similar to: Time Series filter help?

I have and XTS time series object that has date and time. I started with 1 minute data and used apply.daily(x, sum) to sum the data to one cumulative value. This function works just fine however it leaves a time for the last summed value which looks like this 2006-07-19 14:58:00. I need to just have the date and to remove the time value of 14:58:00 just leaving the date value of 2006-07-19 .

I used quantmod to pull in price data from the ticker SPY. The data has date and closing price. I would like to show the day of the week for each closing price. Is that possible? Also, I would like to add the back into the data frame in a new column without changing the structure of the data set if possible. SPY 2009-01-02 92.96 2009-01-05 92.85 2009-01-06 93.47

I have a binary matrix of size N x 300. I then create the following: > set.seed(1234) > (key_file <- sample(letters[1:4], 300, replace=TRUE)) [1] "a" "c" "c" "c" "d" "c" "a" "a" "c" "c" "c" "c" "b" "d" "b" "d" "b"

Hi, I'd like some advice on bootstrapping in R. I have a species x with 20 individuals and a factor containing 0 and 1's (in this case 5 zeros and 15 ones). I want to compare the frequency of the occurrence of 1 with a probability value. This code seems to work to do this in R. attach(test) p <- c(0.5272, (1-0.5272)) sp1_1 <- length(subset(x, x==1)) sp1_0 <- length(subset(x,

Dear list, I've had this problem for a while and I'm looking for a more general and robust technique than I've been able to imagine myself. I need to find N (typically N= 3 to 5) zeros in a function that is not a polynomial in a specified interval. The code below illustrates this, by creating a noisy curve with three peaks of different position, magnitude, width and

Hi, I have an integer matrix consisting of 1's and 0's and I would like to convert this to a data.frame where each column of the matrix becomes a factor variable. Now, some columns of the matrix have only 1's or only 0's as a result there is only 1 level for those columns in the data.frame. However it is required that each factor have 2 levels. So my solution is: m <-

Hi, I cannot find a 'vectorized' solution to this 'for loop' kind of problem. Do you see a vectorized, fast-running solution? Objective: Take the value of X at each timepoint and calculate the corresponding value of Y. Leading 0's and all 1's for X are assigned to Y; otherwise Y is incremented by the number of 0's adjacent to the last 1. The frequency and

Hello, I have simulated 30 observations from a binomial(5,0.1) distribution. Now I need to make frequency table( that means I need to tally how many 0's , 1's 2's....... 5's) I know that the simple R function table() will do this, but I am afraid that some times I may get zero frequency for some particular values (for example in the above there are 5-0's 10-1's ,

Hello, I have the following problem. I am running simulations on possible states of a set of agents (1=employed, 0=unemployed). I store these simulated time series in a matrix like the following, where rows indicates time periods, columns the number of agents (4 agents and 8 periods in this case): Atr=[ 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0

x<-matrix(c(1,0,0,0,1,0,0,0,1),nrow=3) > y<-matrix(c(0,0,0,1,0,0,1,1,0),nrow=3) > z<-matrix(c(0,1,0,0,1,0,1,0,0),nrow=3) > x[z]<-y[z] The resultant matrix x is all zeros except for the last two diagonal cells which are 1's. While y is lower triangualr 0's with the remaining cells all ones. I really don't understand how this deceptively simple looking piece of

Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]

Dear R helpers exposure <- data.frame(id = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20), ead = c(9483.686,50000,6843.4968,10509.37125,21297.8905,50000,706152.8354, 62670.5625, 687.801995,50641.4875,59227.125,43818.5778,52887.72534,601788.7937, 56813.14859,4012356.056,1419501.179,210853.4743,749961,6599.0862), pd =

Hi, I'm trying to do a regression like this: wage.r = lm( log(WAGE) ~ log(EXPER) where EXPER is an integer that goes from 0 to about 50. EXPER contains some zeros, so you can't take its log, and the above regression therefore fails. I would like to make R accept log(0) as 0, is that possible? Or do I have first have to turn the 0's into 1's to be able to do the above

Dear R help, I have a dataset with 1's and 0's.? Here, each row is the observation for an insect.? If the animal is present in light area at a particular time, response is 1 and if it is present in dark area, the response is 0.? I would like to do some formating on the data for analysis.? To be specific, I need the rows where the animal starts in the light (1), went to dark (0), and came

Dear R-users, I have some nonstandard data set which I would like to plot but don't know how to do it in R. The data is in a matrix where the rows represent samples and the columns represent locations. The entries of the matrix are 0's and 1's, where 1 represents an event and 0 represents a non-event. e.g. aberrations <- matrix(rbinom(1000, 1, 0.8), nrow=20, ncol=50,

Please use dput() to post your example matrix. Rambler1 wrote > > I have run into a problem in my code. What I want to accomplish is this: > I have a user input stock symbols into a list and from there I run the > quantmod package to get historical data. I compute the correlation matrix > and then turn that matrix into a simple matrix with 1's or 0's depending > on

Hi Nate! I don't see how that would work. Select doesn't work per element. Say we're trying to vectorize the following C++ code: if(v[0] < 0) v[0] += 1.0f; if(v[1] < 0) v[1] += 1.0f; if(v[2] < 0) v[2] += 1.0f; if(v[3] < 0) v[3] += 1.0f; With SSE assembly this would be as simple as: movaps xmm1, xmm0 // v in xmm0 cmpltps xmm1, zero // zero =

I'm calling a list of symbols and then using a function to build a data frame from that symbol list. It works great until I introduce this index symbol from yahoo '^GSPC'. When and index symbol is introduced I get and error which is below. > Data <- symbolFrame(symbols) Error in get(S) : object '^GSPC' not found Since R does not like the ^ in front of a name it

Suppose I have the first vector: c(1, 6, 8, 9) I will like to create a second vector of size 10 composed of 0 and 1's. The second vector will be composed of four 1's and six 0's. The position of the 1's will be specificed by the first vector. So essentially, I want a second vector in the form: c(1, 0, 0, 0, 0, 1, 0, 1, 1, 0) Any help is greatly appreciated! -- View this message

I have a xts object made of daily closing prices I have acquired using quantmod. Here is my code: library(xts) library(quantmod) library(lubridate) # Gets SPY data getSymbols("SPY") # Subset Prices to just closing price SP500 <- Cl(SPY) # Show day of the week for each date using 2-6 for monday-friday SP500wd <- wday(SP500) # Add Price and days of week together