605 House Prices: Advanced Regression
Yuen Chun Wong
May 15, 2018
Data Source: https://www.kaggle.com/c/house-prices-advanced-regression-techniques
#Download required packages
suppressWarnings(suppressMessages(library(RCurl)))
suppressWarnings(suppressMessages(library(dplyr)))
suppressWarnings(suppressMessages(library(XML)))
suppressWarnings(suppressMessages(library(Hmisc))) # correlation
suppressWarnings(suppressMessages(library(ggplot2)))
suppressWarnings(suppressMessages(library(Matrix)))The Data structure
Load data and check the data structure
#Prepare raw data sets: read in csv or txt files into R
housing_data <- read.csv(file="C:/Users/johnw/Dropbox/MSDA/DATA605/final/train.csv", header=TRUE, sep=",")
summary(housing_data)## Id MSSubClass MSZoning LotFrontage
## Min. : 1.0 Min. : 20.0 C (all): 10 Min. : 21.00
## 1st Qu.: 365.8 1st Qu.: 20.0 FV : 65 1st Qu.: 59.00
## Median : 730.5 Median : 50.0 RH : 16 Median : 69.00
## Mean : 730.5 Mean : 56.9 RL :1151 Mean : 70.05
## 3rd Qu.:1095.2 3rd Qu.: 70.0 RM : 218 3rd Qu.: 80.00
## Max. :1460.0 Max. :190.0 Max. :313.00
## NA's :259
## LotArea Street Alley LotShape LandContour
## Min. : 1300 Grvl: 6 Grvl: 50 IR1:484 Bnk: 63
## 1st Qu.: 7554 Pave:1454 Pave: 41 IR2: 41 HLS: 50
## Median : 9478 NA's:1369 IR3: 10 Low: 36
## Mean : 10517 Reg:925 Lvl:1311
## 3rd Qu.: 11602
## Max. :215245
##
## Utilities LotConfig LandSlope Neighborhood Condition1
## AllPub:1459 Corner : 263 Gtl:1382 NAmes :225 Norm :1260
## NoSeWa: 1 CulDSac: 94 Mod: 65 CollgCr:150 Feedr : 81
## FR2 : 47 Sev: 13 OldTown:113 Artery : 48
## FR3 : 4 Edwards:100 RRAn : 26
## Inside :1052 Somerst: 86 PosN : 19
## Gilbert: 79 RRAe : 11
## (Other):707 (Other): 15
## Condition2 BldgType HouseStyle OverallQual
## Norm :1445 1Fam :1220 1Story :726 Min. : 1.000
## Feedr : 6 2fmCon: 31 2Story :445 1st Qu.: 5.000
## Artery : 2 Duplex: 52 1.5Fin :154 Median : 6.000
## PosN : 2 Twnhs : 43 SLvl : 65 Mean : 6.099
## RRNn : 2 TwnhsE: 114 SFoyer : 37 3rd Qu.: 7.000
## PosA : 1 1.5Unf : 14 Max. :10.000
## (Other): 2 (Other): 19
## OverallCond YearBuilt YearRemodAdd RoofStyle
## Min. :1.000 Min. :1872 Min. :1950 Flat : 13
## 1st Qu.:5.000 1st Qu.:1954 1st Qu.:1967 Gable :1141
## Median :5.000 Median :1973 Median :1994 Gambrel: 11
## Mean :5.575 Mean :1971 Mean :1985 Hip : 286
## 3rd Qu.:6.000 3rd Qu.:2000 3rd Qu.:2004 Mansard: 7
## Max. :9.000 Max. :2010 Max. :2010 Shed : 2
##
## RoofMatl Exterior1st Exterior2nd MasVnrType MasVnrArea
## CompShg:1434 VinylSd:515 VinylSd:504 BrkCmn : 15 Min. : 0.0
## Tar&Grv: 11 HdBoard:222 MetalSd:214 BrkFace:445 1st Qu.: 0.0
## WdShngl: 6 MetalSd:220 HdBoard:207 None :864 Median : 0.0
## WdShake: 5 Wd Sdng:206 Wd Sdng:197 Stone :128 Mean : 103.7
## ClyTile: 1 Plywood:108 Plywood:142 NA's : 8 3rd Qu.: 166.0
## Membran: 1 CemntBd: 61 CmentBd: 60 Max. :1600.0
## (Other): 2 (Other):128 (Other):136 NA's :8
## ExterQual ExterCond Foundation BsmtQual BsmtCond BsmtExposure
## Ex: 52 Ex: 3 BrkTil:146 Ex :121 Fa : 45 Av :221
## Fa: 14 Fa: 28 CBlock:634 Fa : 35 Gd : 65 Gd :134
## Gd:488 Gd: 146 PConc :647 Gd :618 Po : 2 Mn :114
## TA:906 Po: 1 Slab : 24 TA :649 TA :1311 No :953
## TA:1282 Stone : 6 NA's: 37 NA's: 37 NA's: 38
## Wood : 3
##
## BsmtFinType1 BsmtFinSF1 BsmtFinType2 BsmtFinSF2
## ALQ :220 Min. : 0.0 ALQ : 19 Min. : 0.00
## BLQ :148 1st Qu.: 0.0 BLQ : 33 1st Qu.: 0.00
## GLQ :418 Median : 383.5 GLQ : 14 Median : 0.00
## LwQ : 74 Mean : 443.6 LwQ : 46 Mean : 46.55
## Rec :133 3rd Qu.: 712.2 Rec : 54 3rd Qu.: 0.00
## Unf :430 Max. :5644.0 Unf :1256 Max. :1474.00
## NA's: 37 NA's: 38
## BsmtUnfSF TotalBsmtSF Heating HeatingQC CentralAir
## Min. : 0.0 Min. : 0.0 Floor: 1 Ex:741 N: 95
## 1st Qu.: 223.0 1st Qu.: 795.8 GasA :1428 Fa: 49 Y:1365
## Median : 477.5 Median : 991.5 GasW : 18 Gd:241
## Mean : 567.2 Mean :1057.4 Grav : 7 Po: 1
## 3rd Qu.: 808.0 3rd Qu.:1298.2 OthW : 2 TA:428
## Max. :2336.0 Max. :6110.0 Wall : 4
##
## Electrical X1stFlrSF X2ndFlrSF LowQualFinSF
## FuseA: 94 Min. : 334 Min. : 0 Min. : 0.000
## FuseF: 27 1st Qu.: 882 1st Qu.: 0 1st Qu.: 0.000
## FuseP: 3 Median :1087 Median : 0 Median : 0.000
## Mix : 1 Mean :1163 Mean : 347 Mean : 5.845
## SBrkr:1334 3rd Qu.:1391 3rd Qu.: 728 3rd Qu.: 0.000
## NA's : 1 Max. :4692 Max. :2065 Max. :572.000
##
## GrLivArea BsmtFullBath BsmtHalfBath FullBath
## Min. : 334 Min. :0.0000 Min. :0.00000 Min. :0.000
## 1st Qu.:1130 1st Qu.:0.0000 1st Qu.:0.00000 1st Qu.:1.000
## Median :1464 Median :0.0000 Median :0.00000 Median :2.000
## Mean :1515 Mean :0.4253 Mean :0.05753 Mean :1.565
## 3rd Qu.:1777 3rd Qu.:1.0000 3rd Qu.:0.00000 3rd Qu.:2.000
## Max. :5642 Max. :3.0000 Max. :2.00000 Max. :3.000
##
## HalfBath BedroomAbvGr KitchenAbvGr KitchenQual
## Min. :0.0000 Min. :0.000 Min. :0.000 Ex:100
## 1st Qu.:0.0000 1st Qu.:2.000 1st Qu.:1.000 Fa: 39
## Median :0.0000 Median :3.000 Median :1.000 Gd:586
## Mean :0.3829 Mean :2.866 Mean :1.047 TA:735
## 3rd Qu.:1.0000 3rd Qu.:3.000 3rd Qu.:1.000
## Max. :2.0000 Max. :8.000 Max. :3.000
##
## TotRmsAbvGrd Functional Fireplaces FireplaceQu GarageType
## Min. : 2.000 Maj1: 14 Min. :0.000 Ex : 24 2Types : 6
## 1st Qu.: 5.000 Maj2: 5 1st Qu.:0.000 Fa : 33 Attchd :870
## Median : 6.000 Min1: 31 Median :1.000 Gd :380 Basment: 19
## Mean : 6.518 Min2: 34 Mean :0.613 Po : 20 BuiltIn: 88
## 3rd Qu.: 7.000 Mod : 15 3rd Qu.:1.000 TA :313 CarPort: 9
## Max. :14.000 Sev : 1 Max. :3.000 NA's:690 Detchd :387
## Typ :1360 NA's : 81
## GarageYrBlt GarageFinish GarageCars GarageArea GarageQual
## Min. :1900 Fin :352 Min. :0.000 Min. : 0.0 Ex : 3
## 1st Qu.:1961 RFn :422 1st Qu.:1.000 1st Qu.: 334.5 Fa : 48
## Median :1980 Unf :605 Median :2.000 Median : 480.0 Gd : 14
## Mean :1979 NA's: 81 Mean :1.767 Mean : 473.0 Po : 3
## 3rd Qu.:2002 3rd Qu.:2.000 3rd Qu.: 576.0 TA :1311
## Max. :2010 Max. :4.000 Max. :1418.0 NA's: 81
## NA's :81
## GarageCond PavedDrive WoodDeckSF OpenPorchSF EnclosedPorch
## Ex : 2 N: 90 Min. : 0.00 Min. : 0.00 Min. : 0.00
## Fa : 35 P: 30 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.00
## Gd : 9 Y:1340 Median : 0.00 Median : 25.00 Median : 0.00
## Po : 7 Mean : 94.24 Mean : 46.66 Mean : 21.95
## TA :1326 3rd Qu.:168.00 3rd Qu.: 68.00 3rd Qu.: 0.00
## NA's: 81 Max. :857.00 Max. :547.00 Max. :552.00
##
## X3SsnPorch ScreenPorch PoolArea PoolQC
## Min. : 0.00 Min. : 0.00 Min. : 0.000 Ex : 2
## 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.000 Fa : 2
## Median : 0.00 Median : 0.00 Median : 0.000 Gd : 3
## Mean : 3.41 Mean : 15.06 Mean : 2.759 NA's:1453
## 3rd Qu.: 0.00 3rd Qu.: 0.00 3rd Qu.: 0.000
## Max. :508.00 Max. :480.00 Max. :738.000
##
## Fence MiscFeature MiscVal MoSold
## GdPrv: 59 Gar2: 2 Min. : 0.00 Min. : 1.000
## GdWo : 54 Othr: 2 1st Qu.: 0.00 1st Qu.: 5.000
## MnPrv: 157 Shed: 49 Median : 0.00 Median : 6.000
## MnWw : 11 TenC: 1 Mean : 43.49 Mean : 6.322
## NA's :1179 NA's:1406 3rd Qu.: 0.00 3rd Qu.: 8.000
## Max. :15500.00 Max. :12.000
##
## YrSold SaleType SaleCondition SalePrice
## Min. :2006 WD :1267 Abnorml: 101 Min. : 34900
## 1st Qu.:2007 New : 122 AdjLand: 4 1st Qu.:129975
## Median :2008 COD : 43 Alloca : 12 Median :163000
## Mean :2008 ConLD : 9 Family : 20 Mean :180921
## 3rd Qu.:2009 ConLI : 5 Normal :1198 3rd Qu.:214000
## Max. :2010 ConLw : 5 Partial: 125 Max. :755000
## (Other): 9Each house is independant from each other. Ground Living area is the most major concern to buyer of house. Sales price and Ground Living Area.
X = Ground living rea Y = Sales price
X = housing_data$GrLivArea
Y = housing_data$SalePricecheck skewness of X skewed at right
library(moments)
label_X = paste("The skewness of X: ", skewness(X)) #get skewness
hist(X, main=label_X)
The data is skewed to the right. https://www.r-bloggers.com/measures-of-skewness-and-kurtosis/
#let x be housing_data$GrLivArea
#histogram :check distribution of x and skewed to the right
hist(housing_data$GrLivArea)
X <- housing_data$GrLivAreaProbability
Calculate as a minimum the below probabilities a through c. Assume the small letter “x” is estimated as the 1st quartile of the X variable, and the small letter “y” is estimated as the 1st quartile of the Y variable. Interpret the meaning of all probabilities. In addition, make a table of counts as shown below.
#first quantile of X and Y
x <- unname(quantile(X))[2]
y <- unname(quantile(Y))[2]
paste("The first quantile of X:", x)## [1] "The first quantile of X: 1129.5"paste("The first quantile of Y:", y)## [1] "The first quantile of Y: 129975"- P(X>x | Y>y): The probablity of X where ground living area is larger than first quantile given the probability of Y where sales price is larger than first quantitle.
# subset of housing data where Y > y
subset_Y_gt_y = subset( housing_data, housing_data$SalePrice > y)
subset_X_gt_x_where_Y_gt_y = subset(subset_Y_gt_y, subset_Y_gt_y$GrLivArea > x)
pct_X_gt_x_where_Y_gt_y = nrow(subset_X_gt_x_where_Y_gt_y)/nrow(housing_data)
paste(" P(X>x | Y>y): ", pct_X_gt_x_where_Y_gt_y)## [1] " P(X>x | Y>y): 0.653424657534247"- P(X>x, Y>y): The probablity of X and Y where ground living area and sales price are larger than first quantile.
pct_X_gt_x_and_Y_gt_y = nrow(subset(housing_data, (housing_data$SalePrice >y) & (housing_data$GrLivArea >x)))/nrow(housing_data)
paste("P(X>x, Y>y)",pct_X_gt_x_and_Y_gt_y
)## [1] "P(X>x, Y>y) 0.653424657534247"- P(X
y): The probablity of X where ground living area is lesser than first quanile given the probability of Y where sales price is larger than first quantitle.
subset_X_lt_x_where_Y_gt_y = subset(subset_Y_gt_y, subset_Y_gt_y$GrLivArea <= x)
pct_X_lt_x_where_Y_gt_y =nrow(subset_X_lt_x_where_Y_gt_y)/nrow(housing_data)
paste(" P(X<x | Y>y): ", pct_X_lt_x_where_Y_gt_y)## [1] " P(X<x | Y>y): 0.0965753424657534"For each case in the table:
P(X
P(X>x, Y<=y)
P(X>x, Y>y)
pct_X_lte_x_and_Y_lte_y = nrow(subset(housing_data, housing_data$GrLivArea <= x & housing_data$SalePrice <=y)) / nrow(housing_data)
pct_X_lte_x_and_Y_gt_y = nrow(subset(housing_data, housing_data$GrLivArea <= x & housing_data$SalePrice >y)) / nrow(housing_data)
pct_X_gt_x_and_Y_lte_y = nrow(subset(housing_data, housing_data$GrLivArea > x & housing_data$SalePrice <=y)) / nrow(housing_data)
pct_X_gt_x_and_Y_gt_y = nrow(subset(housing_data, housing_data$GrLivArea > x & housing_data$SalePrice >y)) / nrow(housing_data)
#Creat and fill the table
cross_table <- matrix(
c(pct_X_lte_x_and_Y_lte_y,pct_X_lte_x_and_Y_gt_y, (pct_X_lte_x_and_Y_lte_y+ pct_X_lte_x_and_Y_gt_y) , pct_X_gt_x_and_Y_lte_y,pct_X_gt_x_and_Y_gt_y,(pct_X_gt_x_and_Y_lte_y+pct_X_gt_x_and_Y_gt_y),
(pct_X_lte_x_and_Y_lte_y + pct_X_gt_x_and_Y_lte_y),(pct_X_lte_x_and_Y_gt_y + pct_X_gt_x_and_Y_gt_y),
(pct_X_lte_x_and_Y_lte_y + pct_X_lte_x_and_Y_gt_y + pct_X_gt_x_and_Y_lte_y + pct_X_gt_x_and_Y_gt_y )
),ncol=3,byrow=TRUE)
colnames(cross_table) <- c("<=1st quartile",">1st quartile","Total")
rownames(cross_table) <- c("<= 1st q",">1st q","Total")
cross_table <- as.table(cross_table)
cross_table## <=1st quartile >1st quartile Total
## <= 1st q 0.15342466 0.09657534 0.25000000
## >1st q 0.09657534 0.65342466 0.75000000
## Total 0.25000000 0.75000000 1.00000000Does splitting the training data in this fashion make them independent? Let A be the new variable counting those observations above the 1st quartile for X, and let B be the new variable counting those observations above the 1st quartile for Y. Does P(AB)=P(A)P(B)? Check mathematically, and then evaluate by running a Chi Square test for association.
P_A = nrow(subset(housing_data, housing_data$GrLivArea > x)) / nrow(housing_data)
P_B = nrow(subset(housing_data, housing_data$SalePrice > y)) / nrow(housing_data)
paste("P(A)P(B)", P_A*P_B)## [1] "P(A)P(B) 0.5625"P(AB) is 0.65 while to P(A)P(B) = 0.5625. P(AB) does not equal to P(A)P(B)
Chi-square test: Null hypothesis(H0): the X and Y variables are independent. Alternative hypothesis(H1): X and Y variables are dependent
housing_data_grLvingarea_salesPrice <- subset(housing_data, select=c("GrLivArea", "SalePrice"))
chisq.test(housing_data_grLvingarea_salesPrice)##
## Pearson's Chi-squared test
##
## data: housing_data_grLvingarea_salesPrice
## X-squared = 200960, df = 1459, p-value < 2.2e-16Since the p-value is very small, we can reject the H0. And X and Y variable are dependent.
Descriptive and Inferential Statistics
Provide univariate descriptive statistics and appropriate plots for the training data set. Provide a scatterplot of X and Y.
plot(X, Y, main="Housing Data",
xlab="GrLivArea", ylab="Sale Price", pch=19)
Derive a correlation matrix for any THREE quantitative variables in the dataset.
TotalBsmtSF: Total square feet of basement area GrLivArea: Above grade (ground) living area square feet SalePrice: Sale Price
https://www.r-bloggers.com/create-a-correlation-matrix-in-r/
housing_data_totalBasementSF_GrLivingArea_SalesPrice = subset(housing_data, select=c("TotalBsmtSF","GrLivArea", "SalePrice" ))
corr_matrix = cor(housing_data_totalBasementSF_GrLivingArea_SalesPrice)
round(corr_matrix,2)## TotalBsmtSF GrLivArea SalePrice
## TotalBsmtSF 1.00 0.45 0.61
## GrLivArea 0.45 1.00 0.71
## SalePrice 0.61 0.71 1.00Test the hypotheses that the correlations between each pairwise set of variables is 0 and provide a 92% confidence interval.
https://www.statmethods.net/stats/correlations.html
#Hypotheses test in 92% confidence interval
#H0: correlations between TotalBsmtSF and GrLivArea set of variables = 0.
#H1: correlations between TotalBsmtSF and GrLivArea set of variables <> 0.
cor.test(~TotalBsmtSF+GrLivArea , data = housing_data_totalBasementSF_GrLivingArea_SalesPrice,conf.level=0.92 )##
## Pearson's product-moment correlation
##
## data: TotalBsmtSF and GrLivArea
## t = 19.503, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 92 percent confidence interval:
## 0.4177447 0.4904754
## sample estimates:
## cor
## 0.4548682#Hypotheses test in 92% confidence interval
#H0: correlations between TotalBsmtSF and SalePrice set of variables = 0.
#H1: correlations between TotalBsmtSF and SalePrice set of variables <> 0.
cor.test(~TotalBsmtSF+SalePrice , data = housing_data_totalBasementSF_GrLivingArea_SalesPrice,conf.level=0.92 )##
## Pearson's product-moment correlation
##
## data: TotalBsmtSF and SalePrice
## t = 29.671, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 92 percent confidence interval:
## 0.5841762 0.6413763
## sample estimates:
## cor
## 0.6135806#Hypotheses test in 92% confidence interval
#H0: correlations between GrLivArea and SalePrice set of variables = 0.
#H1: correlations between GrLivArea and SalePrice set of variables <> 0.
cor.test(~GrLivArea+SalePrice , data = housing_data_totalBasementSF_GrLivingArea_SalesPrice,conf.level=0.92 )##
## Pearson's product-moment correlation
##
## data: GrLivArea and SalePrice
## t = 38.348, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 92 percent confidence interval:
## 0.6850407 0.7307245
## sample estimates:
## cor
## 0.7086245Discuss the meaning of your analysis. Would you be worried about familywise error? Why or why not?
From all the p-values for correlation tests, they are all very small and we can reject the null hypothese.
I am not worry about the familywise error because these variables are all realted to each other.
Linear Algebra and Correlation
Invert your 3 x 3 correlation matrix from above. (This is known as the precision matrix and contains variance inflation factors on the diagonal.) Multiply the correlation matrix by the precision matrix, and then multiply the precision matrix by the correlation matrix. Conduct LU decomposition on the matrix.
https://www.statmethods.net/advstats/matrix.html
#precision matrix = Invert corr matrix
paste("Precision Matrix")## [1] "Precision Matrix"round(solve(corr_matrix),4 )## TotalBsmtSF GrLivArea SalePrice
## TotalBsmtSF 1.6059 -0.0647 -0.9395
## GrLivArea -0.0647 2.0112 -1.3855
## SalePrice -0.9395 -1.3855 2.5582paste("Multiply the correlation matrix by the precision matrix")## [1] "Multiply the correlation matrix by the precision matrix"round(corr_matrix %*% solve(corr_matrix) ,4)## TotalBsmtSF GrLivArea SalePrice
## TotalBsmtSF 1 0 0
## GrLivArea 0 1 0
## SalePrice 0 0 1paste("Multiply the correlation matrix by the precision matrix, and then multiply the precision matrix by the correlation matrix")## [1] "Multiply the correlation matrix by the precision matrix, and then multiply the precision matrix by the correlation matrix"round(corr_matrix %*% solve(corr_matrix) %*% corr_matrix ,4)## TotalBsmtSF GrLivArea SalePrice
## TotalBsmtSF 1.0000 0.4549 0.6136
## GrLivArea 0.4549 1.0000 0.7086
## SalePrice 0.6136 0.7086 1.0000paste("LU decomposition")## [1] "LU decomposition"lum <- lu(corr_matrix)paste("Lower triangle")## [1] "Lower triangle"#as.data.frame(as.table(lum$L))
paste("Upper triangle")## [1] "Upper triangle"#as.data.frame(as.table(lum$U))Calculus-Based Probability & Statistics
Many times, it makes sense to fit a closed form distribution to data. For the first variable that you selected which is skewed to the right, shift it so that the minimum value is above zero as necessary. Then load the MASS package and run fitdistr to fit an exponential probability density function. (See https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/fitdistr.html ). Find the optimal value of λ for this distribution, and then take 1000 samples from this exponential distribution using this value (e.g., rexp(1000, λ)). Plot a histogram and compare it with a histogram of your original variable. Using the exponential pdf, find the 5th and 95th percentiles using the cumulative distribution function (CDF). Also generate a 95% confidence interval from the empirical data, assuming normality. Finally, provide the empirical 5th percentile and 95th percentile of the data. Discuss.
#load MASS
suppressWarnings(suppressMessages(library(MASS)))
#fitting an exponential probability density function
exp_x = fitdistr(X, "exponential")
lamda = exp_x$estimate
sample_exp_x = rexp(1000,lamda)
hist(sample_exp_x)
Modeling
Build some type of multiple regression model and submit your model to the competition board. Provide your complete model summary and results with analysis. Report your Kaggle.com user name and score.
For most of the house buyer, beside location, the size of the house and number of bathroom are the considered the most.
Below are the multiple regression model with living area and bathrooms
lm_multi_regression <- lm(SalePrice~GrLivArea + TotalBsmtSF + FullBath + HalfBath, data = housing_data)
summary(lm_multi_regression)##
## Call:
## lm(formula = SalePrice ~ GrLivArea + TotalBsmtSF + FullBath +
## HalfBath, data = housing_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -629227 -21306 -1461 18224 272570
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -27199.116 4412.919 -6.164 9.19e-10 ***
## GrLivArea 53.277 3.757 14.182 < 2e-16 ***
## TotalBsmtSF 71.975 3.353 21.465 < 2e-16 ***
## FullBath 27706.831 2974.310 9.315 < 2e-16 ***
## HalfBath 20660.030 2906.519 7.108 1.84e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 47830 on 1455 degrees of freedom
## Multiple R-squared: 0.6385, Adjusted R-squared: 0.6375
## F-statistic: 642.4 on 4 and 1455 DF, p-value: < 2.2e-16All the p-values are very small, so that we know they have strong correlation for sales price.
estimate_sales_price <- function ( g,b,fb,hb ) {
if (is.na(b)) { b = 0}
return ((-27199.116 + 53.277*g + 71.975*b + 27706.831*fb + 20660.030*hb))
}
housing_data_test <- read.csv(file="C:/Users/johnw/Dropbox/MSDA/DATA605/final/test.csv", header=TRUE, sep=",")
housing_data_test$SalePrice <- housing_data_test$Id #use id as place holder
for(row in 1:nrow(housing_data_test)) {
housing_data_test[row,]$SalePrice <- estimate_sales_price( housing_data_test[row,]$GrLivArea, housing_data_test[row,]$TotalBsmtSF, housing_data_test[row,]$FullBath, housing_data_test[row,]$HalfBath)
}
predict_id_sales_price <- subset(housing_data_test, select=c("Id", "SalePrice"))
write.csv(predict_id_sales_price, "C:/Users/johnw/Dropbox/MSDA/DATA605/final/predict_saleprice.csv")Kaggle submission.
Record ID of dataframe in the submission file is removed manually before uploading.
https://www.kaggle.com/c/house-prices-advanced-regression-techniques/leaderboard
Username: jwdata