similar to: How can I read the following complicated table

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]]

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

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

Hi, I conduct a panel data estimation and obtain estimators for two of the coefficients beta1 and beta2. R tells me the mean and covariance of the distribution of (beta1, beta2). Now I would like to find the distribution of the quotient beta1/beta2, and one way to do it is to simulate via the joint distribution (beta1, beta2), where both beta1 and beta2 follow t distribution. How could we

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

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 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),

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

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

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-

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 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 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

Dear all, Let say I have following data-frame: Dat <- structure(list(dat = c(-0.387795842956327, -0.23270882099043, -0.89528973290562, 0.95857175595512, 1.61680582493783, -1.17738110289352, 0.210601060411423, -0.827369747447338, -0.36896112964414, 0.440288648776096, 1.28018410608809, -0.897113649961341, 0.342216546981718, -1.17288066266219, -1.57994101992621, -0.913655547602414,

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 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]]

Hi, I am using xylot function in Lattice package. I find it excellently written, but I don't know how to find resources on this function. One thing I would like to change is the line type. To my knowledge, type ="l" means a regular line, type "g" means broken lines, but I can't find a complete list of the choices. The info obtained from the query in R is limited.

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]]

Hi, I have a table which I can read via read.csv: fx1<-read.csv(file="A_FX_M.csv", header=TRUE) TIME REER NTD JPY GBP HKD 1 198001 124.26 36.030 237.96 2.263980 4.8366 2 198002 126.59 36.030 244.05 2.290426 4.8765 3 198003 128.33 36.026 248.62 2.206045 4.9960 4 198004 127.85 36.063 251.67 2.215330 4.9760 5 198005 124.40 36.050 228.35 2.302026 4.8891 6 198006

I have many xls grade report sheets with the same format XXX High School Grade Report Confidential Yes Math English Science John 90 85 90 Mary 75 88 93 …… Since the reports are prepared on a regular basis, I have many reports with identical format. Without the title “XXX High School Grade Report Confidential Yes”, I can just change the xls to csv and read them via read.csv. How