I and my coworkers enter data in turns. One day I do, the next week someone else does and we always enter 50 observations at a time (into an Excel sheet). So I can be pretty sure that I entered the cases from 101 to 150, and 301 to 350. We then read the data into R to work with it. How can I select only the cases I entered?
Now I know that I can do that by copying from the excel sheet, however, I wonder if it is doable in R?
I checked several documents about subsetting data with R, also tried things like
data<-data[101:150 & 301:350,]
but didn't work. I appreciate if someone would guide me to a more comprehensive guide answering this question.
The answer to the specific example you gave is
data[c(100:150,300:350),]
Can you be more specific about which cases you want? Is it the first 50 of each 100, or the first 50 of each 300, or ... ? To get the indices for the first n of each m cases you could use something like
c(outer(0:4,seq(1,100,by=10),"+"))
(here n=5, m=10); outer is a generalized outer product. An alternate (and possibly more intuitive) solution would use rep, e.g.
rep(0:4,10) + rep(seq(1,100,by=10),each=5)
Because R automatically recycles vectors where necessary you could actually shorten this to:
0:4 + rep(seq(1,100,by=10),each=5)
but I would recommend the slightly longer formulation as more understandable.
Related
I recently just started with R a few weeks ago at the Uni. We were given a problem which we had to solve. However in this problem, I find that there are two answers that fit the question:
Verify that you created lo_heval correctly (incl. missing values). Store your verification in the object proof2.
So i find this is correct:
proof2 <- soep[1:100, c("heval", "lo_heval")]
But I think that this answer is also correct:
proof2 <- table(soep$heval, soep$lo_heval, useNA = "always")
Instead of having to decide for one answer, how do I combine them both into the object? I tried to use &, but I get an error. I may be using it wrong.
Prof. if you're seeing this, please don't fail me. I just can't decide between them.
Thanks in advance!
R lists can hold any arbitrary objects in them, so you could use
proof2 <- list(
soep[1:100, c("heval", "lo_heval")],
table(soep$heval, soep$lo_heval, useNA = "always")
)
However, to my mind 100 rows of two columns isn't proof - it's an exercise to look through those and verify things are right. (And what about the rows past 100? It's a decent spot check, but if there are more rows in the data it is more strong evidence than proof.) The table approach, on the other hand, seems succinct and effective.
I previously worked on a project where we examined some sociological data. I did the descriptive statistics and after several months, I was asked to make some graphs from the stats.
I made the graphs, but something seemed odd and when I compared the graph to the numbers in the report, I noticed that they are different. Upon investigating further, I noticed that my cleaning code (which removed participants with duplicate IDs) now results with more rows, e.g. more participants with unique IDs than previously. I now have 730 participants, whereas previously there were 702 I don't know if this was due to updates of some packages and unfortunately I cannot post the actual data here because it is confidential, but I am trying to find out who these 28 participants are and what happened in the data.
Therefore, I would like to know if there is a method that allows the user to filter the cases so that the mean of some variables is a set number. Ideally it would be something like this, but of course I know that it's not going to work in this form:
iris %>%
filter_if(mean(.$Petal.Length) == 1.3)
I know that this was an incorrect attempt but I don't know any other way that I would try this, so I am looking for help and suggestions.
I'm not convinced this is a tractable problem, but you may get somewhere by doing the following.
Firstly, work out what the sum of the variable was in your original analysis, and what it is now:
old_sum <- 702 * old_mean
new_sum <- 730 * new_mean
Now work out what the sum of the variable in the extra 28 cases would be:
extra_sum <- new_sum - old_sum
This allows you to work out the relative proportions of the sum of the variable from the old cases and from the extra cases. Put these proportions in a vector:
contributions <- c(extra_sum/new_sum, old_sum/new_sum)
Now, using the functions described in my answer to this question, you can find the optimal solution to partitioning your variable to match these two proportions. The rows which end up in the "extra" partition are likely to be the new ones. Even if they aren't the new ones, you will be left with a sample that has a mean that differs from your original by less than one part in a million.
I'm creating a Monte Carlo model using R. My model creates matrices that are filled with either zeros or values that fall within the constraints. I'm running a couple hundred thousand n values thru my model, and I want to find the average of the non zero matrices that I've created. I'm guessing I can do something in the last section.
Thanks for the help!
Code:
n<-252500
PaidLoss_1<-numeric(n)
PaidLoss_2<-numeric(n)
PaidLoss_3<-numeric(n)
PaidLoss_4<-numeric(n)
PaidLoss_5<-numeric(n)
PaidLoss_6<-numeric(n)
PaidLoss_7<-numeric(n)
PaidLoss_8<-numeric(n)
PaidLoss_9<-numeric(n)
for(i in 1:n){
claim_type<-rmultinom(1,1,c(0.00166439057698873, 0.000810856947763742, 0.00183509730283373, 0.000725503584841243, 0.00405428473881871, 0.00725503584841243, 0.0100290201433936, 0.00529190850119495, 0.0103277569136224, 0.0096449300102424, 0.00375554796858996, 0.00806589279617617, 0.00776715602594742, 0.000768180266302492, 0.00405428473881871, 0.00226186411744623, 0.00354216456128371, 0.00277398429498122, 0.000682826903379993))
claim_type<-which(claim_type==1)
claim_Amanda<-runif(1, min=34115, max=2158707.51)
claim_Bob<-runif(1, min=16443, max=413150.50)
claim_Claire<-runif(1, min=30607.50, max=1341330.97)
claim_Doug<-runif(1, min=17554.20, max=969871)
if(claim_type==1){PaidLoss_1[i]<-1*claim_Amanda}
if(claim_type==2){PaidLoss_2[i]<-0*claim_Amanda}
if(claim_type==3){PaidLoss_3[i]<-1* claim_Bob}
if(claim_type==4){PaidLoss_4[i]<-0* claim_Bob}
if(claim_type==5){PaidLoss_5[i]<-1* claim_Claire}
if(claim_type==6){PaidLoss_6[i]<-0* claim_Claire}
}
PaidLoss1<-sum(PaidLoss_1)/2525
PaidLoss3<-sum(PaidLoss_3)/2525
PaidLoss5<-sum(PaidLoss_5)/2525
PaidLoss7<-sum(PaidLoss_7)/2525
partial output of my numeric matrix
First, let me make sure I've wrapped my head around what you want to do: you have several columns -- in your example, PaidLoss_1, ..., PaidLoss_9, which have many entries. Some of these entries are 0, and you'd like to take the average (within each column) of the entries that are not zero. Did I get that right?
If so:
Comment 1: At the very end of your code, you might want to avoid using sum and dividing by a number to get the mean you want. It obviously works, but it opens you up to a risk: if you ever change the value of n at the top, then in the best case scenario you have to edit several lines down below, and in the worst case scenario you forget to do that. So, I'd suggest something more like mean(PaidLoss_1) to get your mean.
Right now, you have n as 252500, and your denominator at the end is 2525, which has the effect of inflating your mean by a factor of 100. Maybe that's what you wanted; if so, I'd recommend mean(PaidLoss_1) * 100 for the same reasons as above.
Comment 2: You can do what you want via subsetting. Take a smaller example as a demonstration:
test <- c(10, 0, 10, 0, 10, 0)
mean(test) # gives 5
test!=0 # a vector of TRUE/FALSE for which are nonzero
test[test!=0] # the subset of test which we found to be nonzero
mean(test[test!=0]) # gives 10, the average of the nonzero entries
The middle three lines are just for demonstration; the only necessary lines to do what you want are the first (to declare the vector) and the last (to get the mean). So your code should be something like PaidLoss1 <- mean(PaidLoss_1[PaidLoss_1 != 0]), or perhaps that times 100.
Comment 3: You might consider organizing your stuff into a dataframe. Instead of typing PaidLoss_1, PaidLoss_2, etc., it might make sense to organize all this PaidLoss stuff into a matrix. You could then access elements of the matrix with [ , ] indexing. This would be useful because it would clean up some of the code and prevent you from having to type lots of things; you could also then make use of things like the apply() family of functions to save you from having to type the same commands over and over for different columns (such as the mean). You could also use a dataframe or something else to organize it, but having some structure would make your life easier.
(And to be super clear, your code is exactly what my code looked like when I first started writing in R. You can decide if it's worth pursuing some of that optimization; it probably just depends how much time you plan to eventually spend in R.)
Let's say I have a table with thousands of values (time, value). When I plot them, I can see that the values are "kind of" repetitive (all 24 hours the values roughly repeat).
But there is an clearly visible attack time at the beginning. How can I find out, at which time the repetition is 100% the same (or even 99%) using R?
(For example: after three days, all following cycles are 99% identically)
I assume you mean some of your rows are duplicates. In which case:
df <- df[!duplicated(df), ]
This is a common question. You can find an answer faster by searching other questions on the same topic on SO. Good luck learning R.
I have been trying to produce a command in R that allows me to produce a new vector where each row is the sum of 25 rows from a previous vector.
I've tried making a function to do this, this allows me to produce a result for one data point.
I shall put where I haver got to; I realise this is probably a fairly basic question but it is one I have been struggling with... any help would be greatly appreciated;
example<-c(1;200)
fun.1<-function(x)
{sum(x[1:25])}
checklist<-sapply(check,FUN=fun.1)
This then supplies me with a vector of length 200 where all values are NA.
Can anybody help at all?
Your example is a bit noisy (e.g., c(1;200) has no meaning, probably you want 1:200 there, or, if you would like to have a list of lists then something like rep, there is no check variable, it should have been example, etc.).
Here's the code what I think you need probably (as far as I was able to understand it):
x <- rep(list(1:200), 5)
f <- function(y) {y[1:20]}
sapply(x, f)
Next time please be more specific, try out the code you post as an example before submitting a question.