similar to: Setting up a blank table with column names in the hard drive

Dear friends. With file( ) to obtain a pointer of a file, every time we use scan ( ) to read one row of it, the pointer will point to the next row of the file. In the following example, d1 and d2 are obtained the same way but they correspond to different rows of the same file because the pointer of the file moves down a row when a row of the file is read. The following is an example: a1

Dear friends. Every loop of my program will result in a list that is very long, with a structure similar to the one below: Lst <- list(name="Fred", wife="Mary", daily.incomes=c(1:850)) Please notice the large size of "daily.incomes". I need to store all such lists in a csv file so that I can easily view them in Excel. Excel cannot display a row of more than 300

Dear Friends. Greetings! I have asked the question of how to set up a blank file and write a list to it as a row for many times, with the number of lists unknown. I have received many beautiful solutions. Thanks go to Professor *Murdoch, Professor *Menne, Professor Grothendieck and Dr. Olshansky. I have organized the solutions below: ########################################## *Set up a

Hallo, I build a list by the following way: Lst = list(name="Fred", wife="Mary", no.children=3, cild.ages=c(4,7,9)) I know how I can extract the information one by one. But now I want to add a new entry which looks like name="Barney", wife="Liz", no.children=2, cild.ages=c(3,5) How can I add this information to Lst without overwriting the first entry?

Dear Friends. I tried to use nls.control() to change the 'minFactor' in nls( ), but it does not seem to work. I used nls( ) function and encountered error message "step factor 0.000488281 reduced below 'minFactor' of 0.000976563". I then tried the following: 1) Put "nls.control(minFactor = 1/(4096*128))" inside the brackets of nls, but the same error message

Dear Friends. I have a very long expression and I use function D to find its derivative, which will be even longer. I save the resulting expression in a variavle, say bbb. But when I tried to display bbb on the screen, the R screen is not long enough for me to view it. Is there a way to save the expression to a file? Best Wishes Yuchen Luo [[alternative HTML version deleted]]

Dear Friends. I found a puzzling phenomenon in R when you use 'if' within a function: # defining a function aaa aaa=function(a) {if (a==1) {aaa=1}; if (a!=1) {aaa=2} } # using the function: > b=20 > bbb=aaa(b) > bbb [1] 2 > typeof(bbb) [1] "double" > > > c=1 > ccc=aaa(c) > ccc NULL > typeof(ccc) [1] "NULL" It seems that only the last

I am using a "for" loop to read a table row by row and I have to specify how many records are there in the table. I need to read row by row because the table is huge and the memory not large enough for the whole table.: number.of.records=100 fp=file("abc.csv","r") pos=seek(fp, rw="read") for (i in 1:number.of.record){ current.row=scan(file=fp,

Dear collegues. I am using scan( ) to read from a table (a csv file). I am wondering how to let the program know that the end of the file is reached? Your help will be highly appreciated! Best Wishes Yuchen Luo [[alternative HTML version deleted]]

Dear friends. I am using function constrOptim(c(0.5,0.3,0.5), fit.error, fit.error.grr, ui=-1*ui,ci=-1*ci) and I am confronted with error message "initial value not feasible" I plug in the initial value of (0.5,0.3,0.5) to function fit.error and fit.error.grr and have pretty reasonable result. I inequality "ui %*% theta - ci >= 0" as suggested in the R manual and it is

Dear friends. I use nls() and encounter the following puzzling problem: I have a function f(a,b,c,x), I have a data vector of x and a vectory y of realized value of f. Case1 I tried to estimate c with (a=0.3, b=0.5) fixed: nls(y~f(a,b,c,x), control=list(maxiter = 100000, minFactor=0.5 ^2048),start=list(c=0.5)). The error message is: "number of iterations exceeded maximum of

Dear Colleagues. This should be a very common operation and I believe there should be a nice way in R to handle it. I couldn't find it in the manual or by searching online. I am wondering if I could ask for some help in this community. I have 48 csv files; each stores the data for a specific month. The 48 corresponding months are consecutively from January, 2001 to December, 2004. I name

Hi, my friends. When a data file is large, loading the whole file into the memory all together is not feasible. A feasible way is to read one row, process it, store the result, and read the next row. In Fortran, by default, the 'read' command reads one line of a file, which is convenient, and when the same 'read' command is executed the next time, the next row of the same file

I'm studying lists and I came to an example where > L $name [1] "Fred" $wife [1] "Mary" $no.children [1] 4 $child.ages [1] 4 7 9 then following the instructions to extend the list with a new component, I executed: > L[5] <-list(NewName="something") and the new list I got was: > L $name [1] "Fred" $wife [1] "Mary"

Hi, my friends. When a data file is large, loading the whole file into the memory all together is not feasible. A feasible way is to read one row, process it, store the result, and read the next row. In Fortran, by default, the 'read' command reads one line of a file, which is convenient, and when the same 'read' command is executed the next time, the next row of the same file

Dear friends. I use ConstrOptim( ) and got error message "initial value not available". My understanding of "initial value not available" is that one of the following 3 cases happens: 1.The objective function is not well defined at the point of the initial value. 2. The differentiation of the objective function is not well defined at the point of the initial value. 3. The

Dear Colleagues. I believe this should be a problem encountered by many: nls( ) is a very useful and efficient function to use if we are just to display the estimated value on screen. What if we need R to store the estimated parameter in a variable? For example: x=rnorm(10, mean=1000, sd=10) y=x^2+100+rnorm(10) a=nls(y~(x^2+para),control=list(maxiter = 1000, minFactor=0.5

"Beast" is more appropriate than you know. It''s definately Frankenstein code. (read: it ain''t pretty, but it works - some of the time). Anyway, it entertains my wife; not sure if hardcore programmers will get anything out of it or not. You might find it interesting (or pointless) that in a complete bastardization of rails'' main reason for being it

Dear Friends. I found something very puzzling with constOptim(). When I change the parameters for ConstrOptim, the error messages do not seem to be consistent with each other: > constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.3, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible > constrOptim(c(0.5,0.9,0.5),

I'm using R on Windows v2.0.1 with the nlme package (v3.1-53) and am finding some unexpected discrepancies in the output of intervals.lme and coef.lme. I've included a toy dataset at the end, but briefly, the data are longitudinal data from couples in marital therapy. Each spouse's relationship satisfaction is measured 4 times; I've fit both linear and quadratic models to the