{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Test3: (40 marks)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Your ID: 6204641515" ] }, { "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" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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" ], "text/plain": [ " Y\n", "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", "\n", "[29601 rows x 1 columns]" ] }, "execution_count": 33, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Y = df[['Y']]\n", "Y" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [], "source": [ "from sklearn import (\n", " linear_model, metrics, pipeline, preprocessing, model_selection\n", ")\n", "\n", "X_train1, X_test1, y_train1, y_test1 = model_selection.train_test_split(X, Y, test_size=0.2)" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "name": "stderr", "output_type": "stream", "text": [ "C:\\Users\\Admin\\anaconda3\\lib\\site-packages\\sklearn\\utils\\validation.py:993: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().\n", " y = column_or_1d(y, warn=True)\n", "C:\\Users\\Admin\\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" ] }, { "data": { "text/plain": [ "LogisticRegression()" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "logistic_model1 = linear_model.LogisticRegression()\n", "logistic_model1.fit(X_train1, y_train1)" ] }, { "cell_type": "code", "execution_count": 36, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "(0.7773648648648649, 0.7745313291673703)" ] }, "execution_count": 36, "metadata": {}, "output_type": "execute_result" } ], "source": [ "train_acc = logistic_model1.score(X_train1, y_train1)\n", "test_acc = logistic_model1.score(X_test1, y_test1)\n", "\n", "train_acc, test_acc" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Classification report for testing data :-\n", " precision recall f1-score support\n", "\n", " 0 0.77 1.00 0.87 4587\n", " 1 0.00 0.00 0.00 1334\n", "\n", " accuracy 0.77 5921\n", " macro avg 0.39 0.50 0.44 5921\n", "weighted avg 0.60 0.77 0.68 5921\n", "\n" ] } ], "source": [ "from sklearn.metrics import classification_report\n", "print(\"Classification report for testing data :-\")\n", "print(classification_report(y_test1,logistic_model1.predict(X_test1)))" ] }, { "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": [ "#put your code here" ] }, { "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 }