similar to: How to monthly,daily,yearly average

e.g. a= a b c d e 1 1 1 3 1 1 2 1 2 3 4 5 3 1 3 3 8 3 4 1 4 3 3 5 5 1 1 3 1 1 I want to delete colume a and colume c, because they have the same values in every row, then ,I want to get this data.frame . b= b d e 1 1 1 1 2 2 4 5 3 3 8 3 4 4 3 5 5 1 1 1 the following is my code but it's wrong. rm(list=ls()) a=c(1,1,1,1,1); b=c(1,2,3,4,1); c=c(3,3,3,3,3);

Dear R experts: I have following data structure: student id, exam time and score. I want to create three new columes: 1 st before, 2nd before and 3rd before. For example, for student 1's 4th exam score is assumed to be influenced by his previous three cloest scores, which are 6, 9, and 10 in that order. For student 2's 5th exam score is assumed to be inflenced by her previous cloest

I have some data set which is available from 2005-2010 . I would like plot monthly maps out of it. So how I should write loop that can plot this yearly data to for every month and write title for individual months too. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-monthly-maps-from-yearly-data-tp4391704p4391704.html Sent from the R help mailing list archive at

Dear R-users, I've been using R for a while and I am very satisfied! Unfortunately, I still have not figured out an efficient and general way to construct and use lags of time series, especially when I need to work with different packages. Let me give an example. I have two time series x and y and I want to estimate a variaty of distributed lags models and run different tests

my programe is below: a=c(1,2,1,1,1); b=c(1,2,3,4,1); c=c(3,4,3,3,3); d=c(1,2,3,5,1); e=c(1,5,3,5,1) data.f=data.frame(a,b,c,d,e) origin.data<-data.f cor.matrix<-cor(origin.data) origin.cor<-cor.matrix m<-0 for(i in 1:(cor.matrix[1]-1)) { for(j in (i+1):(cor.matrix[2])) { if (cor.matrix[i,j]>=0.95) { data.f<-data.f[,-i]; i<-i+1 }

I have 365 binary files: https://echange-fichiers.inra.fr/get?k=oy3CN1yV1Um7ouRWm2U ,I want to calculate the monthly average. So from the 365 files, I will get 12 files.I would like also to tell R not to take into account the no-data value (-32765).for example, for the first month, there are 31 records: 3 of these records has the value -32765,I want R to take the average of the rest records(28

Hello all, I don' like to have ( ) around my datetime, when send the csv file to somebody. How to remove it? >library(chron) >mydate <- c("2009-05-03","2009-06-07") >mytime <- c("10:30:00","10:20:00") >mydatetime <- chron(dates=mydate,times=mytime,format=c("y-m-d","h:m:s")) >write.csv(mydatetime)  

how to input multiple .txt files? A data folder has lots of .txt files from different customers. Want to read all these .txt files to different master files: such as: cust1.xx.txt, cust1.xxx.txt, cust1.xxxx.txt,.............. to master file: X.txt cust2.xx.txt, cust2.xxx.txt, cust2.xxxx.txt,.............. to master file: Y.txt Thanks! [[alternative HTML version deleted]]

Use  write.csv to output a zoo object, and found the first column in excel like:  (08-10-16 13:12:00)   How to make it look like : 10/16/2008 13:12:00 ?     write.csv(x=zoo.object)   Thanks! [[alternative HTML version deleted]]

I have searched the forums but I am having trouble trying accomplish two specific plots. I am not sure if this is a very basic task but I am having a lot of trouble Here is my data frame: date prod_daily price 1 2003-08-15 12050 4.83 2 2003-08-18 12050 4.94 3 2004-08-10 12300 5.78 4 2004-08-11 12300 5.64 5 2004-08-12 12300

Hallo, I recently stumple over some discussion about Sieve and what's the advantage to integrate it into the server rather than the LDA. Most of the filtering capabilites local users want to have performed are related to the INBOX or more preciesly the way to filter the message coming in. However, there are few tasks left: 1) Auto-archive old mail. 2) Drop old mail. 3) "Seen"

Hi, I've been running some simulations for a while and the performance of R has been great. However, I've recently changed the code to perform a sort of chi-square goodness-of-fit test. To get the observed values for each cell I've been using table() - specifically I've been using cut2 from Hmisc to divide up the range into a specified number of cells and then using

# This is more about trying to find a more effecient way to code some simple vectorized computations using ifelse(). # Say you have some vector representing a factor with a number of levels (6 in this case), representing the location that samples were collected. Population <- gl( n=6, k=5,length=120, labels =c("CO", "CN","Ga","KO", "Mw",

v="IBM" library(quantmod) v v1=getSymbols(v) to.yearly(v1) =============================== when i pass the value through a variable in to.yearly() function it shows the error msg like "Error in try.xts(x) : Error in UseMethod("as.xts") : no applicable method for 'as.xts' applied to an object of class "character"" i need the result of OHLC

I am a new R-language user. I have set up a data frame "mydata",one of the colume of which is "skill". Now I want to select the observations whose "skill" value is equal to 1,by what command can I get it?

Hi all, I am plotting a financial time series, but I need a more detailed X-Axis. Example: x <- zoo(rnorm(1:6000), as.Date("1992-11-11")+c(1:6000)) plot(x) The X-Axis is labeled "1995", "2000" and "2005". I would need either "1995", "1997", etc. or maybe yearly I used google first, then look at "?plot.zoo" but

Dear all, I want to import just part of an excel data file into R. I would like to have the data imported without rownames or colume names. I used read.delim("clipboard", header=F). Somehow even though I added the argument "header=F", I still have the row names V1, V2, ..., Does anyone know how to fix this? Thanks very much in advance. Hannah [[alternative

Dear all, I hope to remove a whole column from a data frame or matrix (> 2000 columns). All value in the column are same. The first thing is to select those columns. For instance, I hope to remove the V3~6 column, for all the value in those colume is zero. V3 V4 V5 V6 V7 V8 V9 V10 1 0 0 0 0 0.000 0.000 0.000 0.000 2 0 0 0 0 0.000 0.000 0.000 0.000 3 0 0 0

Hi, I have a zoo object with monthly frequency : library(zoo) dat <- zooreg(rnorm(50), as.yearmon("2000-01-01"), frequency=12) Now I want to make a zoo object with daily frequency from "dat" wherein value for a each day for a particular month will be value of "dat" at that particular month. Is there any easy way to do that? Thanks, -- View this message in

Dear R users, I want to subset a daily zoo series according to its month, find % of "NA" in each month. I am finding it difficult to subset the daily dataset into monthly for the given operation.I am planning to do this for a huge dataset. Thanks in advance. Regards Vikram [[alternative HTML version deleted]]