[R] Removing objects except user-defined functions

Dear R users. I would like to remove all object from my workspace except the function I have defined. However, is I use rm(list = ls()) everything is cleared. I was thinking to typeof to get information about objects, but I could not get it working right. Thank in advance, Phil

[R] Evaluating a formula

Hi all. How we evaluate a formula in R? Ex.: params <- list(a = 2, b = 3) x <- seq(1,10, length.out = 100) func1 <- as.formula("y ~ a*x^2 + b*x") ##How to evaluate func1 using x and the params list ??? Thank you in advance, Phil [[alternati

Re: [R] Substitute initial guesses of parameters in a function

Thank you! > Date: Wed, 3 Dec 2014 08:02:17 -0500 > From: murdoch.dun...@gmail.com > To: pmassico...@hotmail.com; r-help@r-project.org > Subject: Re: [R] Substitute initial guesses of parameters in a function > > On 03/12/2014 7:37 AM, philippe massicotte wrote: > &g

[R] Substitute initial guesses of parameters in a function

Hi everyone, I have a formula like this: f <- as.formula(y ~ p0a * exp(-0.5 * ((x - p1a)/p2a)^2)) I would like to "dynamically" provide starting values for p0a, p1a, p2a. Is there a way to do it? #Params estimates p <- c(12, 10, 1) # This is where I have difficulties mystart <- substitute(...)

[R] Formula with dynamic number of variables

Hi everyone. I have a non-linear model specified as this (6 variables, a0, S, K, p0, p1, p2): fit <- nlsLM(formula = dat ~ a0 * exp(-S*(x - 250)) + K + ( C ) ) where C = (p0*exp(-0.5*((x-p1)/p2)^2)) The problem is that I do not know in advance how many component (C) I will have in the model. T

Re: [R] legend position

plot(r, smallplot=c(.15, .17, .5, .85), legend.only=TRUE) > > - > David L Carlson > Department of Anthropology > Texas A&M University > College Station, TX 77840-4352 > > -Original Message- > From: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org]

Re: [R] legend position

Thank you, I'll try to work with lattice. Regards,Phil > Date: Mon, 2 Dec 2013 12:06:50 -0800 > From: c...@witthoft.com > To: r-help@r-project.org > Subject: Re: [R] legend position > > It occurs to me that perhaps you're referring to the 'color bar' on the right > of the plot. AFAIK you cannot

Re: [R] legend position

Thank you for reply. If I'm not wrong, legend(...) will works for discrete elements. I'm not sure hot to use it for a colorbar legend sur as the one in the example bellow. Phil > Date: Mon, 2 Dec 2013 11:49:19 -0800 > From: c...@witthoft.com > To: r-help@r-project.org > Subject: Re: [R] legend po

[R] legend position

Hi all. I'm ploting a raster and I can't find the proper way to move the legend. For example, r = raster(system.file("external/test.grd", package="raster"))plot(r) How can I put the legend at the desired position? Thank in advance,Phil [[alternative HT

Re: [R] data manipulation

<- > > > names(neutral_classes[rep(1:length(neutral_classes), > > > sapply(neutral_classes, length))] > > > > > > # get one or more results with 'match': > > > names(neutralVec[match(c(50, 20, 10, -4), neutralVec)]) > > > > >

Re: [R] data manipulation

eutral_classes) > > > > names(neutralVec) <- > > names(neutral_classes[rep(1:length(neutral_classes), > > sapply(neutral_classes, length))] > > > > # get one or more results with 'match': > > names(neutralVec[match(c(50, 20, 10, -4), neutralVec

[R] data manipulation

Hi everyone. I have a list like this: neutral_classes = list(A = 71:100, B = 46:70, C = 21:45, D = 0:20) and I'm trying to return the letter of the named vector for with an integer belong. For example, B if I use the value 50. Any help would be greatly appreciated. Regards,Phil

Re: [R] Trying to avoid nested loop

Thank you for your answer. This is what I needed. > From: s.elli...@lgcgroup.com > To: r-help@r-project.org > Date: Fri, 4 Oct 2013 15:13:49 +0100 > Subject: Re: [R] Trying to avoid nested loop > > > > I'm trying to avoid using nested loops in the following code but I'm > > not sure how to pro

[R] Trying to avoid nested loop

Dear R users. I'm trying to avoid using nested loops in the following code but I'm not sure how to proceed. Any help would be greatly appreciated. With regards,Phil X = matrix(rnorm(100), 10, 10) ## Version with nested loopsresult = 0 for(m in 1:nrow(X)){ for(n in 1:ncol(X)){if(X[m,n] !=

[R] Combining rasters

Hi everyone. I would like to know if it is possible to combine rasters in R to form a "collage". For example, I would like to place 2 copies of the R logo side by side. r = raster(system.file("external/rlogo.grd", package = "raster")) After reading the help file (maybe I missed it) I did no

Re: [R] Problem with converting F to FALSE

Look at "colClasses" in ?read.csv > Date: Thu, 5 Sep 2013 18:14:49 +0530 > From: kiran4u2...@gmail.com > To: r-help@r-project.org > Subject: [R] Problem with converting F to FALSE > > Hi, > I have a peculier problem in R-Project that is when my CSV file have one > column with all values as 'F' th

Re: [R] Histogram

> > > David Carlson > > > > > > -Original Message- > > From: r-help-boun...@r-project.org > > [mailto:r-help-boun...@r-project.org] On Behalf Of philippe > > massicotte > > Sent: Wednesday, September 4, 2013 3:02 PM > > To: Rui B

Re: [R] Histogram

:10, breaks = 0:10) # all counts are 1 > > > and see the difference between h1 and h2, components 'breaks' and 'counts'. > > Hope this helps, > > Rui Barradas > > Em 04-09-2013 19:34, philippe massicotte escreveu: > > Hi everyone. > > I&#

[R] Histogram

Hi everyone. I'm currently translating some Matlab code into R. However, I realized that the hsit function produce different results in both languages. in Matlab, hist(1:10, 10) will produce 10 bins with a count of 1 in each, but in R it will produce 9 classes with count of 2,1,1,1,1,1,1,1,1. I'm

[R] RCurl and google trends

quot;GALX"\\s*value="([^"]+)"')) galx <- str_replace(string = galx.match, pattern = ignore.case('name="GALX"\\s*value="([^"]+)"'), replacement = "\\1") authenticatePage <- postForm(authenticateURL, .params = list(Email =

Re: [R] Fine control of plot

uot; with layout(), as in: > layout(matrix(c(1,1,1,2), nrow=1)) > > but the better way is to see xpd within ?par as described here: > https://stat.ethz.ch/pipermail/r-help/2009-July/206311.html > > along with par()$mai to set the margins appropriately. > > Sarah > > O

Re: [R] Fine control of plot

Hi and thank you for your answer.  Sorry for the html post, here's the code: (you missed a break line between +x and plot(...)  layout(matrix(c(1,2), 1, 2, byrow = TRUE), widths=c(6,2), heights=c(1,1))  x = 1:100  y = rnorm(x)+x  plot(x,y)  reg = lm(y~x)  abline(reg, col = "red")  plot(1, typ

[R] Fine control of plot

Hi everyone. I'm trying to create a graph where I could plot some lines on the right side. Here an example: layout(matrix(c(1,2), 1, 2, byrow = TRUE), widths=c(6,2), heights=c(1,1)) x = 1:100y = rnorm(x)+xplot(x,y) reg = lm(y~x)abline(reg, col = "red") plot(1, type="n", axes=F, xlab="", ylab="", x

[R] Combining boxplot

Hi everyone. I have two data frames that contain the same variables but with different number of observation. I would like to know it was possible to combine the data so I can have "paired" boxplot on the same figure. For example, df1 = data.frame(x = rnorm(100)) df1$type = ifelse(df1$x

[R] Help with layout

Dear R users. I'm new with layout and I can't figure how to teak my graphs. I have the following code: layout(matrix(c(1,2,3), 3, 1, byrow = TRUE), heights=c(0.3,0.3,0.6)) boxplot(rnorm(100), horizontal=TRUE, axes=FALSE) boxplot(rnorm(100), horizontal=TRUE, axes=FALSE) hist(rnorm(100))