similar to: difficulty in Formatting time series data

GuRus How do I use the write function (or write.table or write.csv) to achieve the following please? age=c(32,37,39) names=c("john","peter","jake") I would like create in a directory 3 files each named as john.csv,peter.csv and jake.csv and each file have data from the age vector. That is jon.csv will contain 32, peter.csv will contain 37 and jake.csv will contain

guRus! I have a function f = exp(x^2-y+(1/z)) Also, x can take values from 1 to 37, y from 2 to 20 and Z from -13 to 51. How can I find the maximum of f using any of the optimization functions please? Is there a way to store the possible values of x, y and Z in a single variable like in a List or in a multi-dimensional array? Thanks for your help Raghu [[alternative HTML version deleted]]

Hello I have a data frame like this: dput(states) structure(list(Date = c("24/07/2012", "25/07/2012", "26/07/2012", "27/07/2012", "28/07/2012", "24/07/2012", "25/07/2012", "26/07/2012", "27/07/2012", "28/07/2012"), State = c(1L, 1L, 1L, 1L, 1L, -1L, -1L, -1L, 1L, -1L)), .Names = c("Date",

Dear guRus Say I have a price vector, P which is P = c(20,50,40,50) for four consecutive days. My initial equity is say $100 and I re-invest the profits made in each transaction so my equity curve increases. If I go Long on day 1, short and short on day 2, long and long on day 3 and finally short on day 4, how do I calculate the total profits using R? Basically I stay invested at all times. I

Hi I wish to read a file from my local directory from inside a function. I am passing the filename as the argument but this does not work. Say for example function(dat) { dat1=read.csv("D:\\dat.csv",header=TRUE) } If I call funtion(dat) I get the following error. 'Intuitively' i understand this is a mistake but how do I overcome this and how can I read a file name passed as an

guRus I have say a dataframe, d and I wish to do the following: 1) For each row, I want to take one particular value of the row and multiply it by 2. How do I do it. Say the data frame is as below: OPEN HIGH LOW CLOSE 1931.2 1931.2 1931.2 1931.2 0 0 0 999.05 0 0 0 1052.5 0 0 0 987.8 0 0 0 925.6 0 0 0 866 0 0 0 1400.2 0 0 0 754.5 0 0 0 702.6 0 0 0 653.25 0 0 0 348 0 0 0 801 866.55 866.55

Gabor, Thanks much. Your solution is elegant. My overall scheme is to take present date, and check whether it is a weekend, if not, then create a string based on the date, to concatenate into a url link for download.file( ). The files I need to download have a part which is in the format: mmddyy. I am working to make myself a system to connect to exchanges, and download end of day files from

Have you noticed any problems with big dates (>=1/1/2040) in R? Here is the bit of code that I'm having trouble with: > test.date <- strptime("1/1/2040",format="%m/%d/%Y") > > unlist(test.date) sec min hour mday mon year wday yday isdst 0 0 0 1 0 140 0 0 0 > > date.plus.one <- as.POSIXct(test.date) +

Hi I have say a large vector of 3500 digits. Initially the digits are 0s and 1s. I need to check for a rule to change some of the 0s to -1s in this vector. But once I change a 0 to -1 then I need to start applying the rule to change the next 0 only after I see the next 1 in the vector. Say for example x = (0,0,0,0,0,0,0,1,0,0,1,1,0,0,1,0,0,0,1) I need to traverse from the 9th element to the last

Hey everybody, I''ve got a model that represents kind of a turn-based games. Certain actions can only be made unders certain conditions. For simplicity, we''ll say that the only condition for a particular action is that the weekday be Tuesday. def add_vote raise ''You cannot vote today'' unless Date.today.wday == 2 ... end In my controller I want to show a

I''m having difficulty figuring out the ruby neccessary to do this so I thought I would try the list. I''m hoping to get a date_select to only display certain days (specifically only fridays) but rails doesn''t seem to have the flexibility so I''ve been going at it in ruby to no avail. I''ve looked on google but couldn''t find it (because it

Hi I have a sorted array ( in ascending order) and I want to find the subscript of a number in the array which is the the next highest number to a given number. For example,if I have 67 as a given number and if I have a vector x=c(23,36,45,62,79,103,109), then how do I get the subscript 5 from x (to get 79 which is the next highest to 67) without using a for loop? Thx -- 'Raghu'

I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the

I know the following may sound too basic but I thought the mailing list is for the benefit of all levels of people. I ran a simple if statement on two numeric vectors (news1o and s2o) which are of equal length. I have done an str on both of them for your kind perusal below. I am trying to compare the numbers in both and initiate a new vector s as 1 or 0 depending on if the elements in the arrays

When I just run a for loop it works. But if I am going to run a for loop every time for large vectors I might as well use C or any other language. The reason R is powerful is becasue it can handle large vectors without each element being manipulated? Please let me know where I am wrong. for(i in 1:length(news1o)){ + if(news1o[i]>s2o[i]) + s[i]<-1 + else + s[i]<--1 + } --

I''m using the date_select and datetime_select helpers in my view, and they return Date classes from the params hash. But how do I work with Date classes, they don''t print human readable dates or times, Time classes work well I can use strftime("%H:%M") to print to the screen. Is it possible to convert a Date to a Time, Ive been tinkering in irb but have got

Hi, R experts. I am a new user of R and trying to learn this program. I have a problem. Here is the code. d<-as.Date(c("2000/01/03","2000/01/05","2000/01/19","2000/01/28")) r<-rnorm(4) da<-data.frame(d,r) a<-as.Date("01/01/2000","%d/%m/%Y") b<-as.Date("30/01/2000","%d/%m/%Y") ab<-seq(a,b,by=1)

Hello I have a vector a =(-2,0,0,0,1,0,0,3,0,0,-4) I want to replace all zeros into previous non-zero state. So for instance the above vector should be converted into: a= (-2,-2,-2,-2,1,1,1,3,3,3,-4) I tried many things and finally concluded that probably(?) rollapply may be the best way? I tried f= function(x){ ifelse(x==0,Lag(x),x) } And then, rollappy(a,1,f) and that

Hi, I am wondering if there is a way to display the full anme of the regression coeffients/factors in the summary? Suppose I have a bogus data set using weekday as factor which has 7 levels such as: mydata <- sample(364) wk <- rep(1:7, 52) weekday <-

Hi, I am wondering if there is a simple way to fix the problem I am having. For unknown reason, I could not get the full name of the factors to be printed in the summary. I have tried to used summary.lm as well but the problem still persists. SJ$Weekday <-