I have a weather dataset, i found a simple linear model for two columns Temperature and Humidity and plotted the histogram of its residuals and calculated the mean and std.
model <- lm(Temperature..C. ~ Humidity, data = inputData)
model.res = resid(model)
hist(model.res)
mean(model.res)
sd(model.res)
I should Plot QQ-plot of residuals versus a zero-mean normal distribution with estimated std. I used Kolmogorov-Smirnov to compare a sample with a reference probability distribution but i don't know how to plot it together:
ks<-ks.test(model.res, "pnorm", mean=0, sd=sd(model.res))
qqnorm(model.res, main="qqnorm")
qqline(model.res)
Data example:
Temperature..C. Humidity
1 9.472222 0.89
2 9.355556 0.86
3 9.377778 0.89
4 8.288889 0.83
5 8.755556 0.83
6 9.222222 0.85
7 7.733333 0.95
8 8.772222 0.89
9 10.822222 0.82
10 13.772222 0.72
11 16.016667 0.67
12 17.144444 0.54
13 17.800000 0.55
14 17.333333 0.51
15 18.877778 0.47
16 18.911111 0.46
17 15.388889 0.60
18 15.550000 0.63
19 14.255556 0.69
20 13.144444 0.70
Here is a solution using ggplot2
ggplot(model, aes(sample = rstandard(model))) +
geom_qq() +
stat_qq_line(dparams=list(sd=sd(model.res)), color="red") +
stat_qq_line()
The red line represents the qqline with your sample sd, the blackline a sd of 1.
You did not ask for that, but you could also add a smoothed qqplot:
data_model <- model
data_model$theo <- unlist(qqnorm(data_model$residuals)[1])
ggplot(data_model, aes(sample = rstandard(data_model))) +
geom_qq() +
stat_qq_line(dparams=list(sd=sd(model.res)), color="red") +
geom_smooth(aes(x=data_model$theo, y=data_model$residuals), method = "loess")
Related
I have plotted the conditional density distribution of my variables by using cdplot (R). My independent variable and my dependent variable are not independent. Independent variable is discrete (it takes only certain values between 0 and 3) and dependent variable is also discrete (11 levels from 0 to 1 in steps of 0.1).
Some data:
dat <- read.table( text="y x
3.00 0.0
2.75 0.0
2.75 0.1
2.75 0.1
2.75 0.2
2.25 0.2
3 0.3
2 0.3
2.25 0.4
1.75 0.4
1.75 0.5
2 0.5
1.75 0.6
1.75 0.6
1.75 0.7
1 0.7
0.54 0.8
0 0.8
0.54 0.9
0 0.9
0 1.0
0 1.0", header=TRUE, colClasses="factor")
I wonder if my variables are appropriate to run this kind of analysis.
Also, I'd like to know how to report this results in an elegant way with academic and statistical sense.
This is a run using the rms-packages `lrm function which is typically used for binary outcomes but also handles ordered categorical variables:
library(rms) # also loads Hmisc
# first get data in the form you described
dat[] <- lapply(dat, ordered) # makes both columns ordered factor variables
?lrm
#read help page ... Also look at the supporting book and citations on that page
lrm( y ~ x, data=dat)
# --- output------
Logistic Regression Model
lrm(formula = y ~ x, data = dat)
Frequencies of Responses
0 0.54 1 1.75 2 2.25 2.75 3 3.00
4 2 1 5 2 2 4 1 1
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 22 LR chi2 51.66 R2 0.920 C 0.869
max |deriv| 0.0004 d.f. 10 g 20.742 Dxy 0.738
Pr(> chi2) <0.0001 gr 1019053402.761 gamma 0.916
gp 0.500 tau-a 0.658
Brier 0.048
Coef S.E. Wald Z Pr(>|Z|)
y>=0.54 41.6140 108.3624 0.38 0.7010
y>=1 31.9345 88.0084 0.36 0.7167
y>=1.75 23.5277 74.2031 0.32 0.7512
y>=2 6.3002 2.2886 2.75 0.0059
y>=2.25 4.6790 2.0494 2.28 0.0224
y>=2.75 3.2223 1.8577 1.73 0.0828
y>=3 0.5919 1.4855 0.40 0.6903
y>=3.00 -0.4283 1.5004 -0.29 0.7753
x -19.0710 19.8718 -0.96 0.3372
x=0.2 0.7630 3.1058 0.25 0.8059
x=0.3 3.0129 5.2589 0.57 0.5667
x=0.4 1.9526 6.9051 0.28 0.7773
x=0.5 2.9703 8.8464 0.34 0.7370
x=0.6 -3.4705 53.5272 -0.06 0.9483
x=0.7 -10.1780 75.2585 -0.14 0.8924
x=0.8 -26.3573 109.3298 -0.24 0.8095
x=0.9 -24.4502 109.6118 -0.22 0.8235
x=1 -35.5679 488.7155 -0.07 0.9420
There is also the MASS::polr function, but I find Harrell's version more approachable. This could also be approached with rank regression. The quantreg package is pretty standard if that were the route you chose. Looking at your other question, I wondered if you had tried a logistic transform as a method of linearizing that relationship. Of course, the illustrated use of lrm with an ordered variable is a logistic transformation "under the hood".
JAGS
I have an intercept-only logistic model in JAGS, defined as follows:
model{
for(i in 1:Ny){
y[i] ~ dbern(mu[s[i]])
}
for(j in 1:Ns){
mu[j] <- ilogit(b0[j])
b0[j] ~ dnorm(0, sigma)
}
sigma ~ dunif(0, 100)
}
When I plot the posterior distribution of b0 collapsing across all subjects (i.e., all b0[j]), my 95% HDI includes 0: -0.55 to 2.13. The Effective Sample Size is way above 10,000 for every b0 (around 18,000 on average). Diagnostics look good.
glmer()
Now, this is the equivalent glmer() model:
glmer(response ~ 1 + (1|subject), data = myData, family = "binomial")
The result of this model, however, is as follows:
Random effects:
Groups Name Variance Std.Dev.
speaker (Intercept) 0.3317 0.576
Number of obs: 1544, groups: subject, 27
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.7401 0.1247 5.935 2.94e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
So here it says my estimate is significantly above 0.
What the data look like
Here are the proportions of 0s and 1s by subject. You can see that, for the vast majority of subjects, the proportion of 1 is above 50%.
Any ideas why JAGS and glmer() are so different here?
0 1
1 0.47 0.53
2 0.36 0.64
3 0.29 0.71
4 0.42 0.58
5 0.12 0.88
6 0.22 0.78
7 0.54 0.46
8 0.39 0.61
9 0.30 0.70
10 0.32 0.68
11 0.36 0.64
12 0.66 0.34
13 0.38 0.62
14 0.49 0.51
15 0.35 0.65
16 0.32 0.68
17 0.12 0.88
18 0.45 0.55
19 0.36 0.64
20 0.36 0.64
21 0.28 0.72
22 0.40 0.60
23 0.41 0.59
24 0.19 0.81
25 0.27 0.73
26 0.08 0.92
27 0.12 0.88
You forgot to include a mean value, so your intercept parameter is fixed to zero. Something like this should work:
model{
for(i in 1:Ny){
y[i] ~ dbern(mu[s[i]])
}
for(j in 1:Ns){
mu[j] <- ilogit(b0[j])
b0[j] ~ dnorm(mu0, sigma)
}
mu0 ~ dnorm(0,0.001)
sigma ~ dunif(0, 100)
}
Now the posterior density of mu0 should match the sampling distribution of the intercept parameter from glmer reasonably well.
Alternatively, if you use response ~ -1 + (1|subject) as your glmer formula, you should get results that match your current JAGS model.
I'm having trouble plotting a 95% confidence band with ggplot.
Here is my code:
> my.data
scar response.rate
1 HTS 0.88
2 HTS 0.56
3 HTS 0.56
4 HTS 0.82
5 HTS 0.10
6 HTS 0.47
7 HTS 0.83
8 HTS 0.60
9 Linear 0.83
10 Linear 0.56
11 Linear 0.79
12 Linear 0.55
13 Linear 0.70
14 Linear 0.50
15 Keloid 1.00
16 Keloid 0.83
17 Keloid 1.00
18 Striae Distensae 0.33
19 Striae Distensae 0.33
ggplot(my.data, aes(scar, response.rate))+geom_point()+geom_smooth()
The output it produces:
When I use numbers for the scars, I am able to produce the following:
Any way I can produce the same graph with the scar type instead of numbers?
This method you create the CI's manually, then plot them using geom_errorbar()
library(ggplot2)
# Creating some data:
my.data = data.frame(c(rep("H",5),rep("L",5),rep("K",5),rep("S",5)), rnorm(20,1,.5) )
names(my.data) = c("scar", "response.rate")
# Standard error function
foo = function(x){sd(x)/sqrt(length(x))}
# Creating CI's manually
my.aggs = cbind(aggregate(response.rate ~ scar, data = my.data, FUN = foo),
aggregate(response.rate ~ scar, data = my.data, FUN = mean))
names(my.aggs) = c("scar","se","","means")
# Plotting
ggplot()+
geom_point(data = my.data, aes(as.factor(scar), response.rate)) +
geom_errorbar(data = my.aggs, aes(scar, ymin=means-1.96*se, ymax=means+1.96*se), width=.1)
# Alternative method that doesn't include points
library(gplots)
plotmeans(response.rate ~ scar, data = my.data)
According to the documentation of the mice package, if we want to impute data when we're interested in interaction terms we need to use passive imputation. Which is done the following way.
library(mice)
nhanes2.ext <- cbind(nhanes2, bmi.chl = NA)
ini <- mice(nhanes2.ext, max = 0, print = FALSE)
meth <- ini$meth
meth["bmi.chl"] <- "~I((bmi-25)*(chl-200))"
pred <- ini$pred
pred[c("bmi", "chl"), "bmi.chl"] <- 0
imp <- mice(nhanes2.ext, meth = meth, pred = pred, seed = 51600, print = FALSE)
It is said that
Imputations created in this way preserve the interaction of bmi with chl
Here, a new variable called bmi.chl is created in the original dataset. The meth step tells how this variable needs to be imputed from the existing ones. The pred step says we don't want to predict bmi and chl from bmi.chl. But now, if we want to apply a model, how do we proceed? Is the product defined by "~I((bmi-25)*(chl-200))" is just a way to control for the imputed values of the main effects, i.e. bmi and chl?
If the model we want to fit is glm(hyp~chl*bmi, family="binomial"), what is the correct way to specify this model from the imputed data? fit1 or fit2?
fit1 <- with(data=imp, glm(hyp~chl*bmi, family="binomial"))
summary(pool(fit1))
Or do we have to use somehow the imputed values of the new variable created, i.e. bmi.chl?
fit2 <- with(data=imp, glm(hyp~chl+bmi+bmi.chl, family="binomial"))
summary(pool(fit2))
With passive imputation, it does not matter if you use the passively imputed variable, or if you re-calculate the product term in your call to glm.
The reason that fit1 and fit2 yield different results in your example is because are not just doing passive imputation for the product term.
Instead you are transforming the two variables befor multiplying (i.e., you calculate bmi-25 and chl-100). As a result, the passively imputed variable bmi.chl does not represent the product term bmi*chl but rather (bmi-25)*(chl-200).
If you just calculate the product term, then fit1 and fit2 yield the same results like they should:
library(mice)
nhanes2.ext <- cbind(nhanes2, bmi.chl = NA)
ini <- mice(nhanes2.ext, max = 0, print = FALSE)
meth <- ini$meth
meth["bmi.chl"] <- "~I(bmi*chl)"
pred <- ini$pred
pred[c("bmi", "chl"), "bmi.chl"] <- 0
pred[c("hyp"), "bmi.chl"] <- 1
imp <- mice(nhanes2.ext, meth = meth, pred = pred, seed = 51600, print = FALSE)
fit1 <- with(data=imp, glm(hyp~chl*bmi, family="binomial"))
summary(pool(fit1))
# > round(summary(pool(fit1)),2)
# est se t df Pr(>|t|) lo 95 hi 95 nmis fmi lambda
# (Intercept) -23.94 38.03 -0.63 10.23 0.54 -108.43 60.54 NA 0.41 0.30
# chl 0.10 0.18 0.58 9.71 0.58 -0.30 0.51 10 0.43 0.32
# bmi 0.70 1.41 0.49 10.25 0.63 -2.44 3.83 9 0.41 0.30
# chl:bmi 0.00 0.01 -0.47 9.67 0.65 -0.02 0.01 NA 0.43 0.33
fit2 <- with(data=imp, glm(hyp~chl+bmi+bmi.chl, family="binomial"))
summary(pool(fit2))
# > round(summary(pool(fit2)),2)
# est se t df Pr(>|t|) lo 95 hi 95 nmis fmi lambda
# (Intercept) -23.94 38.03 -0.63 10.23 0.54 -108.43 60.54 NA 0.41 0.30
# chl 0.10 0.18 0.58 9.71 0.58 -0.30 0.51 10 0.43 0.32
# bmi 0.70 1.41 0.49 10.25 0.63 -2.44 3.83 9 0.41 0.30
# bmi.chl 0.00 0.01 -0.47 9.67 0.65 -0.02 0.01 25 0.43 0.33
This is not surprising because the ~I(bmi*chl) in mice and the bmi*chl in glm do the exact same thing. They merely calculate the product of the two variables.
Remark:
Note that I added a line saying that bmi.chl should be used as a predictor when imputing hyp. Without this step, passive imputation has no purpose because the imputation model would neglect the product term, thus being incongruent with the analysis model.
I have a small table of summary data with the odds ratio, upper and lower confidence limits for four categories, with six levels within each category. I'd like to produce a chart using ggplot2 that looks similar to the usual one created when you specify a lm and it's se, but I'd like R just to use the pre-specified values I have in my table. I've managed to create the line graph with error bars, but these overlap and make it unclear. The data look like this:
interval OR Drug lower upper
14 0.004 a 0.002 0.205
30 0.022 a 0.001 0.101
60 0.13 a 0.061 0.23
90 0.22 a 0.14 0.34
180 0.25 a 0.17 0.35
365 0.31 a 0.23 0.41
14 0.84 b 0.59 1.19
30 0.85 b 0.66 1.084
60 0.94 b 0.75 1.17
90 0.83 b 0.68 1.01
180 1.28 b 1.09 1.51
365 1.58 b 1.38 1.82
14 1.9 c 0.9 4.27
30 2.91 c 1.47 6.29
60 2.57 c 1.52 4.55
90 2.05 c 1.31 3.27
180 2.422 c 1.596 3.769
365 2.83 c 1.93 4.26
14 0.29 d 0.04 1.18
30 0.09 d 0.01 0.29
60 0.39 d 0.17 0.82
90 0.39 d 0.2 0.7
180 0.37 d 0.22 0.59
365 0.34 d 0.21 0.53
I have tried this:
limits <- aes(ymax=upper, ymin=lower)
dodge <- position_dodge(width=0.9)
ggplot(data, aes(y=OR, x=days, colour=Drug)) +
geom_line(stat="identity") +
geom_errorbar(limits, position=dodge)
and searched for a suitable answer to create a pretty plot, but I'm flummoxed!
Any help greatly appreciated!
You need the following lines:
p<-ggplot(data=data, aes(x=interval, y=OR, colour=Drug)) + geom_point() + geom_line()
p<-p+geom_ribbon(aes(ymin=data$lower, ymax=data$upper), linetype=2, alpha=0.1)
Here is a base R approach using polygon() since #jmb requested a solution in the comments. Note that I have to define two sets of x-values and associated y values for the polygon to plot. It works by plotting the outer perimeter of the polygon. I define plot type = 'n' and use points() separately to get the points on top of the polygon. My personal preference is the ggplot solutions above when possible since polygon() is pretty clunky.
library(tidyverse)
data('mtcars') #built in dataset
mean.mpg = mtcars %>%
group_by(cyl) %>%
summarise(N = n(),
avg.mpg = mean(mpg),
SE.low = avg.mpg - (sd(mpg)/sqrt(N)),
SE.high =avg.mpg + (sd(mpg)/sqrt(N)))
plot(avg.mpg ~ cyl, data = mean.mpg, ylim = c(10,30), type = 'n')
#note I have defined c(x1, x2) and c(y1, y2)
polygon(c(mean.mpg$cyl, rev(mean.mpg$cyl)),
c(mean.mpg$SE.low,rev(mean.mpg$SE.high)), density = 200, col ='grey90')
points(avg.mpg ~ cyl, data = mean.mpg, pch = 19, col = 'firebrick')