{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Test3: (40 marks)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Your ID: 6304640094" ] }, { "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": 1, "metadata": {}, "outputs": [], "source": [ "import pandas as pd\n", "import numpy as np\n", "import matplotlib.pyplot as plt\n", "import seaborn as sns\n", "\n", "%matplotlib inline\n", "import qeds\n", "\n", "qeds.themes.mpl_style();\n", "plotly_template = qeds.themes.plotly_template()\n", "colors = qeds.themes.COLOR_CYCLE" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [], "source": [ "df= pd.read_excel(\"default of credit card clients.xls\")" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "scrolled": true }, "outputs": [ { "data": { "text/html": [ "
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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", " 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", "[5 rows x 25 columns]" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "df.head()" ] }, { "cell_type": "code", "execution_count": 39, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "\n", "RangeIndex: 30000 entries, 0 to 29999\n", "Data columns (total 25 columns):\n", " # Column Non-Null Count Dtype\n", "--- ------ -------------- -----\n", " 0 Unnamed: 0 30000 non-null int64\n", " 1 X1 30000 non-null int64\n", " 2 X2 30000 non-null int64\n", " 3 X3 30000 non-null int64\n", " 4 X4 30000 non-null int64\n", " 5 X5 30000 non-null int64\n", " 6 X6 30000 non-null int64\n", " 7 X7 30000 non-null int64\n", " 8 X8 30000 non-null int64\n", " 9 X9 30000 non-null int64\n", " 10 X10 30000 non-null int64\n", " 11 X11 30000 non-null int64\n", " 12 X12 30000 non-null int64\n", " 13 X13 30000 non-null int64\n", " 14 X14 30000 non-null int64\n", " 15 X15 30000 non-null int64\n", " 16 X16 30000 non-null int64\n", " 17 X17 30000 non-null int64\n", " 18 X18 30000 non-null int64\n", " 19 X19 30000 non-null int64\n", " 20 X20 30000 non-null int64\n", " 21 X21 30000 non-null int64\n", " 22 X22 30000 non-null int64\n", " 23 X23 30000 non-null int64\n", " 24 Y 30000 non-null int64\n", "dtypes: int64(25)\n", "memory usage: 5.7 MB\n" ] } ], "source": [ "df.info()" ] }, { "cell_type": "code", "execution_count": 43, "metadata": {}, "outputs": [ { "ename": "SyntaxError", "evalue": "invalid syntax (2392767680.py, line 4)", "output_type": "error", "traceback": [ "\u001b[0;36m Input \u001b[0;32mIn [43]\u001b[0;36m\u001b[0m\n\u001b[0;31m else df.loc[idx, 'X2'] == 2:\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n" ] } ], "source": [ "for idx, _ in df.iterrows():\n", " if df.loc[idx, 'X2'] == 1:\n", " df.loc[idx, 'X2'] = \"male\"\n", " else df.loc[idx, 'X2'] == 2:\n", " df.loc[idx, 'X2'] = \"female\"" ] }, { "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": 34, "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", "import pandas as pd\n", "from sklearn.model_selection import train_test_split\n", "from sklearn import metrics #evaluation model.\n", "import matplotlib " ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "ename": "NameError", "evalue": "name 'linear_model' is not defined", "output_type": "error", "traceback": [ "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m", "\u001b[0;31mNameError\u001b[0m Traceback (most recent call last)", "Input \u001b[0;32mIn [35]\u001b[0m, in \u001b[0;36m\u001b[0;34m()\u001b[0m\n\u001b[0;32m----> 1\u001b[0m \u001b[43mlinear_model\u001b[49m\u001b[38;5;241m.\u001b[39mLogisticRegression()\n", "\u001b[0;31mNameError\u001b[0m: name 'linear_model' is not defined" ] } ], "source": [ "from sklearn import (\n", " linear_model, metrics, pipeline, preprocessing, model_selection)\n", "\n", "from sklearn.model_selection import train_test_split\n", "\n", "X_train, X_test, y_train, y_test = model_selection.train_test_split(X, y, test_size=0.2, random_state=100)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "linear_model.LogisticRegression()" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "logistic_model = linear_model.LogisticRegression(multi_class='multinomial')" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "logistic_model.fit(X_train, y_train)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "print(classification_report(y_test, logistic_model.predict(X_test)))" ] }, { "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": 44, "metadata": {}, "outputs": [], "source": [ "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": 45, "metadata": {}, "outputs": [], "source": [ "kf=KFold(n_splits=5, shuffle= True)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "means = []\n", "\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", " # 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 (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 }