I am using h2o.deeplearning to train a neural network on a classification task.
What I have
Y ~ x1 + x2... where all x variables are continuous and Y is binary.
What I want
To be able to train a deeplearning object to predict the probability of a given row of being true or false. That is, a predicted(Y) restricted to between 0 and 1.
What I've tried
When Y is inputted as a numeric (i.e. 0 or 1), h2o deeplearning automatically treats it as a regression problem. This is fine, except the final layer of the NN is linear, not tanh, and the predicted values can be greater than 1 or less than 0. I've not been able to find a way to get the final layer to be a tanh.
When Y is inputted as categorical (i.e. TRUE or FALSE), h2o deeplearning automatically treats it as a classification problem. Instead of giving me the desired probability of Y being 1 or 0, it gives me its best guess of what Y is.
Is there a way around this? A trick, tweak or an overlooked parameter? I have noticed in the h2o.deeplearning documentation a 'distribution' parameter, but no further information on what that's for. My best guess is that it is some kind of link function in the same vein as GLM, but I'm not sure.
If you treat the problem as a binary classification problem then you not only get the “prediction” of 0 or 1, but also the p0 and p1 probabilities that add up to 1. These are the probabilies that the predicted value is the negative and positive class, respectively.
Then just use p1 directly.
Related
For my class we have to create a model to predict the credit balance of each individuals. Based on observations, many results are zero where the lm tries to calculate them.
To overcome this I created a new variable that results in zero if X and Y are true.
CB$Balzero = ifelse(CB$Rating<=230 & CB$Income<90,0,1)
This resulted in getting 90% of the zero results right. The problem is:
How can I place this variable in the lm so it correctly results in zeros when the proposition is true and the calculation when it is false?
Something like: lm=Balzero*(Balance~.)
I think that
y ~ -1 + Balzero:Balance
might work (you haven't given us a reproducible example to try).
-1 tells R to omit the intercept
: specifies an interaction. If both variables are numeric, then A:B includes the product of A and B as a term in the model.
The second term could also be specified as I(Balzero*Balance) (I means "as is", i.e. interpret * in the usual numerical sense, not in its formula-construction context.)
These specifications should fit the model
Y = beta1*Balzero*Balance + eps
where eps is an error term.
If Balzero == 0, the predicted value will be zero. If Balzero==1 the predicted value will be beta1*Balance.
You might want to look into random forest models, which naturally incorporate the kind of qualitative splitting that you're doing by hand in your example.
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 am very new to R (mostly program in SQL) but was faced with a problem that SQL couldn't help me with. I'll try to simplify the problem below.
Assume I have a set of data with 100 rows where each row has a different weight associated with it. Out of those 100 rows of data, 5 have an X value that is top-coded at 1000. Also assume that X can be represented by the linear equation X ~ Y + Z + U + 0 (want a positive value so I don't want a Y-intercept).
Now, without taking the weights of each row of data into consideration, the formula I used in R was:
fit = censReg(X ~ Y + Z + U + 0, left = -Inf, right = 1000, data = dataset)
If I computed summary(fit) I would get 0 left-censored values, 95 uncensored values, and 5 right censored values which is exactly what I want, minus the fact that the weights haven't been sufficiently added into the mix. I checked the reference manual on the censReg function and it doesn't seem like it accepts a weight argument.
Is there something I'm missing about the censReg function or is there another function that would be of better use to me? My end goal is to estimate X in the cases where it is censored (i.e. the 5 cases where it is 1000).
You should use Tobit regression for this situation, it is designed specifically to linearly model latent variables such as the one you describe.
The regression accounts for your weights and the censored observations, which can be seen in the derivation of the log-likelihood function for the Type I Tobit (upper and lower bounded).
Tobit regression can be found in the VGAM package using the vglm function with a tobit control parameter. An excellent example can be found here:
http://www.ats.ucla.edu/stat/r/dae/tobit.htm
I'm using the plottol function in the tolerance package of R and getting an error / warning after my plot is generated that say "NOTE: A regression through the origin is fitted!"
I've googled it and come up with nothing and I don't know where to start with trying to figure out what it's warning me about.
It in general indicates that a model is fitted whereby the intercept term is forced to be zero (in other words the model has no intercept). The intercept in a model of y ~ x is the expectation of y when x is equal to 0. By setting the intercept to 0 we are explicitly stating that the expectation of y is 0 when x is 0. In other words, the straight line fitted to y ~ x passes through the point (0,0), i.e. the origin.
If you want a less general explanation (more tailored to your specific example) you'll need to provide more details on which of the many functions in tolerance you are using, e.g. via a reproducible example.
I am currently trying to fit an adaBoost model in R using the gbm.fit model. I have tried everything I could but in the end my model keeps giving me prediction values outside of [0,1]. I understand that type = "response" only works for bernoulli but I keep getting values just outside of 0,1. Any thoughts? Thanks!
GBMODEL <- gbm.fit(
x=training.set,
y=training.responses,
distribution="adaboost",
n.trees=5000,
interaction.depth=1,
shrinkage=0.005,
train.fraction=1,
)
predictionvalues = predict(GBMODEL,
newdata=test.predictors,
n.trees=5000,
type="response")
it is correct to obtain y range outside [0,1] by gbm package choosing "adaboost" as your loss function.
After training, adaboost predicts category by the sign of output.
For instance, for binary class problem, y{-1,1}, the class lable will be signed to the sign of output y. So if you got y=0.9 or y=1.9 will give you the same result-observation belongs to y=1 class. However, y=1.9 simply suggests a more confident conclusion than y=0.9. (if you want to know why, I would suggest you to read margin-based explanation of adaboost, you will find very similar result with SVM).
Hope this can help you.
This may not be completely accurate mathematically, but I just did pnorm( predicted values) and you get values from 0 to 1, because the adaboost predicted values appear to be scaled on a Normal(0,1).