Example for Data Analysis

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Example for Data Analysis

Provides a suite of helper functions and a collection of datasets used in the book Data Science Foundations and Machine Learning with R: From Data to Decisions. It is designed to make data science techniques accessible to individuals with minimal coding experience. Here is an example to show how to use the functionality of the package by using the churn_mlc dataset which is available in the package. For more examples and details, please refer to the book Data Science Foundations and Machine Learning with R: From Data to Decisions.

    data(churn_mlc)

    str(churn_mlc)
      'data.frame': 5000 obs. of  20 variables:
       $ state         : Factor w/ 51 levels "AK","AL","AR",..: 17 36 32 36 37 2 20 25 19 50 ...
       $ area_code     : Factor w/ 3 levels "area_code_408",..: 2 2 2 1 2 3 3 2 1 2 ...
       $ account_length: int  128 107 137 84 75 118 121 147 117 141 ...
       $ voice_plan    : Factor w/ 2 levels "yes","no": 1 1 2 2 2 2 1 2 2 1 ...
       $ voice_messages: int  25 26 0 0 0 0 24 0 0 37 ...
       $ intl_plan     : Factor w/ 2 levels "yes","no": 2 2 2 1 1 1 2 1 2 1 ...
       $ intl_mins     : num  10 13.7 12.2 6.6 10.1 6.3 7.5 7.1 8.7 11.2 ...
       $ intl_calls    : int  3 3 5 7 3 6 7 6 4 5 ...
       $ intl_charge   : num  2.7 3.7 3.29 1.78 2.73 1.7 2.03 1.92 2.35 3.02 ...
       $ day_mins      : num  265 162 243 299 167 ...
       $ day_calls     : int  110 123 114 71 113 98 88 79 97 84 ...
       $ day_charge    : num  45.1 27.5 41.4 50.9 28.3 ...
       $ eve_mins      : num  197.4 195.5 121.2 61.9 148.3 ...
       $ eve_calls     : int  99 103 110 88 122 101 108 94 80 111 ...
       $ eve_charge    : num  16.78 16.62 10.3 5.26 12.61 ...
       $ night_mins    : num  245 254 163 197 187 ...
       $ night_calls   : int  91 103 104 89 121 118 118 96 90 97 ...
       $ night_charge  : num  11.01 11.45 7.32 8.86 8.41 ...
       $ customer_calls: int  1 1 0 2 3 0 3 0 1 0 ...
       $ churn         : Factor w/ 2 levels "yes","no": 2 2 2 2 2 2 2 2 2 2 ...

It shows that the ‘churn_mlc’ dataset as a `data.frame` has 20 variables and 5000 observations.

# Partitioning the dataset

We partition the churn_mlc dataset randomly into two groups: train set (80%) and test set (20%). Here, we use the `partition` function from the liver package:

    set.seed(42)

    splits = partition(data = churn_mlc, ratio = c(0.8, 0.2))

    train_set = splits$part1
    test_set  = splits$part2

    test_labels  = test_set$churn

# Classification by kNN algorithm

The churn_mlc dataset has 19 predictors along with the target variable `churn`. Here we use the following predictors:

`account_length`, `voice_plan`, `voice_messages`, `intl_plan`, `intl_mins`, `day_mins`, `eve_mins`, `night_mins`, and `customer_calls`.

First, based on the above predictors, find the k-nearest neighbor for the test set, based on the training dataset, for the k = 8 as follows

    formula = churn ~ account_length + voice_plan + voice_messages + intl_plan + intl_mins +
                      day_mins + eve_mins + night_mins + customer_calls

    predict_knn = kNN(formula, train = train_set, test = test_set, k = 6)

To report Confusion Matrix:

    conf.mat(predict_knn, test_labels)
      Setting levels: reference = "yes", case = "no"
            Predict
      Actual yes  no
         yes  48  97
         no   21 834

    conf.mat.plot(predict_knn, test_labels)
      Setting levels: reference = "yes", case = "no"

To report Mean Squared Error (MSE):

    mse(predict_knn, test_labels)
      [1] 0.118

# Classification by kNN algorithm with data transformation

The predictors that we used in the previous part, do not have the same scale. For example, variable `day_mins` change between 0 and 351.5, whereas variable `voice_plan` is binary. In this case, the values of variable `day_mins` will overwhelm the contribution of `voice_plan`. To avoid this situation we use normalization. So, we use min-max normalization and transfer the predictors as follows:

    predict_knn_trans = kNN(formula, train = train_set, test = test_set, k = 6, scaler = "minmax")

To report Confusion Matrix:

    conf.mat.plot(predict_knn_trans, test_labels)
      Setting levels: reference = "yes", case = "no"

    conf.mat.plot(predict_knn, test_labels)
      Setting levels: reference = "yes", case = "no"

To report the ROC curve, we need the probability of our classification prediction. We can have it by using:

    prob_knn = kNN(formula, train = train_set, test = test_set, k = 6, type = "prob")[, 1]

    prob_knn_trans = kNN(formula, train = train_set, test = test_set, scaler = "minmax", k = 6, type = "prob")[, 1]

To visualize the model performance between the raw data and the transformed data, we could report the ROC curve plot as well as AUC (Area Under the Curve) by using the `plot.roc` function from the pROC package:

    roc_knn = roc(test_labels, prob_knn)
    roc_knn_trans = roc(test_labels, prob_knn_trans)

    ggroc(list(roc_knn, roc_knn_trans), linewidth = 0.8) +
        theme_minimal() + ggtitle("ROC plots with AUC") +
      scale_color_manual(values = c("red", "blue"),
        labels = c(paste("AUC=", round(auc(roc_knn), 3), "; Raw data; "),
                    paste("AUC=", round(auc(roc_knn_trans), 3), "; Transformed data"))) +
      theme(legend.title = element_blank()) +
      theme(legend.position = c(.7, .3), text = element_text(size = 17)) +
        geom_segment(aes(x = 1, xend = 0, y = 0, yend = 1), color = "grey", linetype = "dashed")

# Optimal value of k for the kNN algorithm

To find out the optimal value of `k` based on Accuracy, for the different values of k from 1 to 30, we run the k-nearest neighbor for the test set and compute the Accuracy for these models, by running `kNN.plot()` command

    kNN.plot(formula, train = train_set, ratio = c(0.7, 0.3), scaler = "minmax",
              k.max = 30, set.seed = 3)
      Setting levels: reference = "yes", case = "no"

The plot shows that the maximum value of Accuracy is for the case that k is 6; the higher values of Accuracy indicates better predictions.