When I run the t-test for a numeric and a dichotomous variable there in no problem and I can see the results. The problem is when I run the ggttest of the same t-test. There is an error and says that one of my variable is not found. I do not why that happens. The aml dataset I used is from package boot. Below you can see the code:
https://i.stack.imgur.com/7kuaA.png
library(gginference)
time_group.test16537 = t.test(formula = time~group,
data = aml,
alternative = "two.sided",
paired = FALSE,
var.equal = FALSE,
conf.level = 0.95)
time_group.test16537
ggttest(time_group.test16537,
colaccept="lightsteelblue1",
colreject="gray84",
colstat="navyblue")
The problem comes with these lines of code in ggttest:
datnames <- strsplit(t$data.name, splitter)
len1 <- length(eval(parse(text = datnames[[1]][1])))
len2 <- length(eval(parse(text = datnames[[1]][2])))
It tries to find the len of group and time, but it doesn't see that it came from a data.frame. Pretty bad bug...
For your situation, supposedly you have less than 30 in each group and it plots a t-distribution, so do:
library(gginference)
library(boot)
gginference:::normt(t.test(time~group,data=aml),
colaccept = "lightsteelblue1",colreject = "grey84",
colstat = "navyblue")
t.test doesn't store your data in the output so there is no way that you could extract the data from the list of the output of t.test.
The only way to use formula is:
library(gginference)
t_test <- t.test(questionnaire$pulse ~ questionnaire$gender)
ggttest(t_test)
Original answer here: How to extract the dataset from an "htest" object when using formula in r
Related
I'm working on a project where I need to construct a knn model using R. The professor provided an article with step-by-step instructions (link to article) and some datasets to choose from (link to the data I'm using). I'm getting stuck on step 3 (creating the model from the training data).
Here's my code:
data <- read.delim("data.txt", header = TRUE, sep = "\t", dec = ".")
set.seed(2)
part <- sample(2, nrow(data), replace = TRUE, prob = c(0.65, 0.35))
training_data <- data[part==1,]
testing_data <- data[part==2,]
outcome <- training_data[,2]
model <- knn(train = training_data, test = testing_data, cl = outcome, k=10)
Here's the error message I'm getting:
I checked and found that training_data, testing_data, and outcome all look correct, the issue seems to only be with the knn model.
The issue is with your data and the knn function you are using; it can't handle characters or factor variable
We can force this to work doing something like this first:
library(tidyverse)
data <- data %>%
mutate(Seeded = as.numeric(as.factor(Seeded))-1) %>%
mutate(Season = as.numeric(as.factor(Season)))
But this is a bad idea in general, since Season is not ordered naturally. A better approach would be to instead treat it as a set of dummies.
See this link for examples:
R - convert from categorical to numeric for KNN
there has been a similar question to mine 6 years+ ago and it hasn't been solve (R -- Can I apply the train function in caret to a list of data frames?)
This is why I am bringing up this topic again.
I'm writing my own functions for my big R project at the moment and I'm wondering if there is an opportunity to sum up the model training function train() of the pakage caret for different dataframes with different predictors.
My function should look like this:
lda_ex <- function(data, predictor){
model <- train(predictor ~., data,
method = "lda",
trControl = trainControl(method = "none"),
preProc = c("center","scale"))
return(model)
}
Using it afterwards should work like this:
data_iris <- iris
predictor_iris <- "Species"
iris_res <- lda_ex(data = data_iris, predictor = predictor_iris)
Unfortunately the R formula is not able to deal with a variable as input as far as I tried.
Is there something I am missing?
Thank you in advance for helping me out!
Solving this would help me A LOT to keep my function sheet clean and safe work for sure.
By writing predictor_iris <- "Species", you are basically saving a string object in predictor_iris. Thus, when you run lda_ex, I guess you incur in some error concerning the formula object in train(), since you are trying to predict a string using vectors of covariates.
Indeed, I tried the following toy example:
X = rnorm(1000)
Y = runif(1000)
predictor = "Y"
lm(predictor ~ X)
which gives an error about differences in the lengths of variables.
Let me modify your function:
lda_ex <- function(data, formula){
model <- train(formula, data,
method = "lda",
trControl = trainControl(method = "none"),
preProc = c("center","scale"))
return(model)
}
The key difference is that now we must pass in the whole formula, instead of the predictor only. In that way, we avoid the string-related problem.
library(caret) # Recall to specify the packages needed to reproduce your examples!
data_iris <- iris
formula_iris = Species ~ . # Key difference!
iris_res <- lda_ex(data = data_iris, formula = formula_iris)
New to stackoverflow. I'm working on a project with NHIS data, but I cannot get the svyglm function to work even for a simple, unadjusted logistic regression with a binary predictor and binary outcome variable (ultimately I'd like to use multiple categorical predictors, but one step at a time).
El_under_glm<-svyglm(ElUnder~SO2, design=SAMPdesign, subset=NULL, family=binomial(link="logit"), rescale=FALSE, correlation=TRUE)
Error in eval(extras, data, env) :
object '.survey.prob.weights' not found
I changed the variables to 0 and 1 instead:
Under_narm$SO2REG<-ifelse(Under_narm$SO2=="Heterosexual", 0, 1)
Under_narm$ElUnderREG<-ifelse(Under_narm$ElUnder=="No", 0, 1)
But then get a different issue:
El_under_glm<-svyglm(ElUnderREG~SO2REG, design=SAMPdesign, subset=NULL, family=binomial(link="logit"), rescale=FALSE, correlation=TRUE)
Error in svyglm.survey.design(ElUnderREG ~ SO2REG, design = SAMPdesign, :
all variables must be in design= argument
This is the design I'm using to account for the weights -- I'm pretty sure it's correct:
SAMPdesign=svydesign(data=Under_narm, id= ~NHISPID, weight= ~SAMPWEIGHT)
Any and all assistance appreciated! I've got a good grasp of stats but am a slow coder. Let me know if I can provide any other information.
Using some make-believe sample data I was able to get your model to run by setting rescale = TRUE. The documentation states
Rescaling of weights, to improve numerical stability. The default
rescales weights to sum to the sample size. Use FALSE to not rescale
weights.
So, one solution maybe is just to set rescale = TRUE.
library(survey)
# sample data
Under_narm <- data.frame(SO2 = factor(rep(1:2, 1000)),
ElUnder = sample(0:1, 1000, replace = TRUE),
NHISPID = paste0("id", 1:1000),
SAMPWEIGHT = sample(c(0.5, 2), 1000, replace = TRUE))
# with 'rescale' = TRUE
SAMPdesign=svydesign(ids = ~NHISPID,
data=Under_narm,
weights = ~SAMPWEIGHT)
El_under_glm<-svyglm(formula = ElUnder~SO2,
design=SAMPdesign,
family=quasibinomial(), # this family avoids warnings
rescale=TRUE) # Weights rescaled to the sum of the sample size.
summary(El_under_glm, correlation = TRUE) # use correlation with summary()
Otherwise, looking code for this function's method with 'survey:::svyglm.survey.design', it seems like there may be a bug. I could be wrong, but by my read when 'rescale' is FALSE, .survey.prob.weights does not appear to get assigned a value.
if (is.null(g$weights))
g$weights <- quote(.survey.prob.weights)
else g$weights <- bquote(.survey.prob.weights * .(g$weights)) # bug?
g$data <- quote(data)
g[[1]] <- quote(glm)
if (rescale)
data$.survey.prob.weights <- (1/design$prob)/mean(1/design$prob)
There may be a work around if you assign a vector of numeric values to .survey.prob.weights in the global environment. No idea what these values should be, but your error goes away if you do something like the following. (.survey.prob.weights needs to be double the length of the data.)
SAMPdesign=svydesign(ids = ~NHISPID,
data=Under_narm,
weights = ~SAMPWEIGHT)
.survey.prob.weights <- rep(1, 2000)
El_under_glm<-svyglm(formula = ElUnder~SO2,
design=SAMPdesign,
family=quasibinomial(),
rescale=FALSE)
summary(El_under_glm, correlation = TRUE)
im quite new to R but wanted to use the packages "nls" and "nlstools" since it has nice tools for analysis and evaluation.
the code I use is:
conB1_2015 = read.csv("C:\\Path_to_File\\conB1_2015.csv")
conB1_2015 = na.omit(conB1_2015)
tRef <- mean(conB1_2015$Mean_Soil_Temp_V2..C., na.rm=TRUE)
rRef <- conB1_2015$Lin_Flux..mymol.m.2.s.1.[which.min(abs(conB1_2015$Mean_Soil_Temp_V2..C.-tRef))]
rMax <- max(conB1_2015$Lin_Flux..mymol.m.2.s.1., na.rm=TRUE)
half <- rMax/2
half_SM <- conB1_2015$Soil_Moist_V3[which.min(abs(conB1_2015$Lin_Flux..mymol.m.2.s.1.-half))]
form <- as.formula(Lin_Flux..mymol.m.2.s.1. ~ (rRef)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM)+Soil_Moist_V3)
preview(form, data = conB1_2015, start = c(a = -1.98, b = -0.05), variable = 1)
The Problem is, that i get this Error running this code:
Error in data.frame(value, row.names = rn, check.names = FALSE) :
row names supplied are of the wrong length
When i change the variables in form <- as.formula(Lin_Flux..mymol.m.2.s.1. ~ (rRef)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM)+Soil_Moist_V3)
to form <- as.formula(Lin_Flux..mymol.m.2.s.1.~(rRef<-4.41)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM<-7.19)+Soil_Moist_V3)
the function works fine.
I wanted to automate the script to run over several csv's to test different models on different data. Is it really not possible to pass variables into the preview function or am I missing something? There can't be a problem with headers or the data table since it's working fine in the second example.
First of all, I would like to mention I am just a beginner in R.
I have encountered a problem when trying to predict data from a model generated by nls(). I fitted the exponential decay function into my data and everything seems to be fine, e.g. I got a decent regression line. However, when I use predict() on a new data set, it returns only fitted values.
My code is:
df = data.frame(Time = c(0,5,15,30), Value = c(1, 0.38484677,0.18679383, 0.06732328))
model <- nls(Value~a*exp(-b*Time), start=list(a=1, b=0.15), data = df)
plot(Value~Time, data = df)
lines(df$Time, predict(model))
newtime <- data.frame(Time = seq(1,20, by = 1))
pr = predict(model, newdata = newtime$Time)
pr
[1] 0.979457389 0.450112312 0.095058637 0.009225664
Could someone explain me please, what I am doing wrong? I know there are here some answers to that problem, but none helped me.
Thank you in advance for your help!
The newdata parameter should be a data.frame with the same names as your input data. When you use newdata = newtime$Time you are actually passing in newtime$Time which is not a data.frame anymore since it 'dropped' down to a vector. You can just pass in newtime like so
pr = predict(model, newdata = newtime)