search for: slopes

Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns

I created a small example to show something that I do a lot of. "scale" data by month and return a data.frame with the output. "id" represents repeated observations over "time" and I want to scale the "slope" variable. The "out" variable shows the output I want. My for..loop does the job but is probably very slow versus other methods. ddply

Hello, I'm running into a very strange problem: > xrange <- c(-2.5,2.5) > xdim <- 100 > mobility <- 0.1 > slope <- 1.16 > urange <- slope*xrange > udim <- max(slope*xdim,5) > du <- (urange[2]-urange[1])/udim > uvec <- urange[1]+(1:udim-0.5)*du > # type dependent weight function > ckern <-

Hi there, am trying to run a linear regression with a slope of 0. I have a dataset as follows t d 1 303 2 302 3 304 4 306 5 307 6 303 I would like to test the significance that these points would lie on a horizontal straight line. The standard regression lm(d~t) doesn't seem to allow the slope to be set. Any help very welcome. ed

Hello, I would like to write a loop to 1) run 100 linear regressions, and 2) compile the slopes of all regression into one vector. Sample input data are: y1<-rnorm(100, mean=0.01, sd=0.001) y2<-rnorm(100, mean=0.1, sd=0.01) x<-(c(10,400)) #I have gotten this far with the loop for (i in 1:100) { #create the linear model for each data set model1<-lm(c(y1[i],y2[i])~x) slope1&l...

I am learning ggplot2, and need your help. When I try > p <- ggplot(mtcars, aes(x = wt, y=mpg)) + geom_point() > p + geom_abline(slope=5) (from http://had.co.nz/ggplot2/geom_abline.html) the slope of the abline does not change, but this works: > p + geom_abline(intercept=20) In order to have slope work, I have to use > p + geom_abline(aes(slope=5)) Is it a bug, or is there

Not really r-specific: Z = (b1 - b2) / SQRT ( SEb1^2 + SEb2^2) -------Original Message------- From: Gijsbert Stoet <stoet at volition.wustl.edu> Sent: 07/20/03 09:51 PM To: r-help at stat.math.ethz.ch Subject: [R] how to test whether two slopes are sign. different? > > Hi, suppose I do want to test whether the slopes (e.g. determined with lsfit) of two different population are significantly different, how do I test this (in R). Say for example, I found out what the slope between age and number of books read per year is for two...

...ible values). There are several more variables too, many of which we're using to fit regression models using lm. So I have 3 models fitted (one for each subset of course), each having slope estimates for the predictor variables. What we want to find out, though, is whether or not the overall slopes for the 3 regression lines are significantly different from each other. Is there a way, in R, to calculate the overall slope of each line, and test whether there's homogeneity of regression slopes? (Am I using that phrase in the right context -- comparing the slopes of more than one regressio...

Hello, I have a challenge! I have a large dataset with three columns, "date","temp", "location". "date" is in the format %m/%d/%y %H:%M, with a "temp" recorded every 10 minutes. These temperatures of surface temperatures and so fluctuate during the day, heating up and then cooling down, so the data is a series of peaks and troughs. I would like

Hi All, I am confused with SMATR's test for common slope. My null hypothesis here is that all slopes are parallel (common slopes?), right? So if I get a p value < 0.05 means that we can have confidence to reject it? That slopes are different? Or the other way around? it means that we have statistical confidence that the slopes are parallel? thanks -- Eugenio Larios PhD Student University of Ar...

Hi, I'm trying to get confidence intervals to slopes from a linear model and I can't figure out how to get at them. As a cut 'n' paste example: ################# # dummy dataset - regression data for 3 treatments, each treatment with different (normal) variance x <- rep(1:10, length=30) y <- 10 - (rep(c(0.2,0.5,0.8), each=10)*x)+...

Hi I have the following code which I would like to simplify. Id does linear regressions and returns the r-squares, and the coefficients. It runs slow, as it is doing the regressions for each - is it possible to get the values in a dataframe which looks as follow: expert | xx | seeds | r.squared | slope | intercept Thanks in advance, Rainer library(reshape) rsqs <- as.data.frame(

Dear forum members, How can I force a negative slope in a linear regression even though the slope might be positive? I will need it for the purpose of determining the trend due reasons other than biological because the biological (genetic) trend is not positive for these data. Thanks. Julia Example of the data: [1] 1.254 1.235 1.261 0.952 1.202 1.152 0.801 0.424 0.330 0.251 0.229

...an easy task. alas, the harder task is that I need to scale this by the plot aspect ratio and the axes. How can a function read this from the current plot? (Has someone written such a function, perhaps more embellished, to save me the debugging effort?) Or, is there an alternative to srt, which slopes the text relative to the existing scale? *** come to think of it, what I would really like is the ability of text to 'snake' itself along the line itself. I doubt that this is easily possible, but I just wanted to ask. help appreciated. sincerely, /ivo welch

Dear List, Im new to R - making a transition from SAS. I have a space delimited file with the following structure. Each line in the datafile is identified by the first letter. A = Inventory (Inventory) X = Stratum (Stratum_no Total Ye=year established) P = Plot (Plot_no age slope= species) T = Tree (tree_no frequency) L = Leader (leader diameter height) F = Feature (start_height finish_height

Hi all, I try to test a linear slope using offset. I have: > m2 <- glm(Y~X*V) > summary(m2) Call: glm(formula = Y ~ X * V) Deviance Residuals: Min 1Q Median 3Q Max -2.01688 -0.56028 0.05224 0.53213 3.60216 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.3673 0.8476 1.613 0.119788 X

I am interested in whether the slopes in a linear model are different from 0. I.e. I would like to obtain the slope estimates, and their standard errors, ``relative to 0'' for each group, rather than relative to some baseline. Explicitly I would like to write/represent the model as y = a_i + b_i*x + E i = 1, ..., K, wh...

Hello, I'm a newcomer to R so please forgive me if this is a silly question. It's that I have a linear regression: fm <- lm (x ~ y) and I want to test whether the slope of the regression is significantly less than 1. How can I do this in R? I'm also interested in comparing the slopes of two regressions: fm1 <- lm (x ~ y) fm2 <- lm (a ~ b) and asking if the slope of fm1 is less than the slope of fm2. Is this easy to do in R? I will be very grateful for any help. regards, Avril Coghlan (University College Dublin, Ireland)

Dear R users, I have a question on using weighted.mean() while aggregating a data frame. I have a data frame with columns Sub, Length and Slope: > x[1:5,] Sub Length Slope 1 2 351.547 0.0025284969 2 2 343.738 0.0025859390 3 1 696.659 0.0015948968 4 2 5442.338 0.0026132544 5 1 209.483 0.0005304225 and I would like to calculate the weighted.mean of Slope, using Length

Hello everybody, I have a question. I would like to get a correlation between constitutive and induced plant defence which I messured on 30 plant species. So I have table with Species, Induced defence (ID), and constitutive defence (CD). Since Induced and constitutive defence are not independant (so called spurious correlation) I should do a randomisation test. I have a syntax of my