# Multivariate Linear Regression With R

Hi ML Enthusiasts! Today, We will be learning how to perform multivariate linear regression with R. In this case-study, we will be using a very famous data set “King County House sales” in which we will treat price as the dependent variable and other variables as independent variables. We will use these independent variables to predict the price of houses. Before going further, let’s know about our variables first.

### Dataset

In this data set, we have 19 house features and the number of observations being 21613.
id: notation of the house
date: date on which the house was sold
price: price at which house was sold(target variable)
bedrooms: number of bedrooms per house
bathrooms: number of bathrooms per bedroom
sqft_living: sqft footage of the home
sqft_lot: sqft footage of the lot
floors: total floors of the house
waterfront: house which is having the waterfront view
view: has been viewed
condition: how good is the condition of the house
sqft_above: apart from basement, sqft footage of the house
yr_built: year in which the house was built
yr_renovated: year in which house was renovated
zipcode -lat: latitude coordinate
long: longitude coordinate
sqft_living15: living room area in 2015
sqft_lot15: lot size area in 2015

### Reading and Summarizing the data

Now, let’s read our csv file and store it in house dataframe and see if the data has any missing values.

house = read.csv('kc_house_data.csv')
summary(house)
##        id                         date           price
##  Min.   :1.000e+06   20140623T000000:  142   Min.   :  75000
##  1st Qu.:2.123e+09   20140625T000000:  131   1st Qu.: 321950
##  Median :3.905e+09   20140626T000000:  131   Median : 450000
##  Mean   :4.580e+09   20140708T000000:  127   Mean   : 540088
##  3rd Qu.:7.309e+09   20150427T000000:  126   3rd Qu.: 645000
##  Max.   :9.900e+09   20150325T000000:  123   Max.   :7700000
##                      (Other)        :20833
##     bedrooms        bathrooms      sqft_living       sqft_lot
##  Min.   : 0.000   Min.   :0.000   Min.   :  290   Min.   :    520
##  1st Qu.: 3.000   1st Qu.:1.750   1st Qu.: 1427   1st Qu.:   5040
##  Median : 3.000   Median :2.250   Median : 1910   Median :   7618
##  Mean   : 3.371   Mean   :2.115   Mean   : 2080   Mean   :  15107
##  3rd Qu.: 4.000   3rd Qu.:2.500   3rd Qu.: 2550   3rd Qu.:  10688
##  Max.   :33.000   Max.   :8.000   Max.   :13540   Max.   :1651359
##
##      floors        waterfront            view          condition
##  Min.   :1.000   Min.   :0.000000   Min.   :0.0000   Min.   :1.000
##  1st Qu.:1.000   1st Qu.:0.000000   1st Qu.:0.0000   1st Qu.:3.000
##  Median :1.500   Median :0.000000   Median :0.0000   Median :3.000
##  Mean   :1.494   Mean   :0.007542   Mean   :0.2343   Mean   :3.409
##  3rd Qu.:2.000   3rd Qu.:0.000000   3rd Qu.:0.0000   3rd Qu.:4.000
##  Max.   :3.500   Max.   :1.000000   Max.   :4.0000   Max.   :5.000
##
##  Min.   : 1.000   Min.   : 290   Min.   :   0.0   Min.   :1900
##  1st Qu.: 7.000   1st Qu.:1190   1st Qu.:   0.0   1st Qu.:1951
##  Median : 7.000   Median :1560   Median :   0.0   Median :1975
##  Mean   : 7.657   Mean   :1788   Mean   : 291.5   Mean   :1971
##  3rd Qu.: 8.000   3rd Qu.:2210   3rd Qu.: 560.0   3rd Qu.:1997
##  Max.   :13.000   Max.   :9410   Max.   :4820.0   Max.   :2015
##
##   yr_renovated       zipcode           lat             long
##  Min.   :   0.0   Min.   :98001   Min.   :47.16   Min.   :-122.5
##  1st Qu.:   0.0   1st Qu.:98033   1st Qu.:47.47   1st Qu.:-122.3
##  Median :   0.0   Median :98065   Median :47.57   Median :-122.2
##  Mean   :  84.4   Mean   :98078   Mean   :47.56   Mean   :-122.2
##  3rd Qu.:   0.0   3rd Qu.:98118   3rd Qu.:47.68   3rd Qu.:-122.1
##  Max.   :2015.0   Max.   :98199   Max.   :47.78   Max.   :-121.3
##
##  sqft_living15    sqft_lot15
##  Min.   : 399   Min.   :   651
##  1st Qu.:1490   1st Qu.:  5100
##  Median :1840   Median :  7620
##  Mean   :1987   Mean   : 12768
##  3rd Qu.:2360   3rd Qu.: 10083
##  Max.   :6210   Max.   :871200
## 

As can be seen from above details, there are no missing values, so there is no need to do missing value imputation. We can proceed towards our data relationship analysis. Now, let’s see how our data looks like by using the head() function.

head(house)
##           id            date   price bedrooms bathrooms sqft_living
## 1 7129300520 20141013T000000  221900        3      1.00        1180
## 2 6414100192 20141209T000000  538000        3      2.25        2570
## 3 5631500400 20150225T000000  180000        2      1.00         770
## 4 2487200875 20141209T000000  604000        4      3.00        1960
## 5 1954400510 20150218T000000  510000        3      2.00        1680
## 6 7237550310 20140512T000000 1225000        4      4.50        5420
##   sqft_lot floors waterfront view condition grade sqft_above sqft_basement
## 1     5650      1          0    0         3     7       1180             0
## 2     7242      2          0    0         3     7       2170           400
## 3    10000      1          0    0         3     6        770             0
## 4     5000      1          0    0         5     7       1050           910
## 5     8080      1          0    0         3     8       1680             0
## 6   101930      1          0    0         3    11       3890          1530
##   yr_built yr_renovated zipcode     lat     long sqft_living15 sqft_lot15
## 1     1955            0   98178 47.5112 -122.257          1340       5650
## 2     1951         1991   98125 47.7210 -122.319          1690       7639
## 3     1933            0   98028 47.7379 -122.233          2720       8062
## 4     1965            0   98136 47.5208 -122.393          1360       5000
## 5     1987            0   98074 47.6168 -122.045          1800       7503
## 6     2001            0   98053 47.6561 -122.005          4760     101930

### Datetime manipulation in R

The date variable needs our attention here. It’s in incorrect format. Let’s convert it in a UTC format using lubridate library.

#The date may not be in character data type. So, we use as.character()
#in this case andsubset the date string with start=1 and stop = 8
house$date <- as.character(house$date)
house$date <- substr(house$date, 1, 8)
#Convert the date in year-month-date format
house$date <- as.Date(house$date, , format='%Y%m%d')
#Converting date into number of days with 1st jan 1900 as reference date
house$date <- as.numeric(as.Date(house$date, origin = "1900-01-01"))
head(house)
##           id  date   price bedrooms bathrooms sqft_living sqft_lot floors
## 1 7129300520 16356  221900        3      1.00        1180     5650      1
## 2 6414100192 16413  538000        3      2.25        2570     7242      2
## 3 5631500400 16491  180000        2      1.00         770    10000      1
## 4 2487200875 16413  604000        4      3.00        1960     5000      1
## 5 1954400510 16484  510000        3      2.00        1680     8080      1
## 6 7237550310 16202 1225000        4      4.50        5420   101930      1
##   waterfront view condition grade sqft_above sqft_basement yr_built
## 1          0    0         3     7       1180             0     1955
## 2          0    0         3     7       2170           400     1951
## 3          0    0         3     6        770             0     1933
## 4          0    0         5     7       1050           910     1965
## 5          0    0         3     8       1680             0     1987
## 6          0    0         3    11       3890          1530     2001
##   yr_renovated zipcode     lat     long sqft_living15 sqft_lot15
## 1            0   98178 47.5112 -122.257          1340       5650
## 2         1991   98125 47.7210 -122.319          1690       7639
## 3            0   98028 47.7379 -122.233          2720       8062
## 4            0   98136 47.5208 -122.393          1360       5000
## 5            0   98074 47.6168 -122.045          1800       7503
## 6            0   98053 47.6561 -122.005          4760     101930

### Splitting data into training and test data

Our next step is to split the dataset into training and test data set. We do this by using the sample() function of R. We fix the ratio of training to test data as 80:20. So, we convert 0.2xdataset into test data and use the remaining, i.e., 0.8xdataset as training dataset.

splitRatio = sample(1:nrow(house), 0.2*nrow(house))
testHouseData = house[splitRatio,]
trainingHouseData = house[-splitRatio,]

### Correlation Analysis

Now comes the visualization part. Let’s look at how different variables are connected to price variable.

library(ggplot2)
library(GGally)
relPlot1 <- ggpairs(data = trainingHouseData,
columns = 3:7,
mapping = aes(color = "dark green"),
axisLabels = "show")
relPlot1

One thing to be noted above is that all the columns from 3 to 7 are having continuous numerical data. So, the columns = 3:7 has been fixed. After 7th column, all the columns from 8 to 12 are categorical in nature. So, we use columns = c(3, 8:12) with 3rd column(price) having continuous numerical data.

relPlot2 <- ggpairs(data = trainingHouseData,
columns = c(3, 8:12),
mapping = aes(color = "dark green"),
axisLabels = "show")
relPlot2

Let’s verify the relationship between columns 3, 15, 18 and 19.

relPlot3 <- ggpairs(data = trainingHouseData,
columns = c(3, 15, 18, 19),
mapping = aes(color = "dark green"),
axisLabels = "show")
relPlot3

As you can see from the above plots, -corr between price vs sqft_living: 0.701 -corr between price vs bathrooms: 0.524 -corr between price vs grade: 0.654 -corr between price vs view: 0.406 -corr between price vs waterfront: 0.324 -corr between price vs lat: 0.304 -corr between price vs bedrooms: 0.303 -corr between price vs floors: 0.282 The correlation values between price and other variables are very small so that can be ignored as they are not posing any impact on the prices of the houses.

### Multivariate linear regression with R: Checking multi-collinearity

Let’s now check the multi-collinearity among the variables using variance inflation factor concept.

library(car)
## Loading required package: carData
finalModel <- lm(price~sqft_living + bathrooms + grade + view + waterfront + bedrooms + floors, data=trainingHouseData)
vif(finalModel)
## sqft_living   bathrooms       grade        view  waterfront    bedrooms
##    3.906712    2.858481    2.733885    1.321960    1.214251    1.609194
##      floors
##    1.451042

Since all the vif values are less than 5, we can say that the model doesn’t have multicollinearity among the variables. So, we can consider all the variables of the said model.

step(finalModel)
## Start:  AIC=427290.6
## price ~ sqft_living + bathrooms + grade + view + waterfront +
##     bedrooms + floors
##
##               Df  Sum of Sq        RSS    AIC
##                       9.3237e+14 427291
## - bathrooms    1 7.6493e+11 9.3314e+14 427303
## - floors       1 2.0967e+12 9.3447e+14 427327
## - bedrooms     1 7.3436e+12 9.3971e+14 427424
## - view         1 3.5588e+13 9.6796e+14 427936
## - waterfront   1 3.6058e+13 9.6843e+14 427945
## - grade        1 8.4533e+13 1.0169e+15 428789
## - sqft_living  1 1.3231e+14 1.0647e+15 429583
##
## Call:
## lm(formula = price ~ sqft_living + bathrooms + grade + view +
##     waterfront + bedrooms + floors, data = trainingHouseData)
##
## Coefficients:
## (Intercept)  sqft_living    bathrooms        grade         view
##   -468540.7        189.5     -14629.0      98753.5      68512.7
##  waterfront     bedrooms       floors
##    561511.0     -27997.1     -24612.7

### Multivariate linear regression with R: Final model obtained

According to the above stats, our finalModel equation is price = 191.1(sqft_living) – 13159.1(bathrooms) + 98298.6(grade)+66911.4(view) +580801.6(waterfront) – 27495.8(bedrooms) – 27293.2(floors) – 467519.7. Let’s find out what is the value of R-square of this model by using summary() function.

summary(finalModel)
##
## Call:
## lm(formula = price ~ sqft_living + bathrooms + grade + view +
##     waterfront + bedrooms + floors, data = trainingHouseData)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -1186479  -125387   -18824    95760  4742318
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4.685e+05  1.553e+04 -30.164  < 2e-16 ***
## sqft_living  1.895e+02  3.826e+00  49.523  < 2e-16 ***
## bathrooms   -1.463e+04  3.885e+03  -3.766 0.000167 ***
## grade        9.875e+04  2.495e+03  39.585  < 2e-16 ***
## view         6.851e+04  2.668e+03  25.684  < 2e-16 ***
## waterfront   5.615e+05  2.172e+04  25.853  < 2e-16 ***
## bedrooms    -2.800e+04  2.400e+03 -11.667  < 2e-16 ***
## floors      -2.461e+04  3.948e+03  -6.234 4.64e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 232300 on 17283 degrees of freedom
## Multiple R-squared:  0.5901, Adjusted R-squared:  0.5899
## F-statistic:  3554 on 7 and 17283 DF,  p-value: < 2.2e-16

### R-square obtained by multivariate linear regression with R

From above summary, we can say R-square is 59.72% or about 59.72% of variation in prices is explained by our chosen variables. Let’s work on increasing the accuracy of model by eliminating bedrooms and floors.

finalModel2 <- lm(price~sqft_living + bathrooms + grade + view + waterfront, data=trainingHouseData)
vif(finalModel2)
## sqft_living   bathrooms       grade        view  waterfront
##    3.279054    2.447752    2.482267    1.303474    1.210474

Again, all the vif values are less than 5. So, we incorporate all the variables.

step(finalModel2)
## Start:  AIC=427451.8
## price ~ sqft_living + bathrooms + grade + view + waterfront
##
##               Df  Sum of Sq        RSS    AIC
##                       9.4132e+14 427452
## - bathrooms    1 3.8429e+12 9.4516e+14 427520
## - waterfront   1 3.7003e+13 9.7832e+14 428116
## - view         1 3.9994e+13 9.8131e+14 428169
## - grade        1 9.5435e+13 1.0368e+15 429120
## - sqft_living  1 1.3552e+14 1.0768e+15 429775
##
## Call:
## lm(formula = price ~ sqft_living + bathrooms + grade + view +
##     waterfront, data = trainingHouseData)
##
## Coefficients:
## (Intercept)  sqft_living    bathrooms        grade         view
##   -548057.1        175.7     -30342.4      99983.0      72120.9
##  waterfront
##    567935.6

The equation in this case becomes price = 178.1(sqft_living) – 29722.6(bathrooms) + 98868.0(grade) + 70679.4(view) +588863.0(waterfront)-544112.4. Let’s now find out the R square of this model.

summary(finalModel2)
##
## Call:
## lm(formula = price ~ sqft_living + bathrooms + grade + view +
##     waterfront, data = trainingHouseData)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -1200221  -125937   -18891    97404  4857478
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.481e+05  1.390e+04  -39.44   <2e-16 ***
## sqft_living  1.757e+02  3.522e+00   49.88   <2e-16 ***
## bathrooms   -3.034e+04  3.612e+03   -8.40   <2e-16 ***
## grade        9.998e+04  2.388e+03   41.86   <2e-16 ***
## view         7.212e+04  2.661e+03   27.10   <2e-16 ***
## waterfront   5.679e+05  2.179e+04   26.07   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 233400 on 17285 degrees of freedom
## Multiple R-squared:  0.5861, Adjusted R-squared:  0.586
## F-statistic:  4896 on 5 and 17285 DF,  p-value: < 2.2e-16

As you can see, the R square is nearly the same but our model has less variable dependency now. Also, since the P value of all the variables is less than 0.05, we reject the null hypothesis that there is no relationship among a particular variable and price and accept the alternate hypothesis that the price is dependent upon all the given variables.

### Checking if assumptions of linear regression were satisfied

Now, let’s see if the assumptions of linear regression are satisfied or not.

library(lmtest)
## Loading required package: zoo
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
##     as.Date, as.Date.numeric
par(mfrow=c(2,2))
plot(finalModel2)

According to the assumptions of linear regression, these graphs must be uniformly distributed but that is not so here. So, we need to improve our model. Let’s now use our second model to predict the price for test dataset.

### Multivariate linear regression with R – Predictions

The distribution is uniform in nature which is another positive indication. Now, let’s do the prediction.

testHouseData\$predictedPrice = predict(finalModel3, testHouseData)