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knn algorithm machine learning, in this tutorial we are going to explain classification and regression problems.

Machine learning is a subset of artificial intelligence which provides machines the ability to learn automatically and improve from previous experience without being explicitly programmed.

The major part of machine learning is data. Feed the machine with data and make a model and predict. Feed with more data and the model becomes more accurate accordingly.

Naïve Bayes classification in R

## What is knn algorithm?

K Nearest Neighbour is a supervised learning algorithm that classifies a new data point into the target class, depending on the features of its neighboring data points.

Let’s look at the student dataset with GPA and GRE scores for classification problems and Boston housing data for a regression problem.

The euclidian distance used for calculating the distance between k neighbors and some of the variables have different magnitudes, so standazation is important.

Some of the popular application examples are

• Recommendation system
• Loan Approval
• Anamoly Detection
• Text Categorization
• Finance
• Medicine

Let’s see how do we apply knn algorithm in classification and regression.

Market Basket Analysis in R

## Classification Approach

library(caret)
library(pROC)
library(mlbench)

### Getting Data

data <- read.csv("D:/RStudio/knn/binary.csv", header = T)
str(data)

You can access the dataset from this link

'data.frame': 400 obs. of  4 variables:
$admit: int 0 1 1 1 0 1 1 0 1 0 ...$ gre  : int  380 660 800 640 520 760 560 400 540 700 ...
$gpa : num 3.61 3.67 4 3.19 2.93 3 2.98 3.08 3.39 3.92 ...$ rank : int  3 3 1 4 4 2 1 2 3 2 ...

The data frame contains 400 observations and 4 variables and rank variables stored as integer currently need to convert into factor variable. Admit is the response variable or dependent variable let’s recode 0 and 1 into No and Yes.

data$admit[data$admit == 0] <- 'No'
data$admit[data$admit == 1] <- 'Yes'
data$admit <- factor(data$admit)

### Data Partition

Let’s create independent samples and create training and test dataset for prediction.

set.seed(1234)
ind <- sample(2, nrow(data), replace = T, prob = c(0.7, 0.3))
training <- data[ind == 1,]
test <- data[ind == 2,]
str(training)
'data.frame': 284 obs. of  4 variables:
$admit: Factor w/ 2 levels "No","Yes": 1 2 2 2 2 2 1 2 1 1 ...$ gre  : int  380 660 800 640 760 560 400 540 700 800 ...
$gpa : num 3.61 3.67 4 3.19 3 2.98 3.08 3.39 3.92 4 ...$ rank : int  3 3 1 4 2 1 2 3 2 4 ...

The training dataset contains now 284 observations with 4 variables and the test dataset contains 116 observations and 4 variables.

Cluster optimization in R

### KNN Model

Before making knn model we need to create train control. Let’s create train control based on below code.

trControl <- trainControl(method = "repeatedcv",
number = 10,
repeats = 3,
classProbs = TRUE,
summaryFunction = twoClassSummary)

trainControl is from caret package

number of iteration is 10 times.

Repeat the cross validation is 3 times.

set.seed(222)
fit <- train(admit ~ .,
data = training,
method = 'knn',
tuneLength = 20,
trControl = trControl,
preProc = c("center", "scale"),
metric = "ROC",
tuneGrid = expand.grid(k = 1:60))

### Model Performance

fit

k-Nearest Neighbors

284 samples
3 predictor
2 classes: 'No', 'Yes'
Pre-processing: centered (3), scaled (3)
Resampling: Cross-Validated (10 fold, repeated 3 times)
Summary of sample sizes: 256, 256, 256, 256, 255, 256, ...
Resampling results across tuning parameters:
k   ROC   Sens  Spec
1  0.54  0.71  0.370
2  0.56  0.70  0.357
3  0.58  0.80  0.341
4  0.56  0.78  0.261
5  0.59  0.81  0.285
6  0.59  0.82  0.277
7  0.59  0.86  0.283
8  0.59  0.86  0.269
9  0.60  0.87  0.291
10  0.59  0.87  0.274
11  0.60  0.88  0.286
12  0.59  0.87  0.277
13  0.59  0.87  0.242
14  0.59  0.89  0.257
15  0.60  0.88  0.228
16  0.61  0.90  0.221
17  0.63  0.90  0.236
18  0.63  0.90  0.215
19  0.63  0.90  0.229
20  0.64  0.90  0.222
21  0.64  0.91  0.211
22  0.64  0.91  0.225
23  0.64  0.92  0.214
24  0.64  0.93  0.217
25  0.65  0.92  0.200
26  0.65  0.93  0.199
27  0.66  0.93  0.203
28  0.66  0.94  0.210
29  0.67  0.94  0.199
30  0.67  0.94  0.199
.....................
59  0.66  0.96  0.096
60  0.67  0.96  0.100

ROC was used to select the optimal model using the largest value.

The final value used for the model was k = 30.

We have carried out 10 cross-validations and the best ROC we got at k=30

Decision Trees in R

plot(fit)

varImp(fit)

ROC curve variable importance

     Importance
gpa       100.0
rank       25.2
gre         0.0

gpa is more important followed by rank and gre is not important.

pred <- predict(fit, newdata = test)
confusionMatrix(pred, test$admit) Confusion Matrix and Statistics  Reference Prediction No Yes No 79 29 Yes 3 5 Accuracy : 0.724 95% CI : (0.633, 0.803) No Information Rate : 0.707 P-Value [Acc > NIR] : 0.385 Kappa : 0.142 Mcnemar's Test P-Value : 9.9e-06 Sensitivity : 0.963 Specificity : 0.147 Pos Pred Value : 0.731 Neg Pred Value : 0.625 Prevalence : 0.707 Detection Rate : 0.681 Detection Prevalence : 0.931 Balanced Accuracy : 0.555 'Positive' Class : No  Model accuracy is 72% with 84 correct classifications out of 116 classifications. Regression analysis in R ## Regression Let’s look at the Bostonhousing data data("BostonHousing") data <- BostonHousing) str(data) 'data.frame': 506 obs. of 14 variables:$ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
$zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...$ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
$chas : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...$ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
$rm : num 6.58 6.42 7.18 7 7.15 ...$ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
$dis : num 4.09 4.97 4.97 6.06 6.06 ...$ rad    : num  1 2 2 3 3 3 5 5 5 5 ...
$tax : num 296 242 242 222 222 222 311 311 311 311 ...$ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
$b : num 397 397 393 395 397 ...$ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
$medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ... Medv is the response or dependent variable with numeric values. The data frame contains a total of 506 observations and 14 variables. ### Data Partition Let’s do the data partition for prediction. Timeseries analysis in R set.seed(1234) ind <- sample(2, nrow(data), replace = T, prob = c(0.7, 0.3)) training <- data[ind == 1,] test <- data[ind == 2,] ### KNN Model trControl <- trainControl(method = 'repeatedcv', number = 10, repeats = 3) set.seed(333) Let’s fit the regression model fit <- train(medv ~., data = training, tuneGrid = expand.grid(k=1:70), method = 'knn', metric = 'Rsquared', trControl = trControl, preProc = c('center', 'scale')) ### Model Performance fit k-Nearest Neighbors 355 samples 13 predictor Pre-processing: centered (13), scaled (13) Resampling: Cross-Validated (10 fold, repeated 3 times) Summary of sample sizes: 320, 320, 319, 320, 319, 319, ... Resampling results across tuning parameters: k RMSE Rsquared MAE 1 4.2 0.78 2.8 2 4.0 0.81 2.7 3 4.0 0.82 2.6 4 4.1 0.81 2.7 ...................... 70 5.9 0.72 4.1 R-squared was used to select the optimal model using the largest value. The final value used for the model was k = 3. This model is based on 10 fold cross-validation with 3 repeats. Self Organizing Maps plot(fit)  varImp(fit) loess r-squared variable importance  Overall rm 100.0 lstat 98.0 indus 87.1 nox 82.3 tax 68.4 ptratio 50.8 rad 41.3 dis 41.2 zn 37.9 crim 34.5 b 24.2 age 22.4 chas 0.0 rm is the most important variable and followed by lstat, indus, nox etc.. pred <- predict(fit, newdata = test) RMSE(pred, test$medv)
6.1
plot(pred ~ test\$medv)


## Conclusion

Based on the knn machine algorithm we can make insights for classification and regression problems.

Read Logistic Regression in R

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