I am trying to plot a ROC curve of an identifier used to determine positive incidences against background dataset. The identifier is a list of probability scores with some overlap between the two groups.
FG BG
0.02 0.10
0.03 0.25
0.02 0.12
0.04 0.16
0.05 0.45
0.12 0.31
0.13 0.20
(where FG = Positive and BG = Negative.)
I am plotting a ROC curve using PRROC in R to assess how well the identifier classifies the data into the correct group. Although there is a clear distinction between the classifier values produced between the positive and negative datasets, but my current ROC plot in R shows a low AUC value. My probability scores for the positive data are lower than the background so if I switch the classification around and have the background as the foreground points, I get a high scoring AUC curve and I am not 100% clear why this is the case, which plot is the best to use or whether there was an additional step I have missed before analysing my data.
roc <- roc.curve(scores.class0 = FG, scores.class1 = BG, curve = T)
ROC curve
Area under curve:
0.07143
roc2 <- roc.curve(scores.class0 = BG, scores.class1 = FG, curve = T)
ROC curve
Area under curve:
0.92857
As you have indeed noticed, most ROC analysis tools assume that the scores in your positive class are higher than those of the negative class. More formally, an instance is classified as "positive" if X > T, where T is the decision threshold, and negative otherwise.
There is no fundamental reason for it to be so. It is perfectly valid to have a decision such as X < T, however most ROC software don't have that option.
Using your first option resulting in AUC = 0.07143 would imply that your classifier performs worse than random. This is not correct.
As you noticed, swapping the class labels yields the correct curve value.
This is possible because ROC curves are insensitive to class distributions - and the classes can be reverted without a problem.
However I wouldn't personally recommend that option. I can see two cases where this can be misleading:
to someone else looking at the code, or yourself in a few months; figuring the classes are wrong and "fixing" them
or if you want to apply the same code to PR curves, which are sensitive to class distributions and where you cannot swap the classes.
An alternative and preferable approach would be to invert your scores for this analysis, so that the positive class effectively has higher scores:
roc <- roc.curve(scores.class0 = -FG, scores.class1 = -BG, curve = T)
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 have a continuous independent variable (let's say 'height') and a binary independent variable (let's say 'gets a job'). I want to see at what cutoff value for height best predicts one's ability to get a job. I also want to see how accurate this model is. I assumed a multinomial logistic model. I wanted a ROC curve so I used the ROCR package in R. This was my code:
mymodel <- multinom(job~height, data = dataset)
pred <- predict(mymodel,dataset,type = 'prob')
roc_pred <- prediction(pred,dataset$job)
roc <- performance(roc_pred,"tpr","fpr")
plot(roc,colorize=T)
Now, this is my question. When I colorize the plot, it gives me the range of cut-off values used to make the plot. I'm a little confused as to what the cutoff values actually are though. Are the cutoff values the heights? Or the probability that a certain data point (person) with a certain height is able to get a job? I have a feeling it's the latter, but I am interested in the former. If it is the latter, how do I obtain the cutoff value for the height??
I found a video that explains the cutoffs you see: https://www.youtube.com/watch?v=YdNhNfJ4Vl8
There are many different ways to estimate optimal cutoffs: Youden Index, Sensitivity + Specificity,Distance to Corner and many others (see this article)
I suggest you use a pROC library to do so
library(pROC)
roc <- roc(fit, obs, percent = TRUE)
roc.out <- coords(roc, "best", ret = c("threshold", "sens", "spec"), transpose = TRUE)
method "best" uses the Younden index (J- index) The maximum value of the Youden index is 1 (perfect test) and the minimum is 0 when the test has no diagnostic value. The minimum occurs when sensitivity=1−specificity, i.e., represented by the equal line (the diagonal) in the ROC diagram. The vertical distance between the equal line and the ROC curve is the J-index for that particular cutoff. The J-index is represented by the ROC-curve itself.
I have tried to add some weights to the FP, TN in the ROCR package and evaluate the Area Under the Curve AUC.
Here is the code:
auc.tmp <- performance(pred,"auc",cost.fp=10, cost.tn=100);
auc <- as.numeric(auc.tmp#y.values)
print(paste0("AUC is: ",auc))
and the output is:
"AUC is: 0.834187518842327"
Then i repeated the commands without the weights:
auc.tmp <- performance(pred,"auc");
auc <- as.numeric(auc.tmp#y.values)
print(paste0("AUC is: ",auc))
Again i received the same output as:
"AUC is: 0.834187518842327"
So, the output with cost and without cost are the same. When i refereed to the ROCR package manual, with ?performance about auc it is stated that:
Since the output of auc is cutoff-independent, this measure cannot be combined with other measures into a parametric curve. The partial area under the ROC curve up to a given false positive rate can be calculated by passing the optional parameter fpr.stop=0.5 (or any other value between 0 and 1) to performance.
The result of codes verifies the ROCR manual claim, about indipendence of AUC output from cutoffs
But now let's look at this image.
A very basic interpretation is that, when i see the TP i go up, and when i see FP i go to left.
Therefore, the area under the curve is hardly dependents on cutoffs?
My data is in the following format and includes a particular statistic
site LRStat
1 3.580728
2 2.978038
3 5.058644
4 3.699278
5 4.349046
This is just a sample of the data.
I then obtained the null LR distribution as well by permuting random pairs of data. I used this to plot a histogram with frequency in the y-axes and LR statistic in the x-axes. How is it possible to determine the critical p-value cut-off points based on the null distribution (as shown in the below figure)?
You now have a sampling distribution of LR values. The quantile function in R will give you an estimate of whatever "critical value" you prefer. If, for instance, you decided you wanted the conventional 0.05 "p-value" you could take your dataframe, named LR_df for illustration, and issue this command:
quantile( LR_df[ , 'LRStat'] , 0.95)
If you wanted all of those "probabilities" on the figure, you would use a vector of values complementary to unity. The following code gives you the LSstat values at which a given proportion of the sample are higher than that value.
quantile( LR_df[ , 'LRStat'] , c(0.9, 0.95, 0.99, 0.999, 0.9999) )
The p-values are just a sampling distribution of a test statistic under a null hypothesis. Your null hypothesis in this case is that the LRstats are uniformly distributed. (I know it sounds strange to put it that way, but if you want to argue with the statisticians then get a copy of http://amstat.tandfonline.com/doi/pdf/10.1198/000313008X332421 .) The choice of p-value for cutoff will depend on scientific or business setting. If you were assessing an investment opportunity the cutoff might be 0.15 but if you are trying to find new scientific knowledge, I think it should be smaller (more stringent test). The field of molecular genetics has a lot of junk (i.e. fails to reproduce results) in their literature because they were not strict enough in the statistical methods.
I am doing some analysis regarding a binomial glm model that I have fitted earlier in R. While looking at my data, I figured out that the suitable cutoff point for my binary outcome should be 0.75 instead of 0.5. I am trying to get the cost() function of the cv.glm() {boot package} to use the 0.75 cutoff point, but I have failed to get the right syntax.
I know for the 0.5 cutoff we normally use:
cost <- function(r, pi = 0) mean(abs(r-pi) > 0.5)
Can someone show me what is the right way to change the cutoff point in this function? (let's stick to 0.75 maybe).