{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Test3: (40 marks)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Your ID: 6504930105" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "In this excercise, we aimed at the case of customers default payments in Taiwan and compares the predictive accuracy of probability of default among data mining methods. From the perspective of risk management, the result of predictive accuracy of the estimated probability of default will be more valuable than the binary result of classification - credible or not credible clients. With the real probability of default as the response variable (Y), and the predictive probability of default as the independent variable (X), this research employed a binary variable, default payment (Yes = 1, No = 0), as the response variable. This study reviewed the literature and used the following 23 variables as explanatory variables: \\\\\n", "\n", "X1: Amount of the given credit (NT dollar): it includes both the individual consumer credit and his/her family (supplementary) credit. \\\\\n", "\n", "X2: Gender (1 = male; 2 = female). \\\\\n", "\n", "X3: Education (1 = graduate school; 2 = university; 3 = high school; 4 = others). \\\\\n", "\n", "X4: Marital status (1 = married; 2 = single; 3 = others). \\\\\n", "\n", "X5: Age (year). \\\\\n", "\n", "X6 - X11: History of past payment. We tracked the past monthly payment records (from April to September, 2005) as follows: X6 = the repayment status in September, 2005; X7 = the repayment status in August, 2005; . . .;X11 = the repayment status in April, 2005. The measurement scale for the repayment status is: -1 = pay duly; 1 = payment delay for one month; 2 = payment delay for two months; . . .; 8 = payment delay for eight months; 9 = payment delay for nine months and above. \\\\\n", "\n", "X12-X17: Amount of bill statement (NT dollar). X12 = amount of bill statement in September, 2005; X13 = amount of bill statement in August, 2005; . . .; X17 = amount of bill statement in April, 2005. \\\\\n", "\n", "X18-X23: Amount of previous payment (NT dollar). X18 = amount paid in September, 2005; X19 = amount paid in August, 2005; . . .;X23 = amount paid in April, 2005. \\\\\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Learning the Logistic Regression, KNN, in Python and Scikit-Learn \n", "\n", "There are 4 sub-question:\n", "\n", "1.1 Create the table to report the proportion of case of customers default payments in Taiwan separated by gender, education, and marital status. What is/are the interesting result/s you can draw from this table?[10 Points].\\\\" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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" ], "text/plain": [ " Y\n", "Gender 1 0.241672\n", " 2 0.207763\n", "Education 0 0.000000\n", " 1 0.192348\n", " 2 0.237349\n", " 3 0.251576\n", " 4 0.056911\n", " 5 0.064286\n", " 6 0.156863\n", "Marital Status 0 0.092593\n", " 1 0.234717\n", " 2 0.209283\n", " 3 0.260062" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "pd.concat([table_gen,table_edu,table_mar], keys=[\"Gender\", \"Education\", \"Marital Status\"])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Answer:\n", "There are a few observations. \n", "First the default prevelance is different for gender. Devault payment is more prevelant in men.\n", "Secondly, the \n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Then, partition the data into training and test sets using 80 $\\%$ to be the train data. The model will be fit to the training data and evaluated on the test set. \n", "\n", "1.2 Use the Logistic regression to train the model with all features. Then, use the test data to conduct the confusion matrix. Interpret the results carefully [10 Points]." ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [], "source": [ "#put your code here\n", "#split data into X and Y\n", "X = df.drop(['Y'], axis=1)\n", "y= df['Y']\n", "#split into training and test\n", "\n", "from sklearn.model_selection import train_test_split\n", "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)" ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0 1\n", "1 1\n", "2 0\n", "3 0\n", "4 0\n", " ..\n", "29995 0\n", "29996 0\n", "29997 1\n", "29998 1\n", "29999 1\n", "Name: Y, Length: 30000, dtype: int64" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "y" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "name": "stderr", "output_type": "stream", "text": [ "C:\\Users\\marth\\anaconda3\\lib\\site-packages\\sklearn\\linear_model\\_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):\n", "STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.\n", "\n", "Increase the number of iterations (max_iter) or scale the data as shown in:\n", " https://scikit-learn.org/stable/modules/preprocessing.html\n", "Please also refer to the documentation for alternative solver options:\n", " https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression\n", " n_iter_i = _check_optimize_result(\n" ] } ], "source": [ "#Model development and prediction\n", "# import the class\n", "from sklearn.linear_model import LogisticRegression\n", "#instatiate the model\n", "logreg = LogisticRegression()\n", "#fit the model\n", "logreg.fit(X_train, y_train)\n", "y_pred = logreg.predict(X_test)" ] }, { "cell_type": "code", "execution_count": 42, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " precision recall f1-score support\n", "\n", " No 0.77 1.00 0.87 4647\n", " Yes 0.00 0.00 0.00 1353\n", "\n", " accuracy 0.77 6000\n", " macro avg 0.39 0.50 0.44 6000\n", "weighted avg 0.60 0.77 0.68 6000\n", "\n" ] }, { "name": "stderr", "output_type": "stream", "text": [ "C:\\Users\\marth\\anaconda3\\lib\\site-packages\\sklearn\\metrics\\_classification.py:1318: UndefinedMetricWarning: Precision and F-score are ill-defined and being set to 0.0 in labels with no predicted samples. Use `zero_division` parameter to control this behavior.\n", " _warn_prf(average, modifier, msg_start, len(result))\n", "C:\\Users\\marth\\anaconda3\\lib\\site-packages\\sklearn\\metrics\\_classification.py:1318: UndefinedMetricWarning: Precision and F-score are ill-defined and being set to 0.0 in labels with no predicted samples. Use `zero_division` parameter to control this behavior.\n", " _warn_prf(average, modifier, msg_start, len(result))\n", "C:\\Users\\marth\\anaconda3\\lib\\site-packages\\sklearn\\metrics\\_classification.py:1318: UndefinedMetricWarning: Precision and F-score are ill-defined and being set to 0.0 in labels with no predicted samples. Use `zero_division` parameter to control this behavior.\n", " _warn_prf(average, modifier, msg_start, len(result))\n" ] } ], "source": [ "from sklearn.metrics import classification_report\n", "target_names = ['No', 'Yes']\n", "print(classification_report(y_test, y_pred, target_names = target_names))\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "1.3 Use the KNN to train the model with all features (you should convert the features to be the standarized variables first). Then, use the test data to conduct the confusion matrix. Interpret the results carefully [10 Points]." ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [], "source": [ "#put your code here\n", "from sklearn.neighbors import KNeighborsClassifier\n", "classifier = KNeighborsClassifier(n_neighbors=9)\n", "from sklearn.model_selection import KFold\n", "import numpy as np" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [], "source": [ "kf=KFold(n_splits=5, shuffle= True)" ] }, { "cell_type": "code", "execution_count": 36, "metadata": {}, "outputs": [], "source": [ "\n", "clf = KNeighborsClassifier(p=1)\n", "clf.fit(X_train, y_train)\n", "y_pred_knn = clf.predict(X_test)" ] }, { "cell_type": "code", "execution_count": 48, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " precision recall f1-score support\n", "\n", " No 0.80 0.91 0.85 4647\n", " Yes 0.39 0.19 0.26 1353\n", "\n", " accuracy 0.75 6000\n", " macro avg 0.59 0.55 0.55 6000\n", "weighted avg 0.70 0.75 0.72 6000\n", "\n" ] } ], "source": [ "print(classification_report(y_test, y_pred_knn, target_names = target_names))" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ " 1.4 Compare the predictive performance of two models. Which one you select? And Why? [10 Points]" ] }, { "cell_type": "code", "execution_count": 47, "metadata": {}, "outputs": [], "source": [ "#put your code here\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Comparing the accuracy rate shows that both models have a similar accuracy. The logistic regression shows a accurcacy of 77% while the knn has a accuracy of 75%\n", "Both models predict the the default payment 'No' relatively well, with both high precision and recall rate in both models.The devault payment with answer Yes is not predicted well by both models. But slightly better by the KNN model.\n", "Overall the KNN model seems to make slightly better predictions, so I would chose that one." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.9.12" } }, "nbformat": 4, "nbformat_minor": 4 }