{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Test3: (40 marks)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Your ID: 6204641218" ] }, { "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": 16, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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" ], "text/plain": [ " Unnamed: 0 X1 X2 X3 X4 X5 X6 X7 X8 X9 ... X15 X16 \\\n", "0 1 20000 2 2 1 24 2 2 -1 -1 ... 0 0 \n", "1 2 120000 2 2 2 26 -1 2 0 0 ... 3272 3455 \n", "2 3 90000 2 2 2 34 0 0 0 0 ... 14331 14948 \n", "3 4 50000 2 2 1 37 0 0 0 0 ... 28314 28959 \n", "4 5 50000 1 2 1 57 -1 0 -1 0 ... 20940 19146 \n", "... ... ... .. .. .. .. .. .. .. .. ... ... ... \n", "29995 29996 220000 1 3 1 39 0 0 0 0 ... 88004 31237 \n", "29996 29997 150000 1 3 2 43 -1 -1 -1 -1 ... 8979 5190 \n", "29997 29998 30000 1 2 2 37 4 3 2 -1 ... 20878 20582 \n", "29998 29999 80000 1 3 1 41 1 -1 0 0 ... 52774 11855 \n", "29999 30000 50000 1 2 1 46 0 0 0 0 ... 36535 32428 \n", "\n", " X17 X18 X19 X20 X21 X22 X23 Y \n", "0 0 0 689 0 0 0 0 1 \n", "1 3261 0 1000 1000 1000 0 2000 1 \n", "2 15549 1518 1500 1000 1000 1000 5000 0 \n", "3 29547 2000 2019 1200 1100 1069 1000 0 \n", "4 19131 2000 36681 10000 9000 689 679 0 \n", "... ... ... ... ... ... ... ... .. \n", "29995 15980 8500 20000 5003 3047 5000 1000 0 \n", "29996 0 1837 3526 8998 129 0 0 0 \n", "29997 19357 0 0 22000 4200 2000 3100 1 \n", "29998 48944 85900 3409 1178 1926 52964 1804 1 \n", "29999 15313 2078 1800 1430 1000 1000 1000 1 \n", "\n", "[30000 rows x 25 columns]" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" } ], "source": [ "import pandas as pd\n", "df = pd.read_excel('defaults.xlsx')\n", "df" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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" ], "text/plain": [ " Y X2\n", "Unnamed: 0 \n", "1 1 2\n", "2 1 2\n", "3 0 2\n", "4 0 2\n", "5 0 1\n", "... .. ..\n", "29996 0 1\n", "29997 0 1\n", "29998 1 1\n", "29999 1 1\n", "30000 1 1\n", "\n", "[30000 rows x 2 columns]" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "df_group = df[['Y','X2']] #age\n", "df_group" ] }, { "cell_type": "code", "execution_count": 38, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "X2\n", "2 18112\n", "1 11888\n", "dtype: int64\n", "X4\n", "2 15964\n", "1 13659\n", "3 323\n", "0 54\n", "dtype: int64\n", "X3\n", "2 14030\n", "1 10585\n", "3 4917\n", "5 280\n", "4 123\n", "6 51\n", "0 14\n", "dtype: int64\n" ] } ], "source": [ "for i in {'X2','X3','X4'}:\n", " print(df[{i}].value_counts())" ] }, { "cell_type": "code", "execution_count": 43, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "Unnamed: 0\n", "1 1\n", "2 1\n", "3 2\n", "4 2\n", "5 1\n", " ..\n", "29996 1\n", "29997 1\n", "29998 1\n", "29999 1\n", "30000 1\n", "Name: Y, Length: 30000, dtype: int64" ] }, "execution_count": 43, "metadata": {}, "output_type": "execute_result" } ], "source": [ "import numpy as np\n", "df['Y']=np.where(df['Y'] == 0, df['X2'] , df['Y'])\n", "df['Y']" ] }, { "cell_type": "code", "execution_count": 48, "metadata": {}, "outputs": [ { "ename": "NameError", "evalue": "name 'SHOW_FIGURE' is not defined", "output_type": "error", "traceback": [ "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m", "\u001b[1;31mNameError\u001b[0m Traceback (most recent call last)", "\u001b[1;32m\u001b[0m in \u001b[0;36m\u001b[1;34m\u001b[0m\n\u001b[0;32m 3\u001b[0m \u001b[1;32mfrom\u001b[0m \u001b[0mmatplotlib\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mfont_manager\u001b[0m \u001b[1;32mimport\u001b[0m \u001b[0mFontProperties\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 4\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m----> 5\u001b[1;33m \u001b[1;32mif\u001b[0m \u001b[0mSHOW_FIGURE\u001b[0m \u001b[1;33m:\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 6\u001b[0m \u001b[1;31m# 1=graduate school, 2=university, 3=high school 4=others\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 7\u001b[0m \u001b[0mdata\u001b[0m\u001b[1;33m[\u001b[0m\u001b[1;34m'X2'\u001b[0m\u001b[1;33m]\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mvalue_counts\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mplot\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mkind\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;34m'bar'\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mfigsize\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m10\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;36m6\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n", "\u001b[1;31mNameError\u001b[0m: name 'SHOW_FIGURE' is not defined" ] } ], "source": [ "import matplotlib\n", "import matplotlib.pyplot as plt\n", "from matplotlib.font_manager import FontProperties\n", "\n", "if SHOW_FIGURE :\n", " # 1=graduate school, 2=university, 3=high school 4=others\n", " data['X2'].value_counts().plot(kind='bar', figsize=(10,6))\n", " # plt.title(\"Number of cars by make\")\n", " # plt.xlabel('Education level', fontproperties=font)\n", " plt.ylabel('Y', fontproperties=font)\n", " # plt.show()\n", " plt.savefig('gender_default.png')" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "gender = 194/(194+444)*100\n", "edu = 170/(170+468)*100\n", "marstat = 514/(514+124)*100\n", "\n", "print(f\"The percentage of Southwest Airlines that serves the route is {round(SW,2)} %.\")\n", "print(f\"The percentage of being a vacation route is {round(VACATION,2)} %.\")\n", "print(f\"The percentage of having gate constraints is {round(GATE,2)} %.\")\n", "print(f\"The percentage of having a slot-controlled at the endpoint airport is {round(SLOT,2)} %.\")" ] }, { "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": null, "metadata": {}, "outputs": [], "source": [ "y =df['Y']\n", "x =df['X2', 'X3', 'X4']" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) " ] }, { "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": null, "metadata": {}, "outputs": [], "source": [ "from sklearn.preprocessing import StandardScaler\n", "scaler = StandardScaler()\n", "scaled_data = scaler.fit_transform(X_train)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "from sklearn.pipeline import Pipeline\n", "from sklearn.preprocessing import StandardScaler\n", "\n", "classifier = KNeighborsClassifier(n_neighbors=9)\n", "classifier = Pipeline([('norm', StandardScaler()), ('knn', classifier)]) # once get the data then convert into z-score first ('norm', standardscaler), then KNN\n", "\n", "means = []\n", "for training,testing in kf.split(features):\n", " # We learn a model for this fold with `fit` and then apply it to the\n", " # testing data with `predict`:\n", " classifier.fit(features[training], labels[training])\n", " prediction = classifier.predict(features[testing])\n", "\n", " # np.mean on an array of booleans returns fraction\n", " # of correct decisions for this fold:\n", " curmean = np.mean(prediction == labels[testing])\n", " means.append(curmean)\n", "print('Mean accuracy: {:.1%}'.format(np.mean(means)))" ] }, { "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": null, "metadata": {}, "outputs": [], "source": [ "#put your code here" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "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": "3.8.8" } }, "nbformat": 4, "nbformat_minor": 4 }