I'm doing some work with arithmetic sequences modulo P, in which the sequences become periodic under the modulo. My worksheet generates a sequence mod P with the first term being 0, the second term being a number K (referencing another cell), and the following terms following the recurrence relation. The period of the sequence (number of values before it repeats itself) is related to the ratio P/K, s, for example, if P=2 and K=1, I get the sequence {0,1,1,0,1,1,0,1,1,...}, which has a period of 3, so when P/K=2, the period is 3.
I currently have a formula which uses the COUNTIF function to count the number of zeroes in the range, which is then divided out of the total range, currently an arbitrary size of 120, and this gives me the correct period for many ratios of P/K. Most of the time, however, the sequence generated exhibits semi-periodicity and sometimes even quasi-periodicity, such as in the case of K=1 and modulo 9: {0,1,1,2,3,5,8,4,3,7,1,8,0,8,8,7,6,4,1,5,6,2,8,1,...}, where P/K=9, the period is 24, and the semi-period is 12 (because of the 0,8,8,... part of the sequence). In such cases, my current COUNTIF formula thinks the full period is 12, even though it should be 24, because it counts the zeroes which define the semi-period.
What I would like to do is adjust the formula so that instead of the criterion for counting being 0, it would only count triplet sequences of cells in the pattern 0,K,K.
My current formula:
=QUOTIENT(120,(COUNTIF(B2:DQ2,0)))
So if I have =QUOTIENT(120,(COUNTIF(B2:DQ2,*X*))) I want the "X", which is currently 0, to reference a specific sequence of cells, namely the first three of the overall series, so something like: =QUOTIENT(120,(COUNTIF(B2:DQ2,(0,C2,D2)))) although obviously that criterion is not in remotely the correct syntax.
I'm not well-versed in writing macros, so that would probably be out of the question.
I would do this with four helper rows plus the final formula. Someone more clever than I am might be able to do it in one cell with an array formula; but compared to array formulas I think the helper rows are easier to understand and, if desired, tweak.
Once this is set up, if you're always going to use three as your criterion, you can hide the helper rows (to hide a row, right-click on the gray number label on the left side of the spreadsheet, and choose "hide").
So your sequence is in row 2, starting in column B. We'll set up the first helper row in row 3, starting in column C. In cell C3 put the formula =C2=$B$2. This will evaluate to FALSE, which is equivalent to 0. Copy and paste that formula all the way to cell DQ3 (or however many columns you want to run it). Cells below a sequence number equal to the first number in the sequence will evaluate to TRUE, which is equivalent to 1.
The next two helper rows are very similar. In cell D4 put the formula =D2=$C$2 and copy and paste to cell DQ4. This row tests which cells are equal to the second number in the sequence.
In cell E5 put the formula =E2=$D$2 and copy and paste to cell DQ5, showing which cells are equal to the third number in the sequence.
The last helper row is a little different, so I left an empty row after the first three helpers. In cell E7 I put the formula =SUM(C3,D4,E5); copy and paste that over to column DQ. This counts how many matches were found in the previous three helper rows. If all three match, the result of this formula will be 3 and your criterion for determining the period will have been fulfilled.
Now to show the period: in the cell you want to have this number, put the formula =MATCH(3,E7:DQ7,0). This searches the last (fourth) helper row looking for a cell that is equal to 3. (Obviously you could modify this method to match only the first two sequence numbers, or to match more than 3, and then you'd adjust the first parameter in the MATCH formula.) The last parameter in this MATCH formula is 0 because the helper row is not sorted. The return value is the index of the first match: a match in E7 would be index 1, a match in E8 would be index 2, etc.
I tested this in LibreOffice 4.4.4.3.
Related
cv.uk.df$new.d[2:nrow(cv.uk.df)] <- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1) # this line of code works
I wanted to know why do we -1 in the tail and -1 in head to create this new column.
I made an effort to understand by removing the -1 and "R"(The code is in R studio) throws me this error.
Could anyone shed some light on this? I can't explain how much I would appreciate it.
Look at what is being done. On the left-hand side of the assignment operator, we have:
cv.uk.df$new.d[2:nrow(cv.uk.df)] <-
Let's pick this apart.
cv.uk.df # This is the data.frame
$new.d # a new column to assign or a column to reassign
[2:nrow(cv.uk.df)] # the rows which we are going to assign
Specifically, this line of code will assign a new value all rows of this column except the first. Why would we want to do that? We don't have your data, but from your example, it looks like you want to calculate the change from one line to the next. That calculation is invalid for the first row (no previous row).
Now let's look at the right-hand side.
<- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1)
The cv.uk.df$deaths column has the same number of rows as the data.frame. R gets grouchy when the numbers of elements don't follow sum rules. For data.frames, the right-hand side needs to have the same number of elements, or a number that can be recycled a whole-number of times. For example, if you have 10 rows, you need to have a replacement of 10 values. Or you can have 5 values that R will recycle.
If your data.frame has 100 rows, only 99 are being replaced in this operation. You cannot feed 100 values into an operation that expects 99. We need to trim the data. Let's look at what is happening. The tail() function has the usage tail(x, n), where it returns the last n values of x. If n is a negative integer, tail() returns all values but the first n. The head() function works similarly.
tail(cv.uk.df$deaths, -1) # This returns all values but the first
head(cv.uk.df$deaths, -1) # This returns all values but the last
This makes sense for your calculation. You cannot subtract the number of deaths in the row before the first row from the number in the first row, nor can you subtract the number of deaths in the last row from the number in the row after the last row. There are more intuitive ways to do this thing using functions from other packages, but this gets the job done.
I have been studying Linear Algebra and came across gauss-elimination method. While it does give the correct answers , I am not able to understand how it works.
So, we compare each row with rest of the rows to see if any is dependent on the current row. But how does it identify, if One of the rows in the matrix, is a linear combination of 2 or more other rows??
This builds upon the result that elementary row operations doesn't change the rank of the matrix. Hence if we reduce the original matrix to its row echelon form, the two matrices share the same rank.
To see that the non-zero rows are linear independent, suppose r1, r2,...,rk are the non-zero rows (sorted in that order), consider c1r1+...+ckrk = 0. Look at the the position of the leading non-zero entry of r1, for that position, the other entries have value 0 there, hence we conclude that c1 = 0. After that, we can conclude the same argument for c2 and so on and conclude that all f them must be zero and hence the rank is equal to the number of non-zero rows.
I'm trying to make a LibreOffice spreadsheet formula that populates a column based on another input column, comparing each input with a series of range pairs defined in another sheet and finally outputting a symbol based on matched criteria. I have a series of ranges that specify a - output, and another series that corresponds to +, but not all inputs will fall into a category. I am using this trinary output later for another expression, which I already have in place.
My question becomes: how can I test input against each range pair without spelling out the cell coordinates for each individual cell (ie OR(AND(">= $A$1", "< $B$1"), AND(">=$A$2", "<$B$2"), ...))? Ideally I could just specify an array to compare against like $A$1:$B$4. Writing it in a python macro would work, too, since I don't plan on sharing this file.
I wrote a really quick list comp in python to illustrate what I'm after. This snippet would be one half, such as testing - qualification, and these values may be fed into a condition that outputs the symbol:
>>> def cmp(f, r):
... return r[0] <= f < r[1]
>>> f = (1, 2, 3)
>>> ranges = ((2, 5), (4, 6), (3, 8))
>>> [any([cmp(i, r) for r in ranges]) for i in f]
[False, True, True]
Here is a small test example with real input and real ranges.
Change the range pairs so that they are in two columns starting from A13. Be sure that they are in sorted order (Data -> Sort).
A B C
~~~~~~~~ ~~~~~~~~ ~
145.1000 145.5000 -
146.0000 146.4000 +
146.6000 147.0000 -
147.0000 147.4000 +
147.6000 148.0000 -
440.0000 445.0000 +
In each row, specify whether it is negative or positive. To do this, I entered the following formula in C13 and filled down. If the range pairs are not consistent enough then enter values for C13 and below manually.
=IF(ISODD(ROW());"-";"+")
Now, enter the following formula in cell C3 and fill down.
=IFNA(IF(
VLOOKUP(A3;A$13:C$18;2;1) >= A3;
VLOOKUP(A3;A$13:C$18;3;1);
"None");"None")
The formula finds the closest pair and then checks if the number is inside that range or not. For better testing, I would also suggest using 145.7000 as input, which should result in no shift if I understood the question correctly.
The results in column C:
-
+
None
None
Documentation: VLOOKUP, IFNA, ROW.
EDIT:
The following formula produces correct results for the example data you gave, and it works for anything between 144.0 and 148.0.
=IFNA(VLOOKUP(A3;A$13:C$18;3;1); "None")
However, 150.0 produces - and 550.0 produces +. If that is not what you want, then use the formula above that has two VLOOKUP expressions.
In Openoffice, Calc, to index areas under condition, between two e.g. words?
E.g. in a column A, one under another "A,B,D,c,A,B,C,A,C)...
I want each time to index only A to C, and when results are reported with Match functions (of corresponding column B values), to continue to the next region AC. How can i do this in #Calc?
1rst region:ABDC,
2nd:ABC,
3rd:AC
In cell B1 enter 1, then in cell B2 enter =IF(A2="A";B1+1;B1)
Copy cell B2 and paste down however many rows your pattern is. Calc will automatically adjust the formula to increase the row numbers as you paste in lower rows. This formula will go up by 1 whenever the pattern has an A, otherwise it will repeat the number from the previous row.
Note that it is case-sensitive and will not increment for a. If you want either case (A or a) to trigger the increment, use =IF(OR(A2="A";A2="a");B1+1;B1)
Ex: I enter the number 9876543210 in a cell.
I want to create an if then formula to add a sequential number to this but working only off of the last digit. the zero in this example.
If the last digit is >= to 3 than add 5 if the last digit is <=2 than add 15.
Then have this formula repeat for 10 numbers - is that possible?
so i imput the 9876543210
it then show:
9876543225
9876543230
9876543245
and so on
=IF((RIGHT(A1,1)/1)>2,A1+5,A1+15)
Assumed that you update the number in the cell A1. Paste the above formula in A2 and copy paste downwards.
If this is Excel, you may want to use MOD (modulo or remainder) function to get the last digit and then perform an IF-THEN or nested IF-THEN to achieve this.
=IF(MOD(A1,10)=3, A1+15, IF(MOD(A1,10)=5, A1+20, A1+30))
This formula translates to the following decision tree:
IF the last digit of the value in cell A3 is 3 Then
Add 15 to it
ELSEIF the last digit of the value in cell A3 is 5 then
Add 20 to it
ELSE
Add 30 to it
END IF
Repeating the operation may require some VBA. If you already know the number of times you need to repeat the operation, you can pre-populate formulas in subsequent rows/columns, each time refer to the immediately preceding cell. For example, if you want to repeat it 5 times, you should compute the diff of first two cells and then add that diff to the value of immediately preceding row/column like this (assuming A1 had the original value, B1 had the formula I posted above and C1 through G1 are the next 5 cells):
In C1: =B1 + ($B1 - $A1)
In D1: =C1 + ($B1 - $A1)
and so on...
Note the use of absolute and relative addresses in these formulae. You can copy/paste the formula in C1 to the subsequent cells and it will automatically adjust itself to refer to immediately preceding cell.
EDIT
I just realized that you want to evaluate the MOD formula in each subsequent cell. In that case you simply need to copy/paste it to subsequent cells instead of using 2nd and 3rd formulas I posted above.