similar to: Adding legends on plots using Lattice package

Example: Will ts objects be obsolete or modified? > a [,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035> lag(a,4) [,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2

Hello, I am writing codes for time series computation but encountering some problems Given the quarterly data from 1983Q1 to 1984Q2 PI1<-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485, -1.190061246, -0.553031799, 0.686874720, 0.953911035), start=c(1983,1), frequency=4) > PI1 Qtr1 Qtr2 Qtr3 Qtr4 1983 2.747365190 2.791594762

I have a program, when I write if(num!=NA) it yields an error message. However, if I write if(is.na(num)==FALSE) it works. Why doesn't the first statement work? Thanks, Miao [[alternative HTML version deleted]]

Hello, I have a question on how to tabulate the time series data. I use RStudio, but if can be done in any other R editor, it should work in RStudio as well. > a1<-11:22 > a1ts<-ts(a1, frequency=4, start=c(1978,1)) > a1ts Qtr1 Qtr2 Qtr3 Qtr4 1978 11 12 13 14 1979 15 16 17 18 1980 19 20 21 22 If I click the variable "a1ts" on the

Hi, I have a structure, which is the result of a function How can I access the elements in the gradient? > dput(test1) structure(-1.17782911684913, gradient = structure(c(-0.0571065371783791, -0.144708170683529), .Dim = 1:2, .Dimnames = list(NULL, c("x1", "x2")))) > test1[[1]] [1] -1.177829 > test1 [1] -1.177829 attr(,"gradient") x1

Hi, How could we print the frequency table (produced by "table") to an Excel file? Is there an easy way to do so? Thanks, Miao > df <- data.frame(x = 1:3, y = 3:1, z = letters[1:3]) > table(df[,c("y","z")]) z y a b c 1 0 0 1 2 0 1 0 3 1 0 0 > test<-table(df[,c("y","z")]) > as.data.frame(test) y z Freq 1 1 a

Hello, I am using RStudio and have trouble finding out the problematic line in the presence of a bug. Could I view the line NUMBER which contains a bug? Is there any R editor able to do it? Thanks, Miao [[alternative HTML version deleted]]

Hello, I work with time series data. From time to time I run programs to produce results that are in time series form (e.g., quarterly or monthly data). After a few days I might need to access part of the results and to run another program. Is there any function or package (like dataframe or zoo?) that might help so that I don't need to copy the results manually to a csv or xls file?

Taking the Grunfeld data, which is built-in in R, for example, (1)How can I construct a dataset (or dataframe) that consists of the data of all firms in 1951? (2)How can I calculate the average capital in each form over the period 1951-1954? What I can imagine is to categorize the data by firm, and then select the data between 1951 and 1954 for each firm, but how can I do it? Thanks, Miao

Hi, I'd like to make a time series at an annual frequency. > a<-xts(x=c(2,4,5), order.by=c("1991","1992","1993")) Error in xts(x = c(2, 4, 5), order.by = c("1991", "1992", "1993")) : order.by requires an appropriate time-based object > a<-xts(x=c(2,4,5), order.by=1991:1993) Error in xts(x = c(2, 4, 5), order.by =

Hi, I let xzoo be an empty object: > xzoo<-{} and I have an existing zoo object x1zoo_f. I would like to combine the two to make a new zoo object, and continue doing so in a loop, which is not shown here. However, when I type > xzoo<-cbind(xzoo, x1zoo_f) An error message emerges Error in zoo(structure(x, dim = dim(x)), index(x), ...) : “x” : attempt to define

Hi everyone and Achim, Achim, I appreciate your help about the function "NCOL". When I use "NCOL" instead of "ncol", I can find out the number of columns (number of time series) in the presence of only one time series (one variable, one column). Now I want to know how I can find out the column names of the zoo objects? In case of more than one time series, the

Hello, I would like to produce a few time series graphs with R. My data is usually quarterly or monthly. I sometimes need to have two y-axes, one at the left, the other right. Is there any toolbox producing nice looking time series graphs? Thanks, miao [[alternative HTML version deleted]]

Hi, I am graphing with the following command in the Lattice and LatticeExtra package xyplot(xts,lty=c(1,2),col=c("blue","red"),type=c("l","g"),par.settings = list(layout.heights = list(panel = c(2, 2))), aspect="xy",xlab="",ylab="%", key=key1,screen=list(a,a,b,b,c,c,d,d), layout=c(2,2),

HI, I have a dataframe with two variable A, B. I transform the two variable and name them as C, D and save it in a dataframe dfcd. However, I wonder why can't I call them by dfcd$C and dfcd$D? Thanks, Miao > A=c(1,2,3) > B=c(4,6,7) > dfab<-data.frame(A,B) > C=dfab["A"]*2 > D=dfab["B"]*3 > dfcd<-data.frame(C,D) > dfcd A B 1 2 12 2 4 18

Hi, I would like to read several datasets and would like to create a set (list? sequence?) of many empty dataframes. How could this be done? How could I declare a set (list? sequence?) of many empty matrices? Thanks, Miao [[alternative HTML version deleted]]

Dear Daniel and Jorge, Thank you very much and it does help. If I have a string "ABCD", how can I access the second element of the string "B"? Thanks, Miao 2012/7/27 Daniel Nordlund <djnordlund@frontier.com> > > -----Original Message----- > > From: r-help-bounces@r-project.org [mailto:r-help-bounces@r-project.org] > > On Behalf Of jpm miao

It seems to only works for zoo or ts objects? I tried to run it for xts objects, and error message occurs. Once I coerce it to be an zoo object (by as.zoo), it works. Error message: Error in model.frame.default(formula = dynformula(PIh - PI ~ L(X, 0:i) + : variable lengths differ (found for 'L(X, 0:i)') In addition: Warning messages: 1: In zoo(coredata(x), order.by = index(x), ...)

Hi jpm miao, You can get CSV files that can be imported into Excel like this: library(prettyR) sink("excel_table1.csv") delim.table(table(df[,c("y","z")])) sink() sink("excel_table2.csv") delim.table(as.data.frame(table(df[,c("y","z")])),label="") sink() sink("excel_table3.csv")

Hello, I have problem running the histogram function "hist". The area under the histogram is much lower than 1. Could anyone tell me what the problem is? Thanks, (The total number of observation is 992 (close to 1000), so the probability that 0<Y1<35 is approximately 0.277) miao rm(list=ls()) par(mfrow=c(1, 1)) Y <- cbind(matrix(35*0.5,1,277), matrix(35*1.5, 1, 146),