untitiled spreadsheet ☆
File Edit View Insert Format Data Tools Extensions Help Last edit was 2 minutes ago
↶ ↷ 🖶 ⮷ | 100%⯆ | $ % .0 .00 123⯆ | Default(Ro... ⯆ | 10 ⯆ | B | I | S | A |⯐|☰
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
1
2
Contsant:
10
3
4
Number 1:
3
5
Number 2:
=(C2 - C4)
6
7
Sum:
=(C4 + C5)
8
Related
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.
I would like to count if in the column there are N subsequent elements of a certain value (in this example it's 1). So for the below example, it should return true for A, but false for B. Empty cells shall be ignored.
A B
+---+---+
| 1 | 1 |
| 1 | 1 |
| 1 | 1 |
| | |
| 1 | 2 |
| | 1 |
+---+---+
| Y | N | <-- RESULT
+---+---+
Ideally would be if I could mark all of those 4 or more subsequent cells.
Have a look at this one as well - you need to pull it across for columns B, C etc.
=ArrayFormula(max(frequency(if(filter(A:A,A:A<>"")=1,sequence(count(A:A)),""),if(filter(A:A,A:A<>"")<>1,sequence(count(A:A)),"")))>=4)
So how does this work? It goes all the way back to #Barry Houdini, who as far as I know was the first to post this elegant method to find the longest sequence of repeated values within a range in Excel.
It uses Frequency. Frequency counts the number of values which fall within a series of bins defined by cut points
Frequency(<Values>,<Cut points>)
So if your cut points were 10,20,30 and your values were 12,15,25,35,36,37 you would expect the counts to be 2,1 and 3.
#Barry's insight was to realise that if you made the row numbers for values that you didn't want to count be the cut points and the row numbers for values you did want to count (in this case 1) be the values, then you could use Frequency to count up the number of consecutive values of interest. This led him to this formula (in his case looking for consecutive zeroes bounded by ones):
=MAX(FREQUENCY(IF((A2:A100=0)*(A2:A100<>""),ROW(A2:A100)),IF(A2:A100=1,ROW(A2:A100))))
which is easily adapted for the present situation simply by changing 0 to 1 and 1 to <>1, filtering out the blanks and using Sequence instead of row number.
try:
=ARRAYFORMULA(IF(TRANSPOSE(MMULT(TRANSPOSE(N(REGEXMATCH(""&
IF(A1:B6="", 1, A1:B6), "1"))),
SEQUENCE(ROWS(A1:B6), 1)^0))/ROWS(A1:B6)=1, "Y", "N"))
Is it possible to perform an arbitrary calculation (eg. A2*B2) on a set of rows and obtain the cumulative sum along the way using ARRAYFORMULA? For example, in the following sheet we have numbers (column A), multipliers (column B), the result of multiplying them (column C), and a cumulative tally (column D):
| A B C D E F
-------------------------------------------------------------------------------
1 | number multiplier result cumulative array formula array formula sum?
2 | 3 4 12 12 12
3 | 2 4 8 20 8
4 | 10 1 10 30 10
5 | 7 9 63 93 63
I can use ARRAYFORMULA in cell E2 (specifically, ARRAYFORMULA(A2:A5*B2:B5)) to do the multiplication. Is it possible to use ARRAYFORMULA (or alternative tool) in cell F2 to show the cumulative total?
use:
=ARRAYFORMULA(IF(A2:A="",,MMULT(TRANSPOSE((ROW(A2:A)<=
TRANSPOSE(ROW(A2:A)))*A2:A*B2:B), SIGN(B2:B))))
Calculate the cumulative sum with the SCAN and LAMBDA functions:
=SCAN(0, F5:F, LAMBDA(accumulated_value, cell_value, accumulated_value + cell_value))
This will run faster as it runs with linear complexity (O(N)) compared to the ARRAYFORMULA solution, which runs in quadratic time (O(N**2)).
Where:
0 is the initial value of the cumulative sum
F5:F is the range to sum over
LAMBDA(accumulated_value, cell_value, accumulated_value + cell_value)) is the function that calculates the sum at each cell
Sample File
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 am writing a sheet where I am trying to create a multi level Index that searches through 5 different columns with 3 pieces of data. So for example:
x = 40
y = 5000
z = 20000
Column1 | Column2 | Column3 | Column4 | Column5 | Column6
13 | 29 | 0 | 0 | 0 | Yes
30 | 870 | 0 | 0 | 0 | No
10 | 870 | 0 | 30000 | 1 | Blue
10 | 870 | 30001 | 100000 | 1 | Yes
10 | 870 | 100001 | 300000 | 1 | Unknown
Here's a sample set of my data, what I need is to compare
the variable x to columns 1 and then 2 (x must fall between these values)
variable y to columns 3 and 4 (y must fall between these values)
and then finally z to column 5 (z must be above these values)
In each of these cases I need to know if the the variable is either lower than or higher than . Finally, I need the matching data from column 6 to be returned as a result in my sheet. At the moment I have a simply IMMENSE list of nested if statements which consider all of these criteria separately but it doesn't lend itself very well to editing when changes need to be made to the values.
I've looked at every single page on the internet (every... single... page...) and can't seem to find the solution to my issue. Most solutions I have found are either using a single data point, using multiple data points against a single range or simply don't seem to work. The latest iteration I have tried is:
=INDEX('LTV Data'!$N$3:$N$10, MATCH($D$5 & $G$8 & $G$12, ARRAYFORMULA($D$5 <= 'LTV Data'!$H$3:$H$10 & $D$5 >= 'LTV Data'!$I$3:$I$10 & $G$12 <= 'LTV Data'!$J$3:$J$10 & $G$12 >= 'LTV Data'!$K$3:$K$10 & $G$8 <= 'LTV Data'!$L$3:$L$10), 0), 7)
But this only produces an error as the separate values I want to test against are concatenated and the Match can't find that string. I'm also unsure about the greater than and less than symbols as to how valid that syntax is. Is anyone able to shed some light on how I can achieve the result I need in a more elegant way than the mass of IFS, ANDS + ORs I have right now? Like I said, it works but it sure ain't pretty ;)
Thanks a bunch in advance!
ETA: So with the above variables the result I would like would be 'Blue'. This is because x falls between columns 1 and 2, y falls between columns 3 and 4 and z is higher than column 5 on the third row. This is all contained in the MATCH statement in the example code above. Please see the MATCH statement to see the comparisons I am trying to make.
You need to put the different criteria together using multiplication if you want to get the effect of an AND in an array:
=INDEX(F2:F10,MATCH(1,(A2:A10<x)*(B2:B10>x)*(C2:C10<y)*(D2:D10>y)*(E1:E10<z),0))
or
=INDEX(F2:F10,MATCH(1,(A2:A10<=x)*(B2:B10>=x)*(C2:C10<=y)*(D2:D10>=y)*(E1:E10<=z),0))
to include the equality (I have used named ranges for x, y and z).
Works in Google Sheets and (if entered as an array formula) Excel.
In Google Sheets you also have the option of using a filter
=filter(F2:F10,A2:A10<=x,B2:B10>=x,C2:C10<=y,D2:D10>=y,E2:E10<=z)
but then you aren't guaranteed to get just one row.