Binomial Logistic Regression.

Standard

I’m officially a Kaggler.

Cut to the good ol’ Titanic challenge. Ol’ is right – It’s been running since 2012 and ends in 3 months! I showed up late to the party. Oh well, I guess it’s full steam ahead from now on.

The competition  ‘Machine Learning from Disaster’ asks you to apply machine learning to analyse and predict  which passengers survived the Titanic tragedy. It is placed as knowledge competition.

Since I am still inching my way up the learning curve, I tried to see what I could achieve with my current tool set. For my very quick first attempt, it seemed like a no-brainer to apply out of the box logistic regression. For those in the competition, this approach, got me at around 75.something% and  placed me at 4,372 of 5172 entries. I have 3 months to better this score. And better it, I shall!

So essentially how this works is that you download the data from Kaggle. 90% of it (889 rows)  is flagged as training data and the rest is test data(418 rows). You need to build your model, predict survival on the test set and pass the data back to Kaggle which computes a score for you and places you accordingly on the ‘Leaderboard’.

The data:

Since we’re working with real world data, we’ll need to take into account the NAs, improper formatting, missings values et al.

After reading in the data, I ran a check to see how many entries had missing values. The simple sapply() sufficed in this case.

dat<-read.csv("C:/Users/Amita/Downloads/train (1).csv",header=T,sep=","
      ,na.strings = c(""))
sapply(dat,function(x){sum(is.na(x))})

The column ‘Cabin’ seems to have the most missing values – like a LOT, so I ditched it. ‘Age’ had quite a few missing values as well , but it seems like a relevant column.

I went ahead and dealt with the missing values by replacing them with the mean of the present values in that column.Easy peasy.

dat[is.na(dat$Age),][6]<- mean(dat$Age,na.rm=T)
dat <- dat[,-11]

 

Next, I divided the training dataset into two – ‘traindat’ and ‘testdat’. The idea was to train the prospective model on the traindat dataset and the predict using the rows in testdat. Computing the RMSE would then give us an idea about the performance of the model.

set.seed(30)
indices<- rnorm(nrow(dat))>0
traindat<- dat[indices,]
testdat<-dat[!indices,]
dim(traindat)
dim(testdat)

 

Structure wise, except for a couple of columns that had to be converted into factors,  the datatypes were on point.

testdat$Pclass<- factor(testdat$Pclass)
testdat$SibSp <- factor(testdat$SibSp)
testdat$Parch <- factor(testdat$Parch)
testdat$Survived<-factor(testdat$Survived)
traindat$Pclass<- factor(traindat$Pclass)
traindat$SibSp <- factor(traindat$SibSp)
traindat$Parch <- factor(traindat$Parch)
traindat$Survived<-factor(traindat$Survived)

 

The model:

Since, the response variable is a categorical variable with only two outcomes, and the predictors are both continuous and categorical, it makes it a candidate for conducting binomial logistic regression.

mod <- glm(Survived ~ Pclass + Sex + Age + SibSp+ Parch + Embarked + Fare ,
       family=binomial(link='logit'),data=traindat)

require(car)
Anova(mod)

2016-10-02_21-06-32.jpg

The result shows significance values for only ‘Pclass’, ‘Sex’, ‘Age’, ‘SibSp’ , so we’ll build a second model with just these variables and use that for further analysis.

mod2 <- glm(Survived ~ Pclass + Sex + Age + SibSp ,
        family=binomial(link='logit'),data=traindat)
Anova(mod2)

Let’s visualize the relationships between the response and predictor variables.

2016-10-02_21-11-49.jpg

  • The log odds (since we used link= ‘logit’) of survival seems to decline as the passenger’s class decreases.
  • Women have a higher log odds of survival than men.
  • Higher the age gets, lower the log odds of survival get.
  • The number of siblings/spouses aboard* also affects the log odds of survival. The log odds for numbers 5 and 8 can go either way indicated by the wide CIs. (*needs to be explored more).

 

Model Performance:

We will test the peformance of mod2 in predicting ‘Survival’ on a new set of data.

pred <- predict(mod2,newdata=testdat[,-2],type="response")
pred <- ifelse(pred > 0.5,1,0)
testdat$Survived <- as.numeric(levels(testdat$Survived))[testdat$Survived]
rmse <- sqrt((1/nrow(testdat)) * sum( (testdat$Survived - pred) ^ 2))
rmse  #0.4527195

error <- mean(pred != testdat$Survived)
print(paste('Accuracy',1-error)) #81% accuracy

81.171% passenger has been correctly classified.

But when I used the same model to run predictions on Kaggle’s test dataset, the uploaded results fetched me a 75.86 %. I’m guessing the reason could be the arbit split ratio between the ‘traindat’ and ‘testdat’. Maybe next time I’ll employ some sort of bootstrapping.

Well, this is pretty much it for now. I will attempt to better my score in the upcoming weeks (time permitting) and in the event I am successful, I shall add subsequent blog posts and share my learnings (Is that even a word?!).

One thing’s for sure though –  This road is loooooong, is long , is long . . . . .

😀

Later!

 

 

 

 

 

Linear mixed-effects model with bootstrapping.

Standard

Dataset here.

We are going to perform a linear mixed effects analysis of the relationship between height and treatment of trees, as studied over a period of time.

Begin by reading in the data and making sure that the variables have the appropriate datatype.

tree<- read.csv(path, header=T,sep=",")
tree$ID<- factor(tree$ID)
tree$plotnr <- factor(tree$plotnr)

Plot the initial graph to get a sense of the relationship between the variables.

library(lattice)
bwplot(height ~ Date|treat,data=tree)

Rplot.png

There seems to be a linear-ish relationship between height and time . The height increases over time.The effect of the treatments(I,IL,C,F) is not clear at this point.

Based on the linear nature of relationship, we can model time to be numeric, i.e. , number of days.

tree$Date<- as.Date(tree$Date)
tree$time<- as.numeric(tree$Date - min(tree$Date))

The data shows us that the data was collected from the same plot on multiple dates, making it  a candidate for repeated measures.

Let’s visually check for the by-plot variability.

boxplot(height ~ plotnr,data=tree)

Rplot01.png

The variation between plots isn’t big – but there are still noticeable differences which will need to be accounted for in our model.

 

 

 

Let’s fit two models : one with interaction between the predictors and one without.

library(lme4)
library(lmerTest)

#fit a model

mod <- lmer(height ~ time + treat + (1|plotnr),data=tree) #without interaction
mod.int <- lmer(height ~ time * treat + (1|plotnr),data=tree) #with interaction

Upon comparing the two,  it’s evident that the model with the interaction would perform better at predicting responses.

anova(mod,mod.int)
> anova(mod,mod.int)
refitting model(s) with ML (instead of REML)
Data: tree
Models:
object: height ~ time + treat + (1 | plotnr)
..1: height ~ time * treat + (1 | plotnr)
       Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq) 
object 7 22654 22702 -11320 22640 
..1   10 20054 20122 -10017 20034 2605.9 3 < 2.2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>  

So far, we have a random intercept model, we should also check whether adding a random slope can help make the model more efficient. We do this by adding a random slope to the model and comparing that with the  random intercept model from before.

mod.int.r <- lmer(height ~ time * treat + (treat|plotnr),data=tree)
anova(mod.int,mod.int.r)

> anova(mod.int,mod.int.r) # we don't need the random slope
refitting model(s) with ML (instead of REML)
Data: tree
Models:
object: height ~ time * treat + (1 | plotnr)
..1: height ~ time * treat + (treat | plotnr)
 Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
object 10 20054 20122 -10017 20034 
..1    19 20068 20197 -10015 20030 3.6708 9 0.9317 #AIC goes up

Nope. Don’t need the slope. Dropping it.

Continuing with mode.int. Let’s determine the p-values using Anova() from the car package.

2016-09-23_22-25-49.jpg

We can see that the main effects of time and treat are highly significant, as is the interaction between the two.

Running a summary() spews out the following log.

2016-09-23_21-36-23.jpg

As seen,  time:treatF is not  significantly different from the first level , time:treatC , that is, there is no difference between F treatment and C treatment in terms of the interaction with time.Treatments I and IL are significantly different from treatment C in terms of interaction with time.Although there is a significant main effect of treatment, none of the levels are actually different from the first level.

Graphically represented, the affect of time on height, by treatment would like this.

Rplot.png

 

Let’s check the data for adherence to model assumptions.

The plot of residuals versus fitted values doesn’t reflect any obvious pattern in the residuals.So, the assumption of linearity has not been violated and looking at the ‘blob’-like nature of the plot suggests the preservation of Homoskedasticity (One day, I shall pronounce this right).

 

rplot03

We’ll check for the assumption of normality of residuals by plotting a histogram. Looks okay to me.

Rplot04.png

There are several methods for dealing with multicollinearity. In this case, we’ll follow the simplest one, which is to consult the correlation matrix from the output of summary(mod.int) from before. Since none of the correlations between predictors approach 1, we can say that the assumption of  multicollinearity has not been violated.

 

Moving on.

Bootstrapping is a great resampling technique to arrive at CIs when dealing with mixed effect models where the degree of  various output from them is not clearly known.While one would want to aim  for as high a sampling number as possible to get tighter CIs , we also need to make allowance for longer associated processing times.

So here, we want predictions for all values of height. We will create a test dataset, newdat, which will go as input to predict().

newdat <- subset(tree,treat=="C")
bb <- bootMer(mod.int, FUN=function(x)predict(x, newdat, re.form=NA),
 nsim=999)

#extract the quantiles and the fitted values.
lci <- apply(bb$t, 2, quantile, 0.025)   
uci <- apply(bb$t, 2, quantile, 0.975)   
pred <- predict(mod.int,newdat,re.form=NA)

 

Plotting the confidence intervals.

library(scales)
palette(alpha(c("blue","red","forestgreen","darkorange"),0.5))
plot(height~jitter(time),col=treat,data=tree[tree$treat=="C",],pch=16)
lines(pred~time,newdat,lwd=2,col="orange",alpha=0.5)
lines(lci~time,newdat,lty=2,col="orange")
lines(uci~time,newdat,lty=2,col="orange")

Rplot02.png

 

 

 

Conclusion

Using R and lme4 (Bates, Maechler & Bolker, 2012) We performed a linear mixed effects analysis of the relationship between height and treatment of trees, as studied over a period of time. As fixed effects, we entered time and treatment  (with an interaction term) into the model. As random effects, we had intercepts for plotnr (plot numbers). Visual inspection of residual plots did not reveal any obvious deviations from homoscedasticity or normality. P-values were obtained by likelihood ratio tests via anova of the full
model with the effect in question against the model without the effect in question.

 

 

. . ..  . and it’s a wrap,folks!

Thanks again for stopping by, and again, I’d be happy to hear your comments and suggestions. Help me make this content better! 🙂

 

 

 

Stepwise Regression – What’s not to like ?

Standard

 

Plenty, apparently.

Besides encouraging you not to think , it doesn’t exactly do a great job at what it claims to do. Given a set of predictors, there is no guarantee that stepwise regression will find the optimal combination. Many of my statisticians buddies , whom I consult from time to time,  have a  gripe with it because it’s not  sensitive to the context of the research. Seems fair.

 

I built an interactive Shiny app to evaluate results from Stepwise regression (direction = “backward) when applied to different predictors and datasets. What I observed during model building and cross-validation was that the model performed better on the data at hand but performs much worse when subjected to cross-validation.After a lot of different random selections and testing, I eventually did find a model that worked well on both the fitted dataset and the cross-validation set, but it performed poorly when applied to new data.Therein lies at least most of the problem.

 

Shiny app can be found here.

The initial model was built to predict the ‘Life Expectancy’ and it does, to a certain extent, do it’s job . But when generalized, it pretty much turned out to be a bit of an uncertainty ridden damp squib. For example, predictions for the variables from the same dataset , such as, ‘Population’ ,’Frost’, ‘Area’ are nowhere close to the observed values. At the same time, the model did okay for variables such as ‘Illiteracy’ and ‘HS.Grad’.

Given all these drawbacks ( and more! ), people do find the motivation to use stepwise regression to produce a simpler model in terms of number of coefficients. It does not necessarily find the optimal model, but it does give a hunch of the possible combination of predictors.

While no one would conclude a statistical study based on stepwise results or publish a paper with it, some might find uses for it, say, to  verify models already created by software systems. Or as an easy-to-use tool for initial exploratory data analysis (with all the necessary caveats in place !) .

You win some, you lose some.

What do you think ? Leave a comment!

p.s. You can find the (needs-to-be-cleaned-up) code for the Shiny app here.

 

 

 

Multiple Regression (sans interactions) : A case study.

Standard

Dataset:

state.x77 – Standard built-in dataset with 50 rows and 8 columns giving the following statistics in the respective columns.

Populationpopulation estimate as of July 1, 1975

Incomeper capita income (1974)

Illiteracy: illiteracy (1970, percent of population)

Life Exp: life expectancy in years (1969–71)

Murder: murder and non-negligent manslaughter rate per 100,000 population (1976)

HS Grad: percent high-school graduates (1970)

Frost: mean number of days with minimum temperature below freezing (1931–1960) in capital or large city

Area: land area in square miles

 

 

Response:

Conduct multiple regression to arrive at an optimal model which is able to predict the life expectancy .

 

 

We begin this exercise by running a few standard diagnostics on the data including by not limited to  head(), tail() , class(), str(), summary() ,names() etc.

The input dataset turns out to be a matrix which will need to be coerced into a data frame using as.data.frame().

Next,we’ll do a quick exploratory analysis on our data to examine the variables for outliers and distribution before proceeding. One way to do this is to plot qqplots for all the variables in the dataset.

for(i in 1:ncol(st)){
qqnorm(st[,i],main=colnames(st[,i]))
}

2016-09-17_10-45-58.jpg

 

It’s a good idea to check the correlation between the variables to get a sense of the dependencies . We do this here using the cor(dataframe) function.

2016-09-16_22-06-10.jpg

We can see that life expectancy appears to be moderately to strongly related to “Income” (positively), “Illiteracy” (negatively), “Murder” (negatively), “HS.Grad” (positively), and “Frost” (no. of days per year with frost, positively).

 

We can also view this by plotting our correlation matrix using pairs(dataframe)  which corroborates the cor.matrix  as seen above.

2016-09-16_22-09-47.jpg

As a starting point for multiple regression would be to build a “full” model which will have Life Expectancy as the response variable and all other variables as explanatory variables. The summary of that linear model will be our “square one” and we will proceed to whittle it down until we reach our optimal model.

> model1 = lm(Life.Exp ~ ., data=st)
> summary(model1)

Call:
lm(formula = Life.Exp ~ ., data = st)

Residuals:
 Min         1Q     Median   3Q       Max 
-1.47514 -0.45887 -0.06352 0.59362 1.21823 

Coefficients:
             Estimate Std. Error t value Pr(>|t|) 
(Intercept) 6.995e+01  1.843e+00  37.956  < 2e-16 ***
Population  6.480e-05  3.001e-05   2.159   0.0367 * 
Income      2.701e-04  3.087e-04   0.875   0.3867 
Illiteracy  3.029e-01  4.024e-01   0.753   0.4559 
Murder     -3.286e-01  4.941e-02  -6.652   5.12e-08 ***
HS.Grad     4.291e-02  2.332e-02    1.840  0.0730 . 
Frost      -4.580e-03  3.189e-03   -1.436  0.1585 
Area       -1.558e-06  1.914e-06   -0.814  0.4205 
Density    -1.105e-03  7.312e-04   -1.511  0.1385 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7337 on 41 degrees of freedom
Multiple R-squared: 0.7501, Adjusted R-squared: 0.7013 
F-statistic: 15.38 on 8 and 41 DF, p-value: 3.787e-10

We see that this 70% of the variation in Life Expectancy is explained by all the variables together (Adjusted R-squared).

Let’s see if this can be made better. We’ll try to do that by removing one predictor at a time from model1, starting with “Illiteracy” (p-value = 0.4559 i.e. least significance) , until we have a model with all significant predictors.

> model2 = update(model1, .~. -Illiteracy)
> summary(model2)

Call:
lm(formula = Life.Exp ~ Population + Income + Murder + HS.Grad + 
 Frost + Area + Density, data = st)

Residuals:
 Min 1Q Median 3Q Max 
-1.47618 -0.38592 -0.05728 0.58817 1.42334 

Coefficients:
              Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.098e+01  1.228e+00   57.825   < 2e-16 ***
Population   5.675e-05  2.789e-05     2.034    0.0483 * 
Income       1.901e-04  2.883e-04     0.659    0.5133 
Murder       -3.122e-01 4.409e-02     -7.081    1.11e-08 ***
HS.Grad       3.652e-02 2.161e-02      1.690    0.0984 . 
Frost        -6.059e-03 2.499e-03      -2.425    0.0197 * 
Area         -8.638e-07 1.669e-06      -0.518     0.6075 
Density       -8.612e-04 6.523e-04       -1.320   0.1939 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7299 on 42 degrees of freedom
Multiple R-squared: 0.7466, Adjusted R-squared: 0.7044 
F-statistic: 17.68 on 7 and 42 DF, p-value: 1.12e-10

 

Next, Area must go (p-value = 0.6075!)

 

> model3 = update(model2, .~. -Area)
> summary(model3)

Call:
lm(formula = Life.Exp ~ Population + Income + Murder + HS.Grad + 
 Frost + Density, data = st)

Residuals:
 Min 1Q Median 3Q Max 
-1.49555 -0.41246 -0.05336 0.58399 1.50535 

Coefficients:
            Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.131e+01 1.042e+00 68.420 < 2e-16 ***
Population   5.811e-05 2.753e-05 2.110 0.0407 * 
Income       1.324e-04 2.636e-04 0.502 0.6181 
Murder      -3.208e-01 4.054e-02 -7.912 6.32e-10 ***
HS.Grad      3.499e-02 2.122e-02 1.649 0.1065 
Frost       -6.191e-03 2.465e-03 -2.512 0.0158 * 
Density     -7.324e-04 5.978e-04 -1.225 0.2272 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7236 on 43 degrees of freedom
Multiple R-squared: 0.745, Adjusted R-squared: 0.7094 
F-statistic: 20.94 on 6 and 43 DF, p-value: 2.632e-11

We will continue to throw out the predictors in this fashion . . . . .

> model4 = update(model3, .~. -Income)
> summary(model4)
> model5 = update(model4, .~. -Density)
> summary(model5)

. . . .  until we reach this point

> summary(model5)

Call:
lm(formula = Life.Exp ~ Population + Murder + HS.Grad + Frost, 
 data = st)

Residuals:
 Min 1Q Median 3Q Max 
-1.47095 -0.53464 -0.03701 0.57621 1.50683 

Coefficients:
              Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.103e+01 9.529e-01 74.542   < 2e-16 ***
Population   5.014e-05 2.512e-05 1.996     0.05201 . 
Murder      -3.001e-01 3.661e-02 -8.199    1.77e-10 ***
HS.Grad      4.658e-02 1.483e-02 3.142     0.00297 ** 
Frost       -5.943e-03 2.421e-03 -2.455    0.01802 * 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7197 on 45 degrees of freedom
Multiple R-squared: 0.736, Adjusted R-squared: 0.7126 
F-statistic: 31.37 on 4 and 45 DF, p-value: 1.696e-12

Now, here we have “Population” with a borderline p-value of 0.052 . Do we throw it out ? Do we keep it ? One way to know is to test for interactions.

> interact= lm(Life.Exp ~ Population * Murder * HS.Grad * Frost,data=st)
> anova(model5,interact)
Analysis of Variance Table

Model 1: Life.Exp ~ Population + Murder + HS.Grad + Frost
Model 2: Life.Exp ~ Population * Murder * HS.Grad * Frost
 Res.Df RSS Df Sum of Sq F Pr(>F) 
1 45 23.308 
2 34 14.235 11 9.0727 1.9699 0.06422 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

P-value of 0.06422 indicates that there is no significant interaction between the terms so we can drop “Population” as it doesn’t really contribute to anything at this point.BUT, if we drop “Population”, the adj R-squared drops a little as well . So as a trade-off, we’ll just keep it. 

So there you have it,folks.Our minimal optimal model with statistically significant slopes.

 

Run model diagnostics

par(mfrow=c(2,2)) # to view all 4 plots at once
plot(model5)

2016-09-17_11-03-29.jpg

1. Residuals vs Fitted

We find equally spread residuals around a horizontal line without distinct patterns so that is a that is a good indication we don’t have non-linear relationships.

2. Normal Q-Q plot

#This plot shows if residuals are normally distributed. Our residuals more or less follow a straight line well, so that’s an encouraging sign.

3. Scale-location
This plot shows if residuals are spread equally along the ranges of predictors. This is how we can check the assumption of equal variance (homoscedasticity). It’s good that we can see a horizontal(-ish) line with equally (randomly) spread points.

4. Residuals vs Leverage

This plot helps us locate the influential cases that have an effect on the regression line.We look for subjects outside the dashed line , the Cook’s distance.The regression results will be altered if we exclude those cases.
‘Full’ model VS ‘Optimal’ model

> anova(model1,model5)
Analysis of Variance Table

Model 1: Life.Exp ~ Population + Income + Illiteracy + Murder + HS.Grad + 
 Frost + Area + Density
Model 2: Life.Exp ~ Population + Murder + HS.Grad + Frost
 Res.Df RSS Df Sum of Sq F Pr(>F)
1 41 22.068 
2 45 23.308 -4 -1.2397 0.5758 0.6818
>

The p-value of 0.6818 suggests that we haven’t lost any significance from when we started, so we’re good.

Onto some cross-validation

We will perform a 10-fold cross validation (m=10 in CVlm function).

A comparison of the cross- validation plots shows that model5 performs the best out of all other models.The lines of best fit are relatively parallel(-ish) and most big symbols are close to the lines indicating a decent accuracy of prediction.Also, the mean squared error is small , 0.6 .

 

 

cvResults <- suppressWarnings(CVlm(data=st, form.lm=model5, m=10, dots=FALSE, seed=29, legend.pos="topleft", printit=FALSE));

mean_squared_error <- attr(cvResults, 'ms')

2016-09-17_0-13-04

 

The relative importance of predictors

The relaimpo package provides measures of relative importance for each of the predictors in the model.

install.packages("relaimpo")
library(relaimpo)
 

 calc.relimp(model5,type=c("lmg","last","first","pratt"),
 rela=TRUE)
 
 # Bootstrap Measures of Relative Importance (1000 samples) 
 boot <- boot.relimp(model5, b = 1000, type = c("lmg", 
 "last", "first", "pratt"), rank = TRUE, 
 diff = TRUE, rela = TRUE)
 booteval.relimp(boot) 

# print result
 plot(booteval.relimp(boot,sort=TRUE)) # plot result

2016-09-17_0-33-51.jpg

Yup. ‘Murder’ is the most important predictor of Life Expectancy, followed by ‘HS.Grad’ and ‘Frost. ‘Population’  sits smug at the other end of the spectrum with marginal effect and importance.

Well, looks like this is as good as it is going to get with this model.

I will leave you here, folks. Please feel free to work through various datasets (as I will) to get a firm grasp on things.

Did you find this article helpful?  Can it be improved upon ? Please leave me a comment !

Until next time!