similar to: Replacing dot with empty space

Dear all, I have a zoo object that has following structure: > str(bldata) zoo [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ... - attr(*, "index")=Classes 'dates', 'times' atomic [1:5219] 7305 7306 7307 7308 7309 ... .. ..- attr(*, "format")= chr "m/d/y" .. ..- attr(*, "origin")= Named num [1:3] 1 1 1970 .. .. ..- attr(*,

Hello, everyone I have a question. Assume I have the following zoo object: me.la <- structure(c(1524.75, 1554.5, 1532.25, 1587.5, 1575.25, 1535.5, 1550, 1493.5, 1492.5, 1472.25, 1457.5, 1442.75, 1399, 1535.75, 1565.25, 1543.5, 1598.5, 1586.5, 1547, 1561.5, 1504.75, 1503.75, 1483.75, 1468.75, 1453.75, 1410, 1546.75, 1575.25, 1554, 1609, 1597.5, 1558.5, 1573, 1516.25, 1515.5, 1495, 1480, 1465,

Hello I have the following data frame which I read from an EXCEL file, and when i try to call one of its columns with a space in their names I am not being able to. For example if I do EURODOLLAR$ED1.Comdty Date I obtain the following error: Error: inesperado símbolo en "EURODOLLAR$ED1.Comdty Date" I have also tried using . or _ instead of the space and have obtained no succes. How do I

Hello, everyone I have a piece of code that looks like this: mrets <- merge(mrets, BMM.SR=apply(mrets, 1, MyFunc, ret="BMM.AV120", stdev="BMM.SD120")) mrets <- merge(mrets, GM1.SR=apply(mrets, 1, MyFunc, ret="GM1.AV120", stdev="GM1.SD120")) mrets <- merge(mrets, IYC.SR=apply(mrets, 1, MyFunc, ret="IYC.AV120",

Hello, I thought I understood filter() with the help from Prof. Grothendieck, but I guess I did not. For example, how does this work: filter(1:10, c(0.1, 0.5, 1, 0.5), "recursive", init=c(1,2,3,4)) Time Series: Start = 1 End = 10 Frequency = 1 [1] 7.10000 6.71000 9.22100 15.87710 21.45821 28.66037 41.08274 55.83522 74.51437 100.78197 If I understand it correctly, the time

Hello, everybody Say I have nm1 <- c(rep(1,10), rep(0,10)) then I can do: diff(nm1) to see where I have shift in value but what if I have nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10)) how can I produce the same ouput as diff(nm1) does, that is zeros everywhere except for one place where SPZ8 changes to SPX9 (there should be 1 there)? What if I have a matrix of characters

Hello, everyone! I have a problem with RBloomberg and this is not the usual "no administrator rights" problem. I have R 2.7.2, RBloomberg 0.1-10, RDCOMclient 0.92-0 RDCOMClient, chron, zoo, stats: these packages load OK. Then, trying to connect, I get following error message: conn <- blpConnect(show.days="week", na.action="previous.days",

Hello, everyone I run Eclipse Ganymede and R 2.7.2 at work. I have one R script file where I open in memory a new xls file (using xlsReadWritePro), call other R scripts, which are in the same folder as the main R script, which get data from an existing xls file, process data, and output results in the xls file which is in memory. That is the idea. But I cannot make it work. First, I do not

Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't know how to optimize the code, but I need to

Hello everyone, I have a plot and I want to but a (formatted) box containing text and numbers, say: Mean: 0.1 St.Deviation: 1.1 Skewness: 1.1 Kurtosis: 0.5 I know there is a way to do this, there is a function in some library, but it's been years since I used this function, and I do not remember where I found it. Could anyone help out? Thank you in advance! Sergey -- Kniven sk?rpes bara

Hello, everyone I wonder if there is in R somewhere a function similar to cumsum(). The function calculates a statistic (say mean or standard deviation) buy adding consequtively one more data point. So, say I have a timeseries of 100 observations. I start by calculating mean of first 30 observations Then I add one observation and calculate mean of 31 observations Then I add one more observation

Hello, everybody Two questions: 1) I am new to maillists, and particularly to r-help maillist. Where specifically do I go online to see my question and answers to it??? If I use the searchable archives, or archives by Robert King, I see my question but not the answers, though I know that at least one persion posted the answer to r-help. Why do not I see the answers? 2) I need to produce 104

Hello, I have an Excel file on a drive and I extract data from it into R session. Once I have extracted the data, I want to delete that Excel file from the drive. Can I do that from within R, please? Thank you for help! Regards, Sergey

Dear members of R forum, Say I have a list: L <- list(1:3, 1:3, 1:3) that I want to turn into a matrix. I wonder why if I do: do.call(cbind, L) I get the matrix I want, but if I do cbind(L) I get something different from what I want. Why is that? How does do.call() actually work? I've read in do.call() help file this sentence: "The behavior of some functions, such as

Hello, fellow R-users. Let me describe the setup first. I have a data.frame, a sample of which is reported below: Company.Name Periods Returns MFR.Factor 350 Wartsila Oyj A 1996-07-31 6.82 0.02 351 Custodia Holding AG 1996-07-31 4.15 -0.02 352 Wartsila Oyj 1996-07-31 7.73 0.09 353 GEA Group AG

Hello, everyone I wonder if it is possible to PHYSICALLY open an Excel file from R. The reason I ask is, I produce regularly an Excel file in R, and then I want to make it look good, so I have a VBA routine in another Excel file that works on the regular Excel file. This formatting file executes VBA code on open, so all I need to do is physically open it (no reading/writing at all). I wonder if

Hi, fellow R users. I have a question about sapply and split combination. I have a big dataframe (40000 observations, 21 variables). First variable (factor) is "date" and it is in format "8.29.97", that is, I have monthly data. Second variable (also factor) has levels 1 to 6 (fractiles 1 to 5 and missing value with code 6). The other 19 variables are numeric. For each month I

Hello, everyone! Assume you have this data: data <- structure(c(66.609375, 67.09375, 66.40625, 66.734375, 67.109375, 66.875, 66.09375, 65.921875, 66.546875, 66.140625, 66.140625, 65.65625, 65.875, 65.59375, 65.515625, 66.09375, 66.015625, 66.140625, 66.109375, 66.421875, 1702.7, 1647.7, 1649.4, 1639.9, 1696.4, 1710.9, 1690.2, 1677.9, 1694.4, 1713.9, 1713.9, 1705.4, 1708.4, 1692.9, 1689.6,

Hi: I want to obtain a particular value from a data.frame. Following is my dataframe: > Quotes BID ASK Name CT2 GOVT 99.92969 99.9375 CT2 TUM0 COMDTY 108.53125 108.5469 TUM0 CT5 GOVT 100.10156 100.1094 GT5 FVM0 COMDTY 115.56250 115.5703 FVM0 TYM0 COMDTY 116.93750 116.9531 TYM0 If I try to run: QuoteTUM0BID = Quotes[Quotes$Name %in%

Hello everyone I need to extract a vector of (t-3) to (t+3) dates, only working days. How can I do that? For today I need a vector: 10.08.210 09.08.2010 06.08.2010 05.08.2010 04.08.2010 03.08.2010 02.08.2010 Regards, Sergey -- Kniven sk?rpes bara mot stenen.