lm

问题 This is my data frame: data <- structure(list(Y = c(NA, -1.793, -0.642, 1.189, -0.823, -1.715, 1.623, 0.964, 0.395, -3.736, -0.47, 2.366, 0.634, -0.701, -1.692, 0.155, 2.502, -2.292, 1.967, -2.326, -1.476, 1.464, 1.45, -0.797, 1.27, 2.515, -0.765, 0.261, 0.423, 1.698, -2.734, 0.743, -2.39, 0.365, 2.981, -1.185, -0.57, 2.638, -1.046, 1.931, 4.583, -1.276, 1.075, 2.893, -1.602, 1.801, 2.405, -5.236, 2.214, 1.295, 1.438, -0.638, 0.716, 1.004, -1.328, -1.759, -1.315, 1.053, 1.958, -2.034, 2.936,

问题 I've found Error in contrasts when defining a linear model in R and have followed the suggestions there, but none of my factor variables take on only one value and I am still experiencing the same issue. This is the dataset I'm using: https://www.dropbox.com/s/em7xphbeaxykgla/train.csv?dl=0. This is the code I'm trying to run: simplelm <- lm(log_SalePrice ~ ., data = train) #Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : # contrasts can be applied only to factors with 2

问题 I'm trying to do a simple linear regression model in R. there are three factor variables in the model. the model is lm(Exercise ~ Econ + Job + Position) where "Exercise" is numeric dependent variable, the amount of time exercising. "Econ", "Job", "Position" are all factor variables. "Econ" is whether a person is employed or not. (levels = employed / not employed) "Job" is the job type a person has. There are five levels for this variable. "Position" is the position a person has in the

问题 I would like to know how can I come up with a lm formula syntax that would enable me to use paste together with cbind for multiple multivariate regression. Example In my model I have a set of variables, which corresponds to the primitive example below: data(mtcars) depVars <- paste("mpg", "disp") indepVars <- paste("qsec", "wt", "drat") Problem I would like to create a model with my depVars and indepVars . The model, typed by hand, would look like that: modExmple <- lm(formula = cbind(mpg,

问题 I want to write a custom function around plot_ly() in R. That way, I can make a series of scatterplots with the same formatting and style, but not duplicate code. I used this page as a guide. This code reproduces the error: library(plotly) my_plot <- function(x, y, ...) { require(broom) plot_ly(data = mtcars, y = y, x = x, showlegend = FALSE, ...) %>% add_markers(y = y) %>% add_lines(y = ~fitted(lm(y ~ x))) %>% add_ribbons(data = augment(lm(y ~ x, data = mtcars)), ymin = ~.fitted - 1.96 * .se

问题 This question already has answers here : How to debug “contrasts can be applied only to factors with 2 or more levels” error? (2 answers) Closed last year . I get this error depending on which variables I include and the sequence in which I specify them in the formula: Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels I've done a little research on this and it looks like it would be caused by the variable in

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 3 years ago . I'm working with R on a sample of 145 observations. I have created five subsamples each with 29 observations, while the response variable q has been sorted. As a result, subset1 contains the 29 lines of the data frame with the lowest output, subset2 contains the following 29 lines, etc. I am regressing the

问题 I have measurements obtained from 2 groups where each group has the same 3 levels. Here's my example data.frame : df <- data.frame(measurement = c(rnorm(10,1,1),rnorm(10,0.75,1),rnorm(10,1.25,1), rnorm(10,0.5,1),rnorm(10,1.75,1),rnorm(10,0.25,1)), group = as.factor(c(rep("a",30),rep("b",30))), level = as.factor(rep(c(rep("x",10),rep("y",10),rep("z",10)),2))) I'm interested in quantifying how measurement in each level is affected by group . I guess a linear model ( lm ) is the appropriate

问题 Due to the size of my dataset I'm bound to use Speedlm, fastLm or biglm. Unfortunately I'm stuck to using speedlm as fastlm doesn't have an update function, and biglm only supports single core. Using speedlm I want to show all residuals. I know that for lm or fastlm I can simply use the residuals() function. However it turns out speedlm doesn't support this. lmfit <- speedglm(formula , res) print(names(lmfit)) [1] "coefficients" "coef" "df.residual" "XTX" "Xy" "nobs" "nvar" "ok" "A" "RSS"

问题 Here are all the variables I'm working with: str(ad.train) $ Date : Factor w/ 427 levels "2012-03-24","2012-03-29",..: 4 7 12 14 19 21 24 29 31 34 ... $ Team : Factor w/ 18 levels "Adelaide","Brisbane Lions",..: 1 1 1 1 1 1 1 1 1 1 ... $ Season : int 2012 2012 2012 2012 2012 2012 2012 2012 2012 2012 ... $ Round : Factor w/ 28 levels "EF","GF","PF",..: 5 16 21 22 23 24 25 26 27 6 ... $ Score : int 137 82 84 96 110 99 122 124 49 111 ... $ Margin : int 69 18 -56 46 19 5 50 69 -26 29 ... $