I am currently working on particular algorithm, but I face with a problem that I'm not sure what I have to do to resolve it. I appreciate if anyone helps me out.
There are some objects{O1,O2,O3,.....}, each of them has a value that we don't know about its amount, we call them {V1,V2,V3,....} also there is another element we call it w(w1,w2,w3.....) which shows the difference between values, I mean w1=v2-v1, w2=v3-v2,w3=v4-v3 and so on. I'm wondering if there is any way to get value of v1,v2,v3...etc without having the value of V1?
Looking forward for your reply guys,
Thanks.
Not in general. Knowing the differences between successive numbers in a list of numbers under-determines the set of numbers. This is particularly obvious in the case when w1 = w2 = w3 = ... = wk = 1. That would tell you that the viare consecutive numbers, but nothing else could be inferred. You wouldn't be able to distinguish 3,4,5,6,7 from 10,11,12,13,14 (for example).
Having said that, it would of course be possible if you know one of the numbers, and the known number wouldn't need to be the first one. Knowing any single one of the numbers would suffice. Furthermore, knowing something like the sum of the vi would be sufficient since you could express the sum as a function of the unknown number v1 and solve the resulting equation.
Related
I'm making a balance sheet, Sheet1 is for the ins and outs, and most values are added manually or simple formulas, and Sheet2 is where I created a formula, in the hopes of being able to reuse it.
I'm not an accountant to understand how I could make the calculations easier, and I'm a programmer, so I understand that the way I may be imagining the solution is likely impossible with the way Libreoffice Calc's formulas work.
So, to explain a bit.
On Sheet1, each column is a month, and the value is a tax that will appear one time each month, dependent on another value.
So, the base value is on ROW 17, and on 18, I would like that result to be set. For every month, of course
On Sheet2, I have the function, it contains 5 steps, with the values being reused a lot (hence, simplifying everything into one line would be hell).
This is the complex formula in question, D1 is the input, C6 is the output.
The formula below is the one used on C2, and repeated down to C5.
I would like to keep the constants as a table since it would be easier to update it in the future in case it suffer any changes.
I have been searching for a possible solution but found none, and I believe that it's likely because I'm looking for a solution like a programmer (use Sheet as a function), and I should seek sort of way, but I don't know how Calc works.
In regards to the calculation, I don't know the specific name, but the idea is, from 0 to A1, I have to B1% from A1-0, then from A2-A1, remove B2%, and so on.
Of course the formula's complexity comes from treating lower values, so for example, if D1 was 2K, then I would have to take 7.5% of R$ 96.02, and everything beyond is 0, since there is nothing remaining for them to calculate
Most of the descriptions I found on MULTIPLE.OPERATIONS were confusing, but I found one that made it much easier to understand.
The answer was to use this formula on Sheet1:
=MULTIPLE.OPERATIONS('Sheet2'.$C$6, 'Sheet2'.$D$1, C17)
I can just copy paste it to the side and the calculation will be executed properly.
To explain the arguments:
1 - where the result will appear
2 - the location of the main/first formula variable
3 - the location of dynamic variable you want to insert in that formula (So this is from Sheet1)
More arguments could be used if more variables were needed, but I just needed one.
This is the place with the best explanation I found for the function.
https://wiki.documentfoundation.org/Documentation/Calc_Functions/MULTIPLE.OPERATIONS
I have an array (x) in R of size 30x11x10.
x=array(-2:20, c(30,11,10))
Each 'grid' or matrix represents a day of data for a month (30 days represented here). I want to find the index (i,j,k) of when the last occurrence of a number less than 2 occurs. Ideally, I would also like the value returned too. If this was in Matlab, I could just use [i,j,k]=find(x(x<2)) but I don't see an exact equivalent for this in R.
I have looked at 'match' as suggested in other posts here, but it seems to find elements when they are specified, but not when a criteria (x<2) is given?
I tried this:
xxx<-match(x,x<2,0) but it returns a long vector of integers that don't appear to show what I am looking for.
Then I tried:xxx<-match(x,x[x<2],0) which looks a bit more promising, but still isn't what I want (to be honest I'm not sure what the output is indexing).
I think I'm probably asking a foolish question here because if I want 3 indices and the value returned, then I should be assigning them to something preemptively right (which I'm not doing)? Can anyone offer any advice?
I'm working with an address database. For further cleaning I need to identify the leading zeros that are stored in the string containing the door number. So a clean and friendly case would be something like 7/5 - where 7 would be the house number, 5 the door number.
The problem is that in some cases, there are leading zeros involved - not only at the beginning of the string, but also in the middle. And of course, there are also "normal" and necessary zeros. So I could have an address like 7/50, where the zero is totally fine and should stay there. But I could also have 007/05, where all the zeros need to go away. So the only pattern I can think of is to check wheter there is a number greater than zero before that zero. If no, then delete that zero, if yes, keep it.
Is there any function to achieve something like this? Or would this require something custom built? Thanks for any suggestions!
You can try the code below base R option using gsub
> gsub("\\b0+", "", s)
[1] "1/1001001" "7/50" "7/50" "7/5" "7/5"
with given
s <- c("01/1001001", "07/050", "0007/50", "7/5", "007/05")
Maybe a negative look behind will help
x <- c("7/50", "7/5", "007/05")
stringr::str_remove_all(x, "\\b(?<![1-9])0+")
# [1] "7/50" "7/5" "7/5"
Hard to say for sure with such a limited set of test cases.
I have a basic question in regards to the R programming language.
I'm at a beginners level and I wish to understand the meaning behind two lines of code I found online in order to gain a better understanding. Here is the code:
as.data.frame(y[1:(n-k)])
as.data.frame(y[(k+1):n])
... where y and n are given. I do understand that the results are transformed into a data frame by the function as.data.frame() but what about the rest? I'm still at a beginners level so pardon me if this question is off-topic or irrelevant in this forum. Thank you in advance, I appreciate every answer :)
Looks like you understand the as.data.frame() function so let's look at what is happening inside of it. We're looking at y[1:(n-k)]. Here, y is a vector which is a collection of data points of the same type. For example:
> y <- c(1,2,3,4,5,6)
Try running that and then calling back y. What you get are those numbers listed out. Now, consider the case you want to just call out the number 1 in that vector. How would you do that? Well, this is where the brackets come into play. If you wanted to just call the number 1 in y:
> y[1]
[1] 1
Therefore, the brackets are a way of calling out or indexing specific items in the vector. Note that the indexing starts at the value 1 and goes up to the number of items in the vector, or length. One last thing before we go back to the example you gave. What if we want to index the numbers 1, 2, and 3 from the vector but not the rest?
> y[1:3]
[1] 1 2 3
This is where the colon comes into play. It allows us to reference a subset of the numbers. However, it will reference all the numbers between the index left of the colon and right of it. Try this out for yourself in R! Play around and see what happens.
Finally going back to your example:
y[1:(n-k)]
How would this work based on what we discussed? Well, the colon means that we are indexing all values in the vector y from two index values. What are those values? Well, they are the numbers to the left and right of the colon. Therefore, we are asking R to give us the values from the first position (index of 1) to the (n-k) position. Therefore, it's important to know what n and k are. If n is 4 and k is 1 then the command becomes:
y[1:3]
The same logic can apply to the second as.data.frame() command in your question. Essentially, R is picking out different numbers from a vector y and multiplying them together.
Hope this helps. The best way to learn R is to play around with a command, throw different numbers at it, guess what will happen, and then see what happens!
deadcheck<-function(a,t){ #function to check if dead for specific age at a time age sending to function
roe<-which( birthmort$age[i]==fertmortc$min & fertmortc$max) #checks row in fertmortc(hart) to pick an age that meets min and max age requirements I think this could be wrong...
prob<-1-(((1-fertmortc$mortality[roe])^(1/365))^t) #finds the prob for the row that meets the above requirements
if(runif(1,0,1)<=prob) {d<-TRUE} else {d<-FALSE} #I have a row that has the probability of death every 7 days.
return(d) #outputs if dead
Background: I am creating an agent based model that is a population in a dataframe that is simulating how Tuberculosis spreads in a population. ( I know that there are probably 10000 better ways of having done this). I have thus far created a loop that populates my dataframe with people ages etc. I am now trying to create a function that will go to a chart that lists the probability of death per year, based on a age bracket. 0-5,5-10,10-15 etc. (I have math in there b/c I want it to check who lives, dies, makes babies every 7 days). I have a function similar to this that check who is pregnant and it works. However I for the life of me can't figure out why this function is not working. I keep getting the following error.
Error in if (runif(1, 0, 1) <= prob) { : argument is of length zero
I am unsure how to fix this.
I apologize in advanced it this is a dumb question, I have been trying to teach myself to code over the last 4-5 months. If I asked this question in the wrong format or incorrectly then please let me know how to do so correctly.
Value of prob is of length zero. It means
prob = NULL
in this case. Try to print alter your code and add
print(prob)
so you can check partial result.
As you suspected in your comments, the expression
birthmort$age[i]==fertmortc$min & fertmortc$max
is problematic. What this does is evaluate the comparison birthmort$age[i]==fertmortc$min, and then takes the result of that comparison and combines it with fertmortc$max using the and operator. This involves forming the and of a Boolean value and an integer, which is unlikely to make much sense.
Just guessing, you perhaps want:
birthmort$age[i] >= fertmortc$min & birthmort$age[i] <= fertmortc$max
I don't know if this will fix your problem -- you haven't given enough to test it. For optimal help, you should give a reproducible example. See this for how to do so in R