# Data Exploration & Machine Learning, Hands-on

Practical walkthroughs on machine learning, data exploration and insight finding.

# Using Correlations To Understand Your Data

Practical walkthroughs on machine learning, data exploration and finding insight.

**Resources**

**Packages Used in this Walkthrough**

**{RCurl}**- downloads https data**{psych}**- correlation matrix plotting**{caret}**- dummyVars function

A great way to explore new data is to use a pairwise correlation matrix. This will measure the correlation between every combination of your variables. It doesn’t really matter if you have an outcome (or response) variable at this point, it will compare everything against everything else.

For those not familiar with the correlation coefficient, it is simply a measure of similarity between two vectors of numbers. The measure value can range between **1** and **-1**, where **1** is perfectly correlated, -**1** is perfectly inversly correlated, and **0** is not correlated at all:

```
print(cor(1:5,1:5))
## 1
print(cor(1:5,5:1))
## -1
print(cor(1:5,c(1,2,3,4,4)))
## 0.9701425
```

To help us understand this process, let’s download the adult.data set from the UCI Machine Learning Repository. The data is from the 1994 Census and attempts to predict those with income under $50,000 a year:

```
library(RCurl) # download https data
urlfile <- 'https://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data'
x <- getURL(urlfile, ssl.verifypeer = FALSE)
adults <- read.csv(textConnection(x), header=F)
# if the above getURL command fails, try this:
# adults <-read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data', header=F)
```

We fill in the missing headers for the UCI set and cast the outcome variable `Income`

to a binary format of **1** and **0** (here I am reversing the orginal order, if it is under $50k, it is **0**, and above, **1**; this will make the final correlation-matrix plot easier to understand):

```
names(adults)=c('Age','Workclass','FinalWeight','Education','EducationNumber',
'MaritalStatus','Occupation','Relationship','Race',
'Sex','CapitalGain','CapitalLoss','HoursWeek',
'NativeCountry','Income')
adults$Income <- ifelse(adults$Income==' <=50K',0,1)
```

We load the **caret** package to dummify (see my other walkthrough) all factor variables as the `cor`

function only accepts numerical values:

```
dmy <- dummyVars(" ~ .", data = adults)
adultsTrsf <- data.frame(predict(dmy, newdata = adults))
```

We borrow two very useful functions from Stephen Turner: `cor.prob`

and `flattenSquareMatrix`

. **cor.prob** will create a correlation matrix along with *p*-values and **flattenSquareMatrix** will flatten all the combinations from the square matrix into a data frame of 4 columns made up of row names, column names, the correlation value and the *p*-value:

```
corMasterList <- flattenSquareMatrix (cor.prob(adultsTrsf))
print(head(corMasterList,10))
```

```
## i j cor p
## 1 age workclass... 0.042627 1.421e-14
## 2 age workclass..Federal.gov 0.051227 0.000e+00
## 3 workclass... workclass..Federal.gov -0.042606 1.454e-14
## 4 age workclass..Local.gov 0.060901 0.000e+00
## 5 workclass... workclass..Local.gov -0.064070 0.000e+00
## 6 workclass..Federal.gov workclass..Local.gov -0.045682 2.220e-16
## 7 age workclass..Never.worked -0.019362 4.759e-04
## 8 workclass... workclass..Never.worked -0.003585 5.178e-01
## 9 workclass..Federal.gov workclass..Never.worked -0.002556 6.447e-01
## 10 workclass..Local.gov workclass..Never.worked -0.003843 4.880e-01
```

This final format allows you to easily order the pairs however you want - for example, by those with the highest absolute correlation value:

```
corList <- corMasterList[order(-abs(corMasterList$cor)),]
print(head(corList,10))
```

```
## i j cor p
## 1953 sex..Female sex..Male -1.0000000 0
## 597 workclass... occupation... 0.9979854 0
## 1256 maritalStatus..Married.civ.spouse relationship..Husband 0.8932103 0
## 1829 race..Black race..White -0.7887475 0
## 527 maritalStatus..Married.civ.spouse maritalStatus..Never.married -0.6448661 0
## 1881 relationship..Husband sex..Female -0.5801353 0
## 1942 relationship..Husband sex..Male 0.5801353 0
## 1258 maritalStatus..Never.married relationship..Husband -0.5767295 0
## 1306 maritalStatus..Married.civ.spouse relationship..Not.in.family -0.5375883 0
## 497 age maritalStatus..Never.married -0.5343590 0
```

The top correlated pair (sex..Female and sex.Male), as seen above, won’t be of much use as they are the only two levels of the same factor. We need to process this a little further to be of practical use. We create a single vector of variable names by filtering those with an absolute correlation of 0.2 or higher against the outcome variable of `Income`

:

```
selectedSub <- subset(corList, (abs(cor) > 0.2 & j == 'Income'))
print(selectedSub)
i j cor p
## 5811 MaritalStatus..Never.married Income -0.3184403 0
## 5832 Relationship..Own.child Income -0.2285320 0
## 5840 Sex..Female Income -0.2159802 0
## 5818 Occupation..Exec.managerial Income 0.2148613 0
## 5841 Sex..Male Income 0.2159802 0
## 5842 CapitalGain Income 0.2233288 0
## 5844 HoursWeek Income 0.2296891 0
## 5779 Age Income 0.2340371 0
## 5806 Education.Number Income 0.3351540 0
## 5829 Relationship..Husband Income 0.4010353 0
## 5809 MaritalStatus..Married.civ.spouse Income 0.4446962 0
```

We save the most correlated features to the `bestSub`

variable:

```
bestSub <- as.character(selectedSub$i)
```

Finally we plot the highly correlated pairs using the **{psych}** package’s `pair.panels`

plot (this can also be done on the original data as `pair.panels`

can handle factor and character variables):

```
library(psych)
pairs.panels(adultsTrsf[c(bestSub, 'Income')])
```

The pairs plot, and in particular the last `Income`

column, tell us a lot about our data set. Being `never married`

is the most negatively correlated with income over $50,000/year and `Hours Worked`

and `Age`

are the most postively correlated.

**Things to keep in mind**

Full source code (also on GitHub):

```
library(RCurl) # download https data
urlfile <- 'https://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data'
x <- getURL(urlfile, ssl.verifypeer = FALSE)
adults <- read.csv(textConnection(x), header=F)
# if the above getURL command fails, try this:
# adults <-read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data', header=F)
names(adults)=c('Age','Workclass','FinalWeight','Education','EducationNumber',
'MaritalStatus','Occupation','Relationship','Race',
'Sex','CapitalGain','CapitalLoss','HoursWeek',
'NativeCountry','Income')
adults$Income <- ifelse(adults$Income==' <=50K',0,1)
library(caret)
dmy <- dummyVars(" ~ .", data = adults)
adultsTrsf <- data.frame(predict(dmy, newdata = adults))
## Correlation matrix with p-values. See http://goo.gl/nahmV for documentation of this function
cor.prob <- function (X, dfr = nrow(X) - 2) {
R <- cor(X, use="pairwise.complete.obs")
above <- row(R) < col(R)
r2 <- R[above]^2
Fstat <- r2 * dfr/(1 - r2)
R[above] <- 1 - pf(Fstat, 1, dfr)
R[row(R) == col(R)] <- NA
R
}
## Use this to dump the cor.prob output to a 4 column matrix
## with row/column indices, correlation, and p-value.
## See StackOverflow question: http://goo.gl/fCUcQ
flattenSquareMatrix <- function(m) {
if( (class(m) != "matrix") | (nrow(m) != ncol(m))) stop("Must be a square matrix.")
if(!identical(rownames(m), colnames(m))) stop("Row and column names must be equal.")
ut <- upper.tri(m)
data.frame(i = rownames(m)[row(m)[ut]],
j = rownames(m)[col(m)[ut]],
cor=t(m)[ut],
p=m[ut])
}
corMasterList <- flattenSquareMatrix (cor.prob(adultsTrsf))
print(head(corMasterList,10))
corList <- corMasterList[order(-abs(corMasterList$cor)),]
print(head(corList,10))
corList <- corMasterList[order(corMasterList$cor),]
selectedSub <- subset(corList, (abs(cor) > 0.2 & j == 'Income'))
# to get the best variables from original list:
bestSub <- sapply(strsplit(as.character(selectedSub$i),'[.]'), "[", 1)
bestSub <- unique(bestSub)
# or use the variables from top selectedSub:
bestSub <- as.character(selectedSub$i)
library(psych)
pairs.panels(adultsTrsf[c(bestSub, 'Income')])
```