similar to: Simple lm/regression question

I'm sure this must be trivial, but I'm a novice with R and can't work out how to handle the axes when I am constructing multiple plots on a page and try to return to a previous one to put multiple data sets it. A simple example: --- x<- 1:10 y<- (1:100)*3 par(mfcol=c(2,1)) plot(x) plot(y) par(mfg=c(1,1)) lines(x) --- The first 5 lines make two plots with a row of dots along

Using a more stable nonlinear modeling tool will also help, but key is to get the periodicity right. y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156, 161, 169, 176, 182, 190, 197, 209, 218, 232, 240) lidata <- data.frame(y=y, t=t) #I use the

Hi lily, You can get fairly good starting values just by eyeballing the curves: plot(y) lines(supsmu(1:20,y)) lines(0.6*cos((1:20)/3+0.6*pi)+17.2) Jim On Wed, Jun 21, 2017 at 9:17 AM, lily li <chocold12 at gmail.com> wrote: > Hi R users, > > I have a question about fitting a cosine curve. I don't know how to set the > approximate starting values. Besides, does the method

Hi R users, I have a question about fitting a cosine curve. I don't know how to set the approximate starting values. Besides, does the method work for sine curve as well? Thanks. Part of the dataset is in the following: y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96,

What I did was to plot your initial values, then plot the smoothed values and guess the constants. That is, I got an "eyeball" fit to the smoothed values. As I have described this as "gross cheating" in the past, you should either split your data, estimate on one subset and then test on another, or estimate on your data and test on a replication. If you get pretty much the same

Hi R-users: [R 2.2 on OSX 10.4.3] I have a (sparse) vegetation data frame with 500 rows (sampling units) and 177 columns (plant species) where the data represent % cover. I need to summarize the cover data by returning the names of the most dominant and the second most dominant species per plot. I reduced the data frame to omit cover below 5%; this is what it looks like stacked. I have

I compiled R 2.1.0 under Windows XP SP2 as a preliminary to rebuilding a custom package for use with R 2.1.0. The compile completed successfully, and I was able to run demo(graphics) successfully. But make check and make check-recommended fail. > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 1.0 year 2005

--l4Siqd0eqV Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hello, I run the following job. Please, compare the results of summary and table concerning berufl. From similar SPSS/PSPP runs, the result of table is correct. Did I misunderstand anything or is there a bug? What does the difference come from? What does '(other)' mean? What about the strange

Hello all R-users _question 1_ I need to make a statistical model and respective ANOVA table but I get distinct results for the T-test (in summary(lm.object) function) and the F-test (in anova(lm.object) ) shouldn't this two approach give me the same result, i.e to indicate the same significants terms in both tests??????? obs. The system has two restrictions: 1) sum( x_i ) = 1 2) sum(

fitz_ra wrote > I know this is posted a lot, I've been through about 40 messages reading > how to do this so let me apologize in advance because I can't get this > operation to work unlike the many examples shown. > > I have a 2 row matrix >> temp > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] > [,9] [,10] > [1,] 17.000

Hi all, I have a very easy questions (I hope). I had measure a property of plants, growing in three different substrates (A, B and C). The rest of the conditions remained constant. There was very high variation on the results. I want to do address, whether there is any difference in the response (my measurement) from substrate to substrate?

Hi, I'm a bit confused how julian() works. If I understand right, it returns the number of days since the origin. I have a vector: > SLDATX[1:10] [1] "1986-01-06" "1986-01-17" "1986-02-02" "1986-02-04" [5] "1986-02-04" "1986-02-21" "1986-03-06" "1986-03-25" [9] "1986-04-06"

If you know the period and want to fit phase and amplitude, this is equivalent to fitting a * sin + b * cos > >>> > I don't know how to set the approximate starting values. I'm not sure what you meant by that, but I suspect it's related to phase and amplitude. > >>> > Besides, does the method work for sine curve as well? sin is the same as cos with

I'm trying the different parameters, but don't know what the error is: Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates Thanks for any suggestions. On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen <don-r-help at isis.cs3-inc.com> wrote: > > If you know the period and want to fit phase and amplitude, this is > equivalent to

I am trying to use R lm() with quantitative and qualitative predictors, but am getting different results than those that I get in SAS. In the R ANOVA table documentation I see that "Type-II tests corresponds to the tests produced by SAS for analysis-of-variance models, where all of the predictors are factors, but not more generally (i.e., when there are quantitative predictors)." Is

Hi All, I have a function that is used with data frames having multiple id's per row and it aggregates the data down to 1 id per row. It also randomly selects one of the within-id values of a variable (mod), which often differ within-id. Assume this data frame (below) is much larger and I want to repeat this function, say 100 times, and then derive the mean values of r over those 100

Dear all, I have just detected what seems a minor inconsistence with data types. If one unlists a POSIXlt time with GMT zone gets a numeric vector, since the POSIXlt list has no `zone` element, while if one unlists a POSIXlt time with a non GMT zone (also non specifying tz if the Sys.timezone is not GMT) gets a character vector due to including the `zone` element. > x <-

Dear R users, Being a R-novice using version 1.7.0 I have not managed to solve the following problem, so I turn to you: I have 12 pairs nonlinear curves having some parameters (p1,p2,p3) in common and others specific to one of the curves (p4a,p4b). I want to fit a nonlinear mixed model with nmle with the random effects varying over the 12 samples (pairs). In order to be able to estimate the

Ok, Thanks foe the answer, Ken: *1L, 2L etc are integers. (That is, identical to as.integer(1) , as.integer(2) etc) Using integers (instead of "numeric" type) is more efficient as here they're used as indexes and would be converted to integer anyway. Compare > is(1) ... and > is(1L) 1L:p is the sequence 1, 2, 3, ..., p (just like 1:p) Just for curiosity, what is

Dear all, Sorry for bringing up an old issue: >pexp(50, 0.5) [1] 1 In some cases, pexp() gives CDF=1. I read some discussion in 2002 saying it has been patched. However it's not working in "R2.4.1Patched". Could anyone help me out? Thanks a lot, Jeann _________________________________________________________________ Fine Dining & Fancy Food. Check Out This