I want to permutate a linear regression (to not lose power with random sampling with replacement).
I know how to randomly sample my dataset:
sampled_random <- df[sample(nrow(df), replace = TRUE),]
My regression is like this:
reg <- lm(DV ~ Iv1 + IV2 + IV3, data = df)
Is there a nice built-in function to repeat this regression x times with different sample_random that I have overseen? As outcome I want the average p-values and the other averaged stuff that you get with summary(reg)
I am not experienced enough to write my own function that does all I want. Is there an R package that does this? Or, better, can you recommend a good (handy) one?
You can write your own code.
res <- lapply(1:100, function(i){
sampled_random <- df[sample(nrow(df), replace = TRUE),]
reg <- lm(DV ~ Iv1 + IV2 + IV3, data = sampled_random)
return(c(summary(reg)$residuals, summary(reg)$r.squared))
})
Related
This is my first question here, so if I need to share more information please let me know.
I have done a Cox regression analysis in R in which I am interested in the effect of implant surface on reoperation over 36 months. Here's a reproducible example:
library(survival)
n <- 100
df <- data.frame(id=1:n,
time=sample(1:36, n, replace=TRUE),
event=sample(0:2, n, replace=TRUE),
implantsurface=sample(0:1, n, replace=TRUE),
covariate1=sample(0:1, n, replace=TRUE),
covariate2=sample(0:1, n, replace=TRUE))
df$time <- as.numeric(df$time)
I adjusted for a number of covariates, which showed that the proportional hazards assumption was violated for covariate1. I split my dataset into 0-4 mo and 4-36 mo as follows (simplified code), so that the PH assumption was no longer violated:
fit1 <- survSplit(Surv(time, event == 1) ~
implantsurface + covariate1 + covariate2,
data = df, cut=c(4),
episode= "tgroup")
fit2 <- coxph(Surv(tstart, time, event) ~
implantsurface + strata(tgroup):covariate1 + covariate2,
data = fit1)
Now I would also like to adjust for competing risks with Fine-Gray regression, but I am unable to do this for the split dataset. I have tried the following:
FG <- finegray(Surv(time = time, event = event.competing, type = "mstate") ~
implantsurface + strata(tgroup):covariate1 + covariate2,
data = fit1, etype = "event_of_interest")
FGfit <- coxph(Surv(fgstart, fgstop, fgstatus) ~
implantsurface + strata(tgroup):covariate1 + covariate2,
weights = fgwt, data = FG)
Error in strata(tgroup) : object 'tgroup' not found
Does anyone know how/if Fine-Gray can be applied to a split survival dataset? Many thanks in advance for thinking along!
I have a bit of an issue. I am trying to develop some code that will allow me to do the following: 1) run a logistic regression analysis, 2) extract the estimates from the logistic regression analysis, and 3) use those estimates to create another logistic regression formula that I can use in a subsequent simulation of the original model. As I am, relatively new to R, I understand I can extract these coefficients 1-by-1 through indexing, but it is difficult to "scale" this to models with different numbers of coefficients. I am wondering if there is a better way to extract the coefficients and setup the formula. Then, I would have to develop the actual variables, but the development of these variables would have to be flexible enough for any number of variables and distributions. This appears to be easily done in Mplus (example 12.7 in the Mplus manual), but I haven't figured this out in R. Here is the code for as far as I have gotten:
#generating the data
set.seed(1)
gender <- sample(c(0,1), size = 100, replace = TRUE)
age <- round(runif(100, 18, 80))
xb <- -9 + 3.5*gender + 0.2*age
p <- 1/(1 + exp(-xb))
y <- rbinom(n = 100, size = 1, prob = p)
#grabbing the coefficients from the logistic regression model
matrix_coef <- summary(glm(y ~ gender + age, family = "binomial"))$coefficients
the_estimates <- matrix_coef[,1]
the_estimates
the_estimates[1]
the_estimates[2]
the_estimates[3]
I just cannot seem to figure out how to have R create the formula with the variables (x's) and the coefficients from the original model in a flexible manner to accommodate any number of variables and different distributions. This is not class assignment, but a necessary piece for the research that I am producing. Any help will be greatly appreciated, and please, treat this as a teaching moment. I really want to learn this.
I'm not 100% sure what your question is here.
If you want to simulate new data from the same model with the same predictor variables, you can use the simulate() method:
dd <- data.frame(y, gender, age)
## best practice when modeling in R: take the variables from a data frame
model <- glm(y ~ gender + age, data = dd, family = "binomial")
simulate(model)
You can create multiple replicates by specifying the nsim= argument (or you can simulate anew every time through a for() loop)
If you want to simulate new data from a different set of predictor variables, you have to do a little bit more work (some model types in R have a newdata= argument, but not GLMs alas):
## simulate new model matrix (including intercept)
simdat <- cbind(1,
gender = rbinom(100, prob = 0.5, size = 1),
age = sample(18:80, size = 100, replace = TRUE))
## extract inverse-link function
invlink <- family(model)$linkinv
## sample new values
resp <- rbinom(n = 100, size = 1, prob = invlink(simdat %*% coef(model)))
If you want to do this later from coefficients that have been stored, substitute the retrieved coefficient vector for coef(model) in the code above.
If you want to flexibly construct formulas, reformulate() is your friend — but I don't see how it fits in here.
If you want to (say) re-fit the model 1000 times to new responses simulated from the original model fit (same coefficients, same predictors: i.e. a parametric bootstrap), you can do something like this.
nsim <- 1000
res <- matrix(NA, ncol = length(coef(model)), nrow = nsim)
for (i in 1:nsim) {
## simulate returns a list (in this case, of length 1);
## extract the response vector
newresp <- simulate(model)[[1]]
newfit <- update(model, newresp ~ .)
res[i,] <- coef(newfit)
}
You don't have to store coefficients - you can extract/compute whatever model summaries you like (change the number of columns of res appropriately).
Let’s say your data matrix including age and gender, or whatever predictors, is X. Then you can use X on the right-hand side of your glm formula, get xb_hat <- X %*% the_estimates (or whatever other data matrix replacing X as long as it has same columns) and plug xb_hat into whatever link function you want.
I am new to R and am trying to loop a mixed model across 90 columns in a dataset.
My dataset looks like the following one but has 90 predictors instead of 7 that I need to evaluate as fixed effects in consecutive models.
I then need to store the model output (coefficients and P values) to finally construct a figure summarizing the size effects of each predictor. I know the discussion of P value estimates from lme4 mixed models.
For example:
set.seed(101)
mydata <- tibble(id = rep(1:32, times=25),
time = sample(1:800),
experiment = rep(1:4, times=200),
Y = sample(1:800),
predictor_1 = runif(800),
predictor_2 = rnorm(800),
predictor_3 = sample(1:800),
predictor_4 = sample(1:800),
predictor_5 = seq(1:800),
predictor_6 = sample(1:800),
predictor_7 = runif(800)) %>% arrange (id, time)
The model to iterate across the N predictors is:
library(lme4)
library(lmerTest) # To obtain new values
mixed.model <- lmer(Y ~ predictor_1 + time + (1|id) + (1|experiment), data = mydata)
summary(mixed.model)
My coding skills are far from being able to set a loop to repeat the model across the N predictors in my dataset and store the coefficients and P values in a dataframe.
I have been able to iterate across all the predictors fitting linear models instead of mixed models using lapply. But I have failed to apply this strategy with mixed models.
varlist <- names(mydata)[5:11]
lm_models <- lapply(varlist, function(x) {
lm(substitute(Y ~ i, list(i = as.name(x))), data = mydata)
})
One option is to update the formula of a restricted model (w/o predictor) in an lapply loop over the predictors. Then summaryze the resulting list and subset the coefficient matrix using a Vectorized function.
library(lmerTest)
mixed.model <- lmer(Y ~ time + (1|id) + (1|experiment), data = mydata)
preds <- grep('pred', names(mydata), value=TRUE)
fits <- lapply(preds, \(x) update(mixed.model, paste('. ~ . + ', x)))
extract_coef_p <- Vectorize(\(x) x |> summary() |> coef() |> {\(.) .[3, c(1, 5)]}())
res <- `rownames<-`(t(extract_coef_p(fits)), preds)
res
# Estimate Pr(>|t|)
# predictor_1 -7.177579138 0.8002737
# predictor_2 -5.010342111 0.5377551
# predictor_3 -0.013030513 0.7126500
# predictor_4 -0.041702039 0.2383835
# predictor_5 -0.001437124 0.9676346
# predictor_6 0.005259293 0.8818644
# predictor_7 31.304496255 0.2511275
Let's say I have these two models
dat1 <- data.frame(x=factor(c(1,2,1,1,2,2)),y=c(2,5,2,1,7,9))
dat2 <- data.frame(x=factor(c(1,2,1,1,2,2)),y=c(3,3,4,3,4,2))
mod1 <- lm(y~x,data=dat1)
mod2 <- lm(y~x, data=dat2)
and calculate a t test between the levels of x in each model
t1 <- pairs(emmeans(mod1, ~x))
t2 <- pairs(emmeans(mod2, ~x))
How can I assess whether the two models are significantly different for this contrast using emmeans?
dat1$dataset <- "dat1"
dat2$dataset <- "dat2"
alldat <- rbind(dat1, dat2)
modsame <- lm(y ~ x, data = alldat)
moddiff <- lm(y ~ x * dataset, data = alldat)
anova(modsame, moddiff)
Don't try to use emmeans() to do this; that isn't its purpose. The anova() call above compares the two models: modsame presumes that the x effects are the same in each dataset; moddiff adds two terms, dataset which accounts for the change in overall mean, and x:dataset which accounts for the change in x effects.
The comparison between the two models comprises a joint test of both the dataset and the x:dataset effects -- it is an F test with 2 numerator d.f. -- not a t test.
I performed a regression analyses in R on some dataset and try to predict the contribution of each individual independent variable on the dependent variable for each row in the dataset.
So something like this:
set.seed(123)
y <- rnorm(10)
m <- data.frame(v1=rnorm(10), v2=rnorm(10), v3=rnorm(10))
regr <- lm(formula=y~v1+v2+v3, data=m)
summary(regr)
terms <- predict.lm(regr,m, type="terms")
In short: run a regression and use the predict function to calculate the terms of v1,v2 and v3 in dataset m. But I am having a hard time understanding what the predict function is calculating. I would expect it multiplies the coefficient of the regression result with the variable data. So something like this for v1:
coefficients(regr)[2]*m$v1
But that gives different results compared to the predict function.
Own calculation:
0.55293884 0.16253411 0.18103537 0.04999729 -0.25108302 0.80717945 0.22488764 -0.88835486 0.31681455 -0.21356803
And predict function calculation:
0.45870070 0.06829597 0.08679724 -0.04424084 -0.34532115 0.71294132 0.13064950 -0.98259299 0.22257641 -0.30780616
The prediciton function is of by 0.1 or so Also if you add all terms in the prediction function together with the constant it doesn’t add up to the total prediction (using type=”response”). What does the prediction function calculate here and how can I tell it to calculate what I did with coefficients(regr)[2]*m$v1?
All the following lines result in the same predictions:
# our computed predictions
coefficients(regr)[1] + coefficients(regr)[2]*m$v1 +
coefficients(regr)[3]*m$v2 + coefficients(regr)[4]*m$v3
# prediction using predict function
predict.lm(regr,m)
# prediction using terms matrix, note that we have to add the constant.
terms_predict = predict.lm(regr,m, type="terms")
terms_predict[,1]+terms_predict[,2]+terms_predict[,3]+attr(terms_predict,'constant')
You can read more about using type="terms" here.
The reason that your own calculation (coefficients(regr)[2]*m$v1) and the predict function calculation (terms_predict[,1]) are different is because the columns in the terms matrix are centered around the mean, so their mean becomes zero:
# this is equal to terms_predict[,1]
coefficients(regr)[2]*m$v1-mean(coefficients(regr)[2]*m$v1)
# indeed, all columns are centered; i.e. have a mean of 0.
round(sapply(as.data.frame(terms_predict),mean),10)
Hope this helps.
The function predict(...,type="terms") centers each variable by its mean. As a result, the output is a little difficult to interpret. Here's an alternative where each variable (constant, x1, and x2) is multiplied to its coefficient.
TLDR: pred_terms <- model.matrix(formula(mod$terms), testData) %*% diag(coef(mod))
library(tidyverse)
### simulate data
set.seed(123)
nobs <- 50
x1 <- cumsum(rnorm(nobs) + 3)
x2 <- cumsum(rnorm(nobs) * 3)
y <- 2 + 2*x1 -0.5*x2 + rnorm(nobs,0,50)
df <- data.frame(t=1:nobs, y=y, x1=x1, x2=x2)
train <- 1:round(0.7*nobs,0)
rm(x1, x2, y)
trainData <- df[train,]
testData <- df[-train,]
### linear model
mod <- lm(y ~ x1 + x2 , data=trainData)
summary(mod)
### predict test set
test_preds <- predict(mod, newdata=testData)
head(test_preds)
### contribution by predictor
test_contribution <- model.matrix(formula(mod$terms), testData) %*% diag(coef(mod))
colnames(test_contribution) <- names(coef(mod))
head(test_contribution)
all(round(apply(test_contribution, 1, sum),5) == round(test_preds,5)) ## should be true
### Visualize each contribution
test_contribution_df <- as.data.frame(test_contribution)
test_contribution_df$pred <- test_preds
test_contribution_df$t <- row.names(test_contribution_df)
test_contribution_df$actual <- df[-train,"y"]
test_contribution_df_long <- pivot_longer(test_contribution_df, -t, names_to="variable")
names(test_contribution_df_long)
ggplot(test_contribution_df_long, aes(x=t, y=value, group=variable, color=variable)) +
geom_line() +
theme_bw()