Linear Search and Reverse Linear Search dont have same execution time
By : Justin West
Date : March 29 2020, 07:55 AM
it fixes the issue It completely depends on your computer architecture and operating system. However, an overview can be done. CPU run lots of processes in a time and CPU is only capable of executing commands. Most basic operations are ALU operations inside CPU.

pvalues of linear combination of coefficients in linear mixed model
By : user2010463
Date : March 29 2020, 07:55 AM
Any of those help It would be nice to have a reproducible example. It's not very hard to set up your own test for this: code :
ct < c(0,1,3) ## set up contrast (linear comb. of coefficients)
m < fixef(m) %*% ct ## mean of contrast
v < c(t(ct) %*% vcov(m) %*% ct) ## variance of contrast
stder < sqrt(v) ## standard error
tstat < m/stder ## t statistic
2*pt(abs(tstat),df=594,lower.tail=FALSE)

loop for linear model execution with changing vars names and data sets
By : user3521535
Date : March 29 2020, 07:55 AM
To fix the issue you can do If I understand correctly, the lm objects can be kept in a list as following. The column that is not included in the formula is removed to keep it simple. code :
# way 1 with loop
f < list()
dat < list(dat1, dat2)
for(i in 1:2) f[[i]] < lm(target ~ ., data = dat[[i]][,3])
# way 2 with lapply
dat < list(dat1[,3], dat2[,3])
lapply(dat, lm, formula = target ~ .)
[[1]]
Call:
FUN(formula = ..1, data = X[[1L]])
Coefficients:
(Intercept) birds_1 snakes_1
4.3333 0.6667 0.3333
[[2]]
Call:
FUN(formula = ..1, data = X[[2L]])
Coefficients:
(Intercept) birds_2 snakes_2
3 0 1

R: Fit curve to points: what linear/nonlinear model to use?
By : igotsqledbythelastad
Date : March 29 2020, 07:55 AM
hope this fix your issue I personally think this question a dupe of this: `nls` fails to estimate parameters of my model but I would be coldblooded if I close it (as OP put a bounty). Anyway, bounty question can not be closed. So the best I could think of, is to post a community wiki answer (I don't want to get this bounty). code :
fit < lm(log(y) ~ log(x))

Estimating bias in linear regression and linear mixed model in R simulation
By : Kapil Jain
Date : March 29 2020, 07:55 AM
To fix the issue you can do There are a couple of methods of simulating bias. I'll take an easy example using a linear model. A linear mixed model could likely use a similar approach, however i am not certain it would go well for a generalized linear mixed model (I am simply not certain). A simple method for estimating bias, when working with a simple linear model, is to 'choose' which model to estimate ones bias from. Lets say for example Y = 3 + 4 * X + e. I have chosen beta < c(3,4), and as such i need to only simulate my data. For a linear model, the model assumptions are code :
set.seed(1)
xseq < seq(10,10)
xlen < length(xseq)
nrep < 100
#Simulate X given a flat prior (uniformly distributed. A normal distribution would likely work fine as well)
X < sample(xseq, size = xlen * nrep, replace = TRUE)
beta < c(3, 4)
esd = 1
emu < 0
e < rnorm(xlen * nrep, emu, esd)
Y < cbind(1, X) %*% beta + e
fit < lm(Y ~ X)
bias < coef(fit) beta
>bias
(Intercept) X
0.0121017239 0.0001369908
#Simulate linear model many times
model_frame < cbind(1,X)
emany < matrix(rnorm(xlen * nrep * 1000, emu, esd),ncol = 1000)
#add simulated noise. Sweep adds X %*% beta across all columns of emany
Ymany < sweep(emany, 1, model_frame %*% beta, "+")
#fit many models simulationiously (lm is awesome!)
manyFits < lm(Y~X)
#Plot density of fitted parameters
par(mfrow=c(1,2))
plot(density(coef(manyFits)[1,]), main = "Density of intercept")
plot(density(coef(manyFits)[2,]), main = "Density of beta")
#Calculate bias, here i use sweep to substract beta across all rows of my coefficients
biasOfMany < rowMeans(sweep(coef(manyFits), 1, beta, ""))
>biasOfMany
(Intercept) X
5.896473e06 1.710337e04

