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I want to generate 95% confidence intervals from the R2 of a linear model. While developing the code and using the same seed for both approaches, I figured it out that doing the bootstrap manually doesn't give me the same results as using the boot function from the boot package. I am wondering now if I am doing something wrong? or why is this happening?
On the other hand, in order to calculate the 95% CI I was trying to use the confint function, but I'm getting an error "$ operator is invalid for atomic vectors". Any solution to avoid this error?
Here is a reproducible example to explain my concerns
#creating the dataframe
a <- rpois(n = 100, lambda = 10)
b <- rnorm(n = 100, mean = 5, sd = 1)
DF<- data.frame(a,b)
#bootstrapping manually
set.seed(123)
x=length(DF$a)
B_manually<- data.frame(replicate(100, summary(lm(a~b, data = DF[sample(x, replace = T),]))$r.squared))
names(B_manually)[1]<- "r_squared"
#Bootstrapping using the function "Boot" from Boot library
set.seed(123)
library(boot)
B_boot <- boot(DF, function(data,indices)
summary(lm(a~b, data[indices,]))$r.squared,R=100)
head(B_manually) == head(B_boot$t)
r_squared
1 FALSE
2 FALSE
3 FALSE
4 FALSE
5 FALSE
6 FALSE
#Why does the results of the manually vs boot function approach differs if I'm using the same seed?
# 2nd question (Using the confint function to determine the 95 CI gives me an error)
confint(B_manually$r_squared, level = 0.95, method = "quantile")
confint(B_boot$t, level = 0.95, method = "quantile")
#Error: $ operator is invalid for atomic vectors
#NOTE: I already used the boot.ci to determine the 95 confidence interval, as well as the
#quantile function to determine the CI, but the results of these CI differs from each others
#and just wanted to compare with the confint function.
quantile(B_function$t, c(0.025,0.975))
boot.ci(B_function, index=1,type="perc")
Thanks in advance for any help!
The boot package does not use replicate with sample to generate the indices. Check the importance.array function under the source code for boot. It basically generates all the indices at one go. So there's no reason to assume that you will end up with the same indices or same result. Take a step back, the purpose of bootstrap is to use random sampling methods to obtain a estimate of your parameters, you should get similar estimates from different implementation of bootstrap.
For example, you can see the distribution of R^2 is very similar:
set.seed(111)
a <- rpois(n = 100, lambda = 10)
b <- rnorm(n = 100, mean = 5, sd = 1)
DF<- data.frame(a,b)
set.seed(123)
x=length(DF$a)
B_manually<- data.frame(replicate(999, summary(lm(a~b, data = DF[sample(x, replace = T),]))$r.squared))
library(boot)
B_boot <- boot(DF, function(data,indices)
summary(lm(a~b, data[indices,]))$r.squared,R=999)
par(mfrow=c(2,1))
hist(B_manually[,1],breaks=seq(0,0.4,0.01),main="dist of R2 manual")
hist(B_boot$t,breaks=seq(0,0.4,0.01),main="dist of R2 boot")
The function confint you are using, is meant for a lm object, and works on estimating a confidence interval for the coefficient, see help page. It takes the standard error of the coefficient and multiply it by the critical t-value to give you confidence interval. You can check out this book page for the formula. The objects from your bootstrapping are not lm objects and this function doesn't work. It is not meant for any other estimates.
Here is the problem I am currently facing: I have a data frame (let's call A) of 200 observations (rows) and 12 variables (columns). where I am, trying to find out the confidence interval using Bootstrap based on Correlation between two variables in the data frame.
My Data:
library(boot)
library(tidyverse)
library(psychometric)
hsb2 <- read.table("https://stats.idre.ucla.edu/stat/data/hsb2.csv", sep=",", header=T)
here I am trying to find out the confidence interval by using bootstrap based correlation formula
I wrote code for that its work.
k<-CIr(r=orig.cor, n = 21, level = .95)
k
n<-length(hsb2$math)
#n
B<-5000
boot.cor.all<-NULL
for (i in 1:B){
index<-sample(1:n, replace=T)
boot.v2<-hsb2$math[index]
boot.v1<-hsb2$write[index]
boot.cor<-cor(boot.v1, boot.v2,method="spearman")
boot.cor.all<-c(boot.cor.all, boot.cor)
}
ci_boot<-quantile(boot.cor.all, prob=c(0.025, 0.975))
ci_boot
Result:
[1] 0.6439442
[1] 0.2939780 0.8416635
2.5% 97.5%
0.5556964 0.7211145
Here is the actual problem I am facing where I have to write a function to get
result for another variable but
this function not working
bo<-function(v1,v2,df){
orig.cor <- cor(df$v1,df$v2,method="spearman")
orig.ci<-CIr(r=orig.cor, n = 21, level = .95)
B<-5000
n<-length(df$v1)
boot.cor.all<-NULL
for (i in 1:B){
index<-sample(1:n, replace=T)
boot.hvltt2<-df$v1[index]
boot.hvltt<-df$v2[index]
boot.cor<-cor(boot.hvltt2, boot.hvltt,method="spearman")
boot.cor.all<-c(boot.cor.all, boot.cor)
}
ci_boot<-quantile(boot.cor.all, prob=c(0.025, 0.975))
return(orig.cor,orig.ci,ci_boot)
}
after calling this function I am getting error
bo(math,write,hsb2)
bo(math,read,hsb2)
bo(female,write,hsb2)
bo(female,read,hsb2)
I am getting this error
Error in cor(df$v1, df$v2, method = "spearman") : supply both 'x' and 'y' or a matrix-like 'x'
how to write a function correctly.
I want the result as each time a call function it needs to be stored in data frame like below
Variable1 variable2 Orig Cor Orig CI bootstrap CI
math wirte 0.643 0.2939780 0.8416635 0.5556964 0.7211145
math read 0.66 0.3242639 0.8511580 0.5736904 0.7400174
female read -0.059 -0.4787978 0.3820967 -0.20432743 0.08176896
female write
science write
science read
The logic was right, I just had to make some changes on how you access the elements on df. R doesn't recognized the objects math and write because they are columns inside the data.frame. One way to pass them as arguments to the function is to define them as strings v1 = "math" and then access them with df[,v1]
bo<-function(v1,v2,df){
orig.cor <- cor(df[,v1],df[,v2],method="spearman")
orig.ci<-CIr(r=orig.cor, n = 21, level = .95)
B<-5000
n<-nrow(df) #Changed length to nrow
boot.cor.all<-NULL
for (i in 1:B){
index<-sample(1:n, replace=T)
boot.hvltt2<-df[index,v1]
boot.hvltt<-df[index,v2]
boot.cor<-cor(boot.hvltt2, boot.hvltt,method="spearman")
boot.cor.all<-c(boot.cor.all, boot.cor)
}
ci_boot<-quantile(boot.cor.all, prob=c(0.025, 0.975))
return(list(orig.cor,orig.ci,ci_boot)) #wrap your returns in a list
}
bo("math","write",hsb2)
I'm using frbs package in R on my data set using 5-fold stratified cross validation. I've implemented stratified CV. I use GFS.GCCL method for frbs.learn function in each fold and predict the result using test data. I get this error as well as 30 equal warning messages:
Error: object 'temp.rule.degree' not found
Warning: In max(MF.temp[m, ], na.rm = TRUE) :
no non-missing arguments to max; returning -Inf
My code is written in below:
library(frbs)
data<-read.csv(file.address)
data[,30] <- unclass(data[,30]) #column 30 has the class of samples
data <- data[,c(1,14,20,26,27, 30)] # I choose to have 5 attr. since
#my data is high dimensional
k <- 5 # 5-fold
seed <- 1
folds <- strf.cv(data, k, seed) #stratification function for CV
range.data.inp <- matrix(apply(data[,-ncol(data)], 2, range), nrow=2)
data<-norm.data(as.matrix(data[,-ncol(data)]),range.data.
inp,min.scale = 0.1, max.scale = 1)
ctrl <- list(popu.size = 30, num.class = 2, num.labels= 3,
persen_cross = 0.9, max.gen = 200, persen_mutant = 0.3,
name="sim-1")
for(i in 1:k){
str <- paste("fold",i)
print(str)
test.ind <- folds[[str]]
test.data <- data[test.ind,]
train.data <- data[-test.ind,]
obj <- frbs.learn(train.data , method.type="GFS.GCCL",
range.data.inp , ctrl)
pred <- predict(obj, test.data)
print("Predicted classes:")
print(pred)
}
I don't have any idea about error and warnings. Please let me know what I should do.
I've had similar problem (and others) trying to reproduce the SLAVE learning starting with the iris example data. I had 2 format items to solve before being able to run this with my artifical data:
my dataframe import was giving me integer, where the learn needs at least numeric.
my distribution of criteria was not flat. When I flattened the distribution (3 values so n/3 samples per value) everything went fine.
That's all I know.
Hope it helps.
I encountered the same issue when I was running SLAVE and GFS.GCCL. When I was looking at the source code of the library. I found that in frbs.learn(), each method has an implementation to calculate the range of input data. So, I think it might be a problem with the range of input data. For example, in GFS.GCCL, in the source code, for setting the parameters, it looks like this:
range.data.input <- range.data
data.train.ori <- data.train
popu.size <- control$popu.size
persen_cross <- control$persen_cross
persen_mutant <- control$persen_mutant
max.gen <- control$max.gen
name <- control$name
n.labels <- control$num.labels
n.class <- control$num.class
num.labels <- matrix(rep(n.labels, ncol(range.data)), nrow = 1)
num.labels <- cbind(num.labels, n.class)
## normalize range of data and data training
range.data.norm <- range.data.input
range.data.norm[1, ] <- 0
range.data.norm[2, ] <- 1
range.data.input.ori <- range.data.input
data.tra.norm <- norm.data(data.train[, 1 : ncol(data.train) - 1], range.data.input, min.scale = 0, max.scale = 1)
data.train <- cbind(data.tra.norm, matrix(data.train[, ncol(data.train)], ncol = 1))
in the first line, range.data is either coming from your specification nor the default setting of frbs.learn(). For the default setting, it gets the max and min for each row. In the source code:
range.data <- rbind(dt.min, dt.max)
After that, the range of data taken by the GFS.GCCL is
range.data.norm <- range.data.input
range.data.norm[1, ] <- 0
range.data.norm[2, ] <- 1
which is between 0 and 1. The GFS.GCCL is also taken the range.data.input as parameter. So, it takes both range.data.norm and range.data.input.
Therefore, I think if internally, there are some calculation corresponding to range.data.input (it needs to be set as min, max for each row), but the setting for this is actually not min and max for each row. The error is generated.
But, in summary, after I remove "range.data"from frbs.learn(), both GFS.GCCL and SLAVE work for me.
You can download the source code from here:
https://cran.r-project.org/web/packages/frbs/index.html
You can find the code for GFS.GCCL and SLAVE in:
FRBS.MainFunction.R
GFS.Methods.R
In addition to #Pilip38's good advice, I have three other ideas that have fixed similar errors for me while working with the frbs package.
Most important: Make sure your output variable is never equal to 0. It looks like you have a binary output variable so I am hoping just adding 1 to it so it is 1/2 instead of 0/1 will work.
Try setting your range.data.inp matrix to be all 0's in the first row and all 1's in the second. Naturally it's better to have a tighter range but it may be causing your bug.
Try decreasing the number of labels to 2.
It's can be a brittle procedure.
I am working with the cumulative emergence of flies over time (taken at irregular intervals) over many summers (though first I am just trying to make one year work). The cumulative emergence follows a sigmoid pattern and I want to create a maximum likelihood estimation of a 3-parameter Weibull cumulative distribution function. The three-parameter models I've been trying to use in the fitdistrplus package keep giving me an error. I think this must have something to do with how my data is structured, but I cannot figure it out. Obviously I want it to read each point as an x (degree days) and a y (emergence) value, but it seems to be unable to read two columns. The main error I'm getting says "Non-numeric argument to mathematical function" or (with slightly different code) "data must be a numeric vector of length greater than 1". Below is my code including added columns in the df_dd_em dataframe for cumulative emergence and percent emergence in case that is useful.
degree_days <- c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94)
emergence <- c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0)
cum_em <- cumsum(emergence)
df_dd_em <- data.frame (degree_days, emergence, cum_em)
df_dd_em$percent <- ave(df_dd_em$emergence, FUN = function(df_dd_em) 100*(df_dd_em)/46)
df_dd_em$cum_per <- ave(df_dd_em$cum_em, FUN = function(df_dd_em) 100*(df_dd_em)/46)
x <- pweibull(df_dd_em[c(1,3)],shape=5)
dframe2.mle <- fitdist(x, "weibull",method='mle')
Here's my best guess at what you're after:
Set up data:
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd <- transform(dd,cum_em=cumsum(emergence))
We're actually going to fit to an "interval-censored" distribution (i.e. probability of emergence between successive degree day observations: this version assumes that the first observation refers to observations before the first degree-day observation, you could change it to refer to observations after the last observation).
library(bbmle)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun <- function(scale,shape,x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull(c(-Inf,x), ## or (c(x,Inf))
shape=shape,scale=scale)),1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
library(bbmle)
I should probably have used something more systematic like the method of moments (i.e. matching the mean and variance of a Weibull distribution with the mean and variance of the data), but I just hacked around a bit to find plausible starting values:
## preliminary look (method of moments would be better)
scvec <- 10^(seq(0,4,length=101))
plot(scvec,sapply(scvec,NLLfun,shape=1))
It's important to use parscale to let R know that the parameters are on very different scales:
startvals <- list(scale=1000,shape=1)
m1 <- mle2(NLLfun,start=startvals,
control=list(parscale=unlist(startvals)))
Now try with a three-parameter Weibull (as originally requested) -- requires only a slight modification of what we already have:
library(FAdist)
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
Looks like the three-parameter fit is much better:
library(emdbook)
AICtab(m1,m2)
## dAIC df
## m2 0.0 3
## m1 21.7 2
And here's the graphical summary:
with(dd,plot(cum_em~degree_days,cex=3))
with(as.list(coef(m1)),curve(sum(dd$emergence)*
pweibull(x,shape=shape,scale=scale),col=2,
add=TRUE))
with(as.list(coef(m2)),curve(sum(dd$emergence)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
(could also do this more elegantly with ggplot2 ...)
These don't seem like spectacularly good fits, but they're sane. (You could in principle do a chi-squared goodness-of-fit test based on the expected number of emergences per interval, and accounting for the fact that you've fitted a three-parameter model, although the values might be a bit low ...)
Confidence intervals on the fit are a bit of a nuisance; your choices are (1) bootstrapping; (2) parametric bootstrapping (resample parameters assuming a multivariate normal distribution of the data); (3) delta method.
Using bbmle::mle2 makes it easy to do things like get profile confidence intervals:
confint(m1)
## 2.5 % 97.5 %
## scale 1576.685652 1777.437283
## shape 4.223867 6.318481
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd$cum_em <- cumsum(dd$emergence)
dd$percent <- ave(dd$emergence, FUN = function(dd) 100*(dd)/46)
dd$cum_per <- ave(dd$cum_em, FUN = function(dd) 100*(dd)/46)
dd <- transform(dd)
#start 3 parameter model
library(FAdist)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$percent) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
summary(m2)
#graphical summary
windows(5,5)
with(dd,plot(cum_per~degree_days,cex=3))
with(as.list(coef(m2)),curve(sum(dd$percent)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
I need to do some robust data-fitting operation.
I have bunch of (x,y) data, that I want to fit to a Gaussian (aka normal) function.
The point is, I want to remove the ouliers. As one can see on the sample plot below, there is another distribution of data thats pollutting my data on the right, and I don't want to take it into account to do the fitting (i.e. to find \sigma, \mu and the overall scale parameter).
R seems to be the right tool for the job, I found some packages (robust, robustbase, MASS for example) that are related to robust fitting.
However, they assume the user already has a strong knowledge of R, which is not my case, and the documentation is only provided as a sort of reference manual, no tutorial or equivalent. My statistical background is rather low, I attempted to read reference material on fitting with R, but it didn't really help (and I'm not even sure thats the right way to go).
But I have the feeling that this is actually a quite simple operation.
I have checked this related question (and the linked ones), however they take as input a single vector of values, and I have a vector of pairs, so I don't see how to transpose.
Any help on how to do this would be appreciated.
Fitting a Gaussian curve to the data, the principle is to minimise the sum of squares difference between the fitted curve and the data, so we define f our objective function and run optim on it:
fitG =
function(x,y,mu,sig,scale){
f = function(p){
d = p[3]*dnorm(x,mean=p[1],sd=p[2])
sum((d-y)^2)
}
optim(c(mu,sig,scale),f)
}
Now, extend this to two Gaussians:
fit2G <- function(x,y,mu1,sig1,scale1,mu2,sig2,scale2,...){
f = function(p){
d = p[3]*dnorm(x,mean=p[1],sd=p[2]) + p[6]*dnorm(x,mean=p[4],sd=p[5])
sum((d-y)^2)
}
optim(c(mu1,sig1,scale1,mu2,sig2,scale2),f,...)
}
Fit with initial params from the first fit, and an eyeballed guess of the second peak. Need to increase the max iterations:
> fit2P = fit2G(data$V3,data$V6,6,.6,.02,8.3,0.10,.002,control=list(maxit=10000))
Warning messages:
1: In dnorm(x, mean = p[1], sd = p[2]) : NaNs produced
2: In dnorm(x, mean = p[4], sd = p[5]) : NaNs produced
3: In dnorm(x, mean = p[4], sd = p[5]) : NaNs produced
> fit2P
$par
[1] 6.035610393 0.653149616 0.023744876 8.317215066 0.107767881 0.002055287
What does this all look like?
> plot(data$V3,data$V6)
> p = fit2P$par
> lines(data$V3,p[3]*dnorm(data$V3,p[1],p[2]))
> lines(data$V3,p[6]*dnorm(data$V3,p[4],p[5]),col=2)
However I would be wary about statistical inference about your function parameters...
The warning messages produced are probably due to the sd parameter going negative. You can fix this and also get a quicker convergence by using L-BFGS-B and setting a lower bound:
> fit2P = fit2G(data$V3,data$V6,6,.6,.02,8.3,0.10,.002,control=list(maxit=10000),method="L-BFGS-B",lower=c(0,0,0,0,0,0))
> fit2P
$par
[1] 6.03564202 0.65302676 0.02374196 8.31424025 0.11117534 0.00208724
As pointed out, sensitivity to initial values is always a problem with curve fitting things like this.
Fitting a Gaussian:
# your data
set.seed(0)
data <- c(rnorm(100,0,1), 10, 11)
# find & remove outliers
outliers <- boxplot(data)$out
data <- setdiff(data, outliers)
# fitting a Gaussian
mu <- mean(data)
sigma <- sd(data)
# testing the fit, check the p-value
reference.data <- rnorm(length(data), mu, sigma)
ks.test(reference.data, data)