I am trying to perform a linear regression on experimental data consisting of replicate measures of the same condition (for several conditions) to check for the reliability of the experimental data. For each condition I have ~5k-10k observations stored in a data frame df:
[1] cond1 repA cond1 repB cond2 repA cond2 repB ...
[2] 4.158660e+06 4454400.703 ...
[3] 1.458585e+06 4454400.703 ...
[4] NA 887776.392 ...
...
[5024] 9571785.382 9.679092e+06 ...
I use the following code to plot scatterplot + lm + R^2 values (stored in rdata) for the different conditions:
for (i in seq(1,13,2)){
vec <- matrix(0, nrow = nrow(df), ncol = 2)
vec[,1] <- df[,i]
vec[,2] <- df[,i+1]
vec <- na.exclude(vec)
plot(log10(vec[,1]),log10(vec[,2]), xlab = 'rep A', ylab = 'rep B' ,col="#00000033")
abline(fit<-lm(log10(vec[,2])~log10(vec[,1])), col='red')
legend("topleft",bty="n",legend=paste("R2 is",rdata[1,((i+1)/2)] <- format(summary(fit)$adj.r.squared,digits=4)))
}
However, the lm seems to be shifted so that it does not fit the trend I see in the experimental data:
It consistently occurs for every condition. I unsuccesfully tried to find an explanation by looking up the scource code and browsing different forums and posts (this or here).
Would have like to simply comment/ask a few questions, but can't.
From what I've understood, both repA and repB are measured with error. Hence, you cannot fit your data using an ordinary least square procedure, which only takes into account the error in Y (some might argue a weighted OLS may work, however I'm not skilled enough to discuss that). Your question seem linked to this one.
What you can use is a total least square procedure: it takes into account the error in X and Y. In the example below, I've used a "normal" TLS assuming there is the same error in X and Y (thus error.ratio=1). If it is not, you can specify the error ratio by entering error.ratio=var(y1)/var(x1) (at least I think it's var(Y)/var(X): check on the documentation to ensure that).
library(mcr)
MCR_reg=mcreg(x1,y1,method.reg="Deming",error.ratio=1,method.ci="analytical")
MCR_intercept=getCoefficients(MCR_reg)[1,1]
MCR_slope=getCoefficients(MCR_reg)[2,1]
# CI for predicted values
x_to_predict=seq(0,35)
predicted_values=MCResultAnalytical.calcResponse(MCR_reg,x_to_predict,alpha=0.05)
CI_low=predicted_values[,4]
CI_up=predicted_values[,5]
Please note that, in Deming/TLS regressions, your x- and y-errors are supposed to follow normal distribution, as explained here. If it's not the case, go for a Passing-Bablok regressions (and the R code is here).
Also note that the R2 isn't defined for Deming nor Passing Bablok regressions (see here). A correlation coefficient is a good proxy, although it does not exactly provide the same information. Since you're studying a linear correlation between two factors, see Pearson's product moment correlation coefficient, and use e.g. the rcorrfunction.
Related
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
I would like to get the bootstrapped t-value and the bootstrapped p-value of a lm.
I have the following code (basically copied from a paper) which works.
# First of all you need the following packages
install.packages("car")
install.packages("MASS")
install.packages("boot")
library("car")
library("MASS")
library("boot")
boot.function <- function(data, indices){
data <- data[indices,]
mod <- lm(prestige ~ income + education, data=data) # the liear model
# the first element of the following vector contains the t-value
# and the second element is the p-value
c(summary(mod)[["coefficients"]][2,3], summary(mod)[["coefficients"]][2,4])
}
Now, I compute the bootstrapping model, which gives me the following:
duncan.boot <- boot(Duncan, boot.function, 1999)
duncan.boot
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = Duncan, statistic = boot.function, R = 1999)
Bootstrap Statistics :
original bias std. error
t1* 5.003310e+00 0.288746545 1.71684664
t2* 1.053184e-05 0.002701685 0.01642399
I have two questions:
My understanding is that the bootsrapped value is the original plus the bias, which means that both bootstrapped values (the bootstrapped t-value as well as the bootstrapped p-value) are greater than the original values. This in turn is not possible, because if the t-value rises (which means more significance) the p-values MUST be lower, right? Therefore I think that I have not yet really understood the output of the boot function (here: duncan.boot). How do I compute the bootstrapped values?
I do not understand how the boot() works. If you look at duncan.boot <- boot(Duncan, boot.function, 1999) you see that I have not passed any arguments for the function "boot.function". I suppose that R sets data <- Duncan. But since I have not passed anything for the argument "indices", I do not understand how the following line in the function "boot.function" works data <- data[indices,]
I hope the questions make sense!??
The boot function is "expecting" to get a function that has two arguments: the first being a data.frame and the second being an "indices" vector (possibly with duplicate entries and probably not using all the indices) to use in selecting rows and probably having some duplicate or triplicates.) It then samples with replacement determined by the pattern of duplicates and triplicates from the original dataframe (multiple times determined by "R" with different "choice sets"), passes those to the indices argument in the boot.function, and then collects the results of the R number of function applications.
Regarding what is reported by the print method for boot objects, take a look at this (done after examining the returned object with str()
> duncan.boot$t0
[1] 5.003310e+00 1.053184e-05
> apply(duncan.boot$t, 2, mean)
[1] 5.342895220 0.002607943
> apply(duncan.boot$t, 2, mean) - duncan.boot$t0
[1] 0.339585441 0.002597411
It becomes more obvious that the T0 value is from the original data while the bias is the difference between the mean of the boot()-ed values and the T0 values. I don't think it makes a lot of sense to be asking why p-values based on parametric considerations are increasing in association with an increase in estimated t-statistics. You are really in two disparate regions of statistical thought when you do that. I would have interpreted the increase in p-values as an effect of the sampling process, which does not take into account the Normal distribution assumptions. It is simply saying something about the sampling distribution of the p-value (which is really just another sample statistic).
(Comment: The sourcebook used at the time of R development was Davison and Hinkley's "Bootstrap Methods and their Applications". I'm no claiming any support for my answer above, but I thought to put it in as a reference after Hagen Brenner asked about sampling with two indices in the comments below. There are many unexpected aspects of bootstrapping that arise after one goes beyond the simple parametric estimation and I would first turn to that reference if I were tackling more complex sampling situations.)
Calling all experts on local regression and/or R!
I have run into a limitation of the standard loess function in R and hope you have some advice. The current implementation supports only 1-4 predictors. Let me set out our application scenario to show why this can easily become a problem as soon as we want to employ globally fit parametric covariables.
Essentially, we have a spatial distortion s(x,y) overlaid over a number of measurements z:
z_i = s(x_i,y_i) + v_{g_i}
These measurements z can be grouped by the same underlying undistorted measurement value v for each group g. The group membership g_i is known for each measurement, but the underlying undistorted measurement values v_g for the groups are not known and should be determined by (global, not local) regression.
We need to estimate the two-dimensional spatial trend s(x,y), which we then want to remove. In our application, say there are 20 groups of at least 35 measurements each, in the most simple scenario. The measurements are randomly placed. Taking the first group as reference, there are thus 19 unknown offsets.
The below code for toy data (with a spatial trend in one dimension x) works for two or three offset groups.
Unfortunately, the loess call fails for four or more offset groups with the error message
Error in simpleLoess(y, x, w, span, degree, parametric, drop.square,
normalize, :
only 1-4 predictors are allowed"
I tried overriding the restriction and got
k>d2MAX in ehg136. Need to recompile with increased dimensions.
How easy would that be to do? I cannot find a definition of d2MAX anywhere, and it seems this might be hardcoded -- the error is apparently triggered by line #1359 in loessf.f
if(k .gt. 15) call ehg182(105)
Alternatively, does anyone know of an implementation of local regression with global (parametric) offset groups that could be applied here?
Or is there a better way of dealing with this? I tried lme with correlation structures but that seems to be much, much slower.
Any comments would be greatly appreciated!
Many thanks,
David
###
#
# loess with parametric offsets - toy data demo
#
x<-seq(0,9,.1);
x.N<-length(x);
o<-c(0.4,-0.8,1.2#,-0.2 # works for three but not four
); # these are the (unknown) offsets
o.N<-length(o);
f<-sapply(seq(o.N),
function(n){
ifelse((seq(x.N)<= n *x.N/(o.N+1) &
seq(x.N)> (n-1)*x.N/(o.N+1)),
1,0);
});
f<-f[sample(NROW(f)),];
y<-sin(x)+rnorm(length(x),0,.1)+f%*%o;
s.fs<-sapply(seq(NCOL(f)),function(i){paste('f',i,sep='')});
s<-paste(c('y~x',s.fs),collapse='+');
d<-data.frame(x,y,f)
names(d)<-c('x','y',s.fs);
l<-loess(formula(s),parametric=s.fs,drop.square=s.fs,normalize=F,data=d,
span=0.4);
yp<-predict(l,newdata=d);
plot(x,y,pch='+',ylim=c(-3,3),col='red'); # input data
points(x,yp,pch='o',col='blue'); # fit of that
d0<-d; d0$f1<-d0$f2<-d0$f3<-0;
yp0<-predict(l,newdata=d0);
points(x,y-f%*%o); # spatial distortion
lines(x,yp0,pch='+'); # estimate of that
op<-sapply(seq(NCOL(f)),function(i){(yp-yp0)[!!f[,i]][1]});
cat("Demo offsets:",o,"\n");
cat("Estimated offsets:",format(op,digits=1),"\n");
Why don't you use an additive model for this? Package mgcv will handle this sort of model, if I understand your Question, just fine. I might have this wrong, but the code you show is relating x ~ y, but your Question mentions z ~ s(x, y) + g. What I show below for gam() is for response z modelled by a spatial smooth in x and y with g being estimated parametrically, with g stored as a factor in the data frame:
require(mgcv)
m <- gam(z ~ s(x,y) + g, data = foo)
Or have I misunderstood what you wanted? If you want to post a small snippet of data I can give a proper example using mgcv...?
Using predict() one can obtain the predicted value of the dependent variable (y) for a certain value of the independent variable (x) for a given model. Is there any function that predicts x for a given y?
For example:
kalythos <- data.frame(x = c(20,35,45,55,70),
n = rep(50,5), y = c(6,17,26,37,44))
kalythos$Ymat <- cbind(kalythos$y, kalythos$n - kalythos$y)
model <- glm(Ymat ~ x, family = binomial, data = kalythos)
If we want to know the predicted value of the model for x=50:
predict(model, data.frame(x=50), type = "response")
I want to know which x makes y=30, for example.
Saw the previous answer is deleted. In your case, given n=50 and the model is binomial, you would calculate x given y using:
f <- function (y,m) {
(logit(y/50) - coef(m)[["(Intercept)"]]) / coef(m)[["x"]]
}
> f(30,model)
[1] 48.59833
But when doing so, you better consult a statistician to show you how to calculate the inverse prediction interval. And please, take VitoshKa's considerations into account.
Came across this old thread but thought I would add some other info. Package MASS has function dose.p for logit/probit models. SE is via delta method.
> dose.p(model,p=.6)
Dose SE
p = 0.6: 48.59833 1.944772
Fitting the inverse model (x~y) would not makes sense here because, as #VitoshKa says, we assume x is fixed and y (the 0/1 response) is random. Besides, if the data weren’t grouped you’d have only 2 values of the explanatory variable: 0 and 1. But even though we assume x is fixed it still makes sense to calculate a confidence interval for the dose x for a given p, contrary to what #VitoshKa says. Just as we can reparameterize the model in terms of ED50, we can do so for ED60 or any other quantile. Parameters are fixed, but we still calculate CI's for them.
The chemcal package has an inverse.predict() function, which works for fits of the form y ~ x and y ~ x - 1
You just have to rearrange the regression equation, but as the comments above state this may prove tricky and not necessarily have a meaningful interpretation.
However, for the case you presented you can use:
(1/coef(model)[2])*(model$family$linkfun(30/50)-coef(model)[1])
Note I did the division by the x coefficient first to allow the name attribute to be correct.
For just a quick view (without intervals and considering additional issues) you could use the TkPredict function in the TeachingDemos package. It does not do this directly, but allows you to dynamically change the x value(s) and see what the predicted y-value is, so it would be fairly simple to move x until the desired Y is found (for given values of additional x's), this will also show possibly problems with multiple x's that would work for the same y.
I know there is COXPHFIT function in MATLAB to do Cox regression, but I have problems understanding how to apply it.
1) How to compare two groups of samples with survival data in days (survdays), censoring (cens) and some predictor value (x)? The groups defined by groups logical variable. Groups have different number of samples.
2) What is the baseline parameter in coxphfit? I did read the docs, but how should I choose the baseline properly?
It would be great if you know a site with good detailed examples on medical survival data. I found only the Mathworks demo that does not even mention coxphfit.
Do you know may be another 3rd party function for Cox regression?
UPDATE: The r tag added since the answer I've got is for R.
With survival analysis, the hazard function is the instantaneous death rate.
In these analyses, you are typically measuring what effect something has on this hazard function. For example, you may ask "does swallowing arsenic increase the rate at which people die?". A background hazard is the level at which people would die anyway (without swallowing arsenic, in this case).
If you read the docs for coxphfit carefully, you will notice that that function tries to calculate the baseline hazard; it is not something that you enter.
baseline The X values at which to
compute the baseline hazard.
EDIT: MATLAB's coxphfit function doesn't obviously work with grouped data. If you are happy to switch to R, then the anaylsis is a one-liner.
library(survival)
#Create some data
n <- 20;
dfr <- data.frame(
survdays = runif(n, 5, 15),
cens = runif(n) < .3,
x = rlnorm(n),
groups = rep(c("first", "second"), each = n / 2)
)
#The Cox ph analysis
summary(coxph(Surv(survdays, cens) ~ x / groups, dfr))
ANOTHER EDIT: That baseline parameter to MATLAB's coxphfit appears to be a normalising constant. R's coxph function doesn't have an equivalent parameter. I looked in Statistical Computing by Michael Crawley and it seems to suggest that the baseline hazard isn't important, since it cancels out when you calculate the likelihood of your individual dying. See Chapter 33, and p615-616 in particular. My knowledge of how the model works isn't deep enough to explain the discrepancy in the MATLAB and R implementations; perhaps you could ask on the Stack Exchange Stats Analysis site.