search for: a3

...83651,1.05384076104419,1.35730105037783,-0.889060319407602,1.29407560544756,-0.0145655801680721,2.62874531083287,-0.188992792468916,-2.36877528363888,0.193706203424581),id=c("a2","a2","a2","a2","a1","a1","a1","a1","a3","a3","a3","a3","a2","a2","a2","a2","a1","a1","a1","a1","a3","a3","a3","a3","a2","a2","a2","a2","a1...

...al","al","al","al","al","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a2","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3","a3") factB<-c("B1","B2","B3","B1","B2&q...

Dear R users, Basically, from the following arbitrary data set: a <- data.frame(id=c(c("A1","A2","A3","A4","A5"),c("A3","A2","A3","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),6),rep("Life",4))) > a id loc clm 1 A1 B1 Gene...

Hi, I apologize if there is a simple answer to this question that I've missed. I did search the mailing list but I might not have used the right keywords. Why does sum(A3^2) give the result of 1, but sum(A3^2)==1 give the result of FALSE? > A3<-matrix(nrow=3,c(1/(2^.5),1/(2^.5),0)) > A3 [,1] [1,] 0.7071068 [2,] 0.7071068 [3,] 0.0000000 > sum(A3^2) [1] 1 > sum(A3^2)^.5 [1] 1 > sum(A3^2)==1 # here's the part I don't understand [1...

...much. --Trey The function and calls below use the data in this Excel file (feel free to access): https://docs.google.com/leaf?id=0B5zZGW2utJN0ZDk1NjA0ZjUtMWU0ZS00ZGQ3LWIxZTUtOWE0NGVmYWMxODJl&hl=en_US ## - plot Siler survival curve ############################## silsurv<-function(a1,b1,a2,a3,b3) { sil=function(t) { h.t<-a1*exp(-b1*t)+a2+a3*exp(b3*t) S.t<-exp(-a1/b1*(1-exp(-b1*t))-a2*t+a3/b3*(1-exp(b3*t))) d.t<-S.t*h.t #return(d.t) return(S.t) #return(h.t) } t<-seq(0,90,1) plot(t,sil(t),ylim=c(0,1),type=&...

...,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4), id=c("a2","a2","a2","a2","a1","a1","a1","a1","a3","a3","a3","a3","a2","a2","a2","a2","a1","a1","a1","a1","a3","a3","a3","a3","a2","a2","a2","a2","a1...

=BE=B4=C6=F4=D5=DF=A3=BA =A1=B0=B5=DA=CB=C4=BD=EC=B1=B1=BE=A9=B3=AF=D1=F4=B9=FA=BC=CA=C9=CC=CE=F1= =BD=DA=A1=B1=BD=AB=D3=DA2003=C4=EA10=D4=C2=D4=DA=B1=B1=BE=A9=BE=D9=B0=EC=A1= =A3=BD=EC=CA=B1=A3=AC=B9=FA=BC=D2=C1=EC=B5=BC=C8=CB=A1=A2=CD=E2=B9=FA=D5=FE= =D2=AA=BA=CD=C0=B4=D7=D4=C3=C0=A1=A2=B5=C2=A1=A2=B7=A8=A1=A2=B0...

I was trying to learn cross building an R pcakage for windows on my linux machine. I picked a relatively small library gee to test. I downloaded the cross-tools and put them in my path, jyan at ludwig:/a3/jyan/src/R-1.2.1/src/gnuwin32$ echo $PATH /home/jyan/cross-tools/bin:/home/jyan/cross-tools/i386-mingw32msvc/bin: /usr/lib:/usr/local/bin:/usr/bin:/bin:/usr/bin/X11:/usr/games Following the instructions, I was able to go through the following: jyan at ludwig:/a3/jyan/src/R-1.2.1/src/gnuwin32$ mak...

Dear sir, I have a matrix like a<-matrix(c(0,2,0,4,0,6,5,8,0),nrow=3) colnames(a)<-c("F1","F2","F3") rownames(a)<-c("A1","A2","A3") a F1 F2 F3 A1 0 4 5 A2 2 0 8 A3 0 6 0 I want to extract all pairs (rownames, columnames) from which the value in the matrix is 0 The result should be something like this A1, F1 A2, F2 A3, F1 A3, F3 how it is possible? thanks for your help.... Best Regards Alberto [[alte...

Dear people from the R help list, I have a question that I can't get my head around to start answering, that is why I am writing to the list. I have data in a format like this (tabs might look weird): John A1 1 0 1 John A2 1 1 1 John A3 1 0 0 Mary A1 1 0 1 Mary A2 0 0 1 Mary A3 1 1 0 Peter A1 1 0 0 Peter A2 0 0 1 Peter A3 1 1 1 Josh A1 1 0 0 Josh A2 Josh A3 0 0 0 I want to convert it into a format where va...

.... I have a 64-bit CentOS 5.6 VM running on a 64-bit FC14 host, libvirt 0.8.3-10.fc14, 64-bit. I am getting messages in my syslog file, 2 about every 20 min. about dnsmasq: Aug 17 04:53:52 Q6600 dnsmasq-dhcp[1969]: DHCPREQUEST(virbr0) 192.168.122.213 52:54:00:a4:98:a3 Aug 17 04:53:52 Q6600 dnsmasq-dhcp[1969]: DHCPACK(virbr0) 192.168.122.213 52:54:00:a4:98:a3 Aug 17 05:20:13 Q6600 dnsmasq-dhcp[1969]: DHCPREQUEST(virbr0) 192.168.122.213 52:54:00:a4:98:a3 Aug 17 05:20:13 Q6600 dnsmasq-dhcp[1969]: DHCPACK(virbr0) 192.168.122.213 52:54:00:a4:98:a3 Aug 17 05:40:57...

..."logit.gee", id = "PlotID", corstr = "unstructured") #Ideally I think I need a correlation matrix structured like below (given for a plot with 3 trees). Here a1 is the correlation within a tree, a2 the correlation between trees from the same plot and the same year and a3 the correlation between trees from the same plot but a different year. Does it make sense to run the model with a unstructured correlation, calculate the average a1, a2 and a3 and use that as a fixed working correlation? matrix(c( 1, "a1", "a2", "a3", "a2&quo...

Hi, I am looking forward to fill the plot using conditions on variables a2 and a3. Whenever variable(a2) goes above variable(a3) i fill it with some color . I am storing the coordinates of a2 and a3 in x and y as well as time where it is occurring . But it is not producing properly. I must be wrong in assigning coordinates. What should be the correct way to produce the desired...

...ake a user defined function so that I can input the desired conditions and just get the results accordingly. Below is an arbitrary data set & sample statements with its outputs in accordance with specified conditions: x <- data.frame(ID=rep(letters[1:5],2), A1=rep(10:14,2), A2=rep(2:6,2), A3=c(101:105,95:99), A4=3*ceil(rt(10,2))) x1 <- subset(x, ID == "a" & A1 <= 10 & A2 > 1 & A3 > 100) #condition 1 x2 <- subset(x, ID == "a" & A1 >= 10 & A2 > 1 & A3 < 100) #condition 2 x3 <- subset(x, ID == "a" & A1...

...n with an old version, but it's in latest version as well. The fix is simple. In the summary.table function, the parameter is calculated incorrectly for a test of independence among all cells when the table is more than 2-way table. Example: Consider X: > X a b c 1 A1 B2 C1 2 A3 BA3 C2 3 A2 B1 C4 4 A1 B2 C3 5 A3 BA3 C2 6 A1 BA3 C1 7 A2 BA3 C2 8 A1 BA3 C3 9 A1 B2 C1 10 A1 BA3 C2 11 A2 B1 C2 12 A1 B2 C3 13 A3 BA3 C4 14 A2 B1 C3 15 A2 B1 C2 16 A2 BA3 C4 17 A3 BA3 C3 18 A3 BA3 C4 19 A2 BA3 C4 20 A1 B2 C3 21 A1 BA3 C1 22 A2 B1 C2 23 A2 BA3 C...

Dear User, thank for the attention. I have a data.frame with 5 columns (ex:ID, a1,a2,a3,a4) and 1000 rows. I wish to find the absolute max value for all data.frame and save a new data.frame with the row where is that value. Ex: ID: 1,2,3,4,5,6,7,8,9,10 a1:1,2,3,4,5,6,7,8,9,10 a2:11,12,13,14,15,16,17,18,19,20 a3:21,22,23,24,25,26,27,28,29,30 a4:31,32,33,34,35,36,37,38,39,40...

...Gompertz-Makeham, with the addition of a juvenile component that includes two parameters---one that describes the initial infant mortality rate, and a negative exponential that describes typical mortality decline over the juvenile period. The entire hazard is expressed as h(x) = a1*exp(-b1*x)+a2+a3*exp(b3*x) I've had success in plotting the curves using the following function: ############################################################ siler<-function(x=c(0.1,0.5,0.001,0.003,0.05)) # model Siler parameters { sil=function(t) { a1 = x[1] b1 = x[2]...

I want to do something like this in R. If I have three vectors > a1 [1] 1 2 3 > a2 [1] 4 5 6 > a3 [1] 9 10 7 I want to compute 1. A vector that is the mean at each serial position of a1, a2, and a3. so in this example it would have the contents 4.667, 5.667, 5.333333 2. A vector that is the SD at each serial position of a1, a2, and a3. so in this example it would have the contents 4.041452,...

...he shape of a data frame using a "pivoting-like" function. I have a dataframe df of observations as follows: ID VALIDITY YEAR PROPERTY PROPERTY VALUE A1 2007 P1 V1 A1 2007 P2 V2 A1 2007 P3 V3 A1 2008 P1 V10 A1 2008 P2 V20 A2 2007 P5 V50 A2 2008 P6 V20 A3 2007 P1 V1 A3 2007 P3 V30 A3 2008 P1 V10 A3 2008 P2 V4 A3 2008 P6 V25 (you can imagine that this data is collected every year from a sample of people with several "measures" - weight, number of children, income... It can happen that some properties could be missi...

Hello, I'm new in R and I want to do one thing that is very easy in excel, however, I cant do it in R. Suppose we have the data frame: data<- data.frame(A=c("a1","a2","a3","a4","a5")) I need to obtain another column in the same data frame (lets say B=c(b1,b2,b3,b4,b5) in the following way: b1=a1/(a1+a2+a3+a4+a5) b2=a2/(a2+a3+a4+a5) b3=a3/(a3+a4+a5) b4=a4/(a4+a5) b5=a5/a5 a1..a5 and b1...b5 are always numeric values (this...