{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Test3: (40 marks)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Your ID: 6304640391"
   ]
  },
  {
   "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": [
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      ],
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       "       Unnamed: 0      X1  X2  X3  X4  X5  X6  X7  X8  X9  ...    X15    X16  \\\n",
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      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import datetime\n",
    "import numpy as np\n",
    "import pandas as pd\n",
    "import seaborn as sns\n",
    "#import pandas_datareader.data as web\n",
    "\n",
    "from sklearn import (\n",
    "    linear_model, metrics, pipeline, preprocessing, model_selection\n",
    ")\n",
    "\n",
    "import matplotlib.pyplot as plt\n",
    "%matplotlib inline\n",
    "\n",
    "df = pd.read_csv(\"default_card.csv\")\n",
    "df\n",
    "# activate plot theme\n",
    "#import qeds\n",
    "#qeds.themes.mpl_style();"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
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      ],
      "text/plain": [
       "       X2  X3  X4  Y\n",
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       "4       1   2   1  0\n",
       "...    ..  ..  .. ..\n",
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       "29998   1   3   1  1\n",
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       "\n",
       "[30000 rows x 4 columns]"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "df1 =df[['X2', 'X3', 'X4', 'Y']]\n",
    "df1 #put your code here"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
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       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>X2</th>\n",
       "      <th>X3</th>\n",
       "      <th>X4</th>\n",
       "      <th>Y</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>X2</th>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.014232</td>\n",
       "      <td>-0.031389</td>\n",
       "      <td>-0.039961</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>X3</th>\n",
       "      <td>0.014232</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>-0.143464</td>\n",
       "      <td>0.028006</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>X4</th>\n",
       "      <td>-0.031389</td>\n",
       "      <td>-0.143464</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>-0.024339</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Y</th>\n",
       "      <td>-0.039961</td>\n",
       "      <td>0.028006</td>\n",
       "      <td>-0.024339</td>\n",
       "      <td>1.000000</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "          X2        X3        X4         Y\n",
       "X2  1.000000  0.014232 -0.031389 -0.039961\n",
       "X3  0.014232  1.000000 -0.143464  0.028006\n",
       "X4 -0.031389 -0.143464  1.000000 -0.024339\n",
       "Y  -0.039961  0.028006 -0.024339  1.000000"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "correlation_data = df1.corr()  #-1<rho<1\n",
    "correlation_data"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### we can see that the 3 independent variables have very little correlation to the independent Y variable "
   ]
  },
  {
   "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": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "X = df[[\"X2\", \"X3\", \"X4\"]]\n",
    "y = df[\"Y\"]\n",
    "\n",
    "X_train, X_test, y_train, y_test = model_selection.train_test_split(X, y, test_size=0.2, random_state=100) \n",
    "#split into train and test"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "LogisticRegression()"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "logistic_model = linear_model.LogisticRegression()\n",
    "logistic_model.fit(X_train, y_train)\n",
    "#train data with logistic model\n",
    "#put your code here"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "              precision    recall  f1-score   support\n",
      "\n",
      "           0       0.77      1.00      0.87      4625\n",
      "           1       0.00      0.00      0.00      1375\n",
      "\n",
      "    accuracy                           0.77      6000\n",
      "   macro avg       0.39      0.50      0.44      6000\n",
      "weighted avg       0.59      0.77      0.67      6000\n",
      "\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "C:\\Users\\KC\\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\\KC\\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\\KC\\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,confusion_matrix\n",
    "print(classification_report(y_test, logistic_model.predict(X_test)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(0.7807916666666667, 0.7708333333333334)"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "train_acc = logistic_model.score(X_train, y_train)\n",
    "test_acc = logistic_model.score(X_test, y_test)\n",
    "\n",
    "train_acc, test_acc"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### The model has a precision rate of 0.77 which is not bad. It means that the model can predict the credit card owner that did not defualt by 77%. The recall rate of 1 meaning that the model predicts relevant data from the sample very well. For the case of defaulting loan, as the data of gender, education, and status have a very low correlation to the defaulting cases. We can expect that the data will generally do badly when it comes to predicting defaulting credit cards since the X's doesn't have a clear effect on Y"
   ]
  },
  {
   "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": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 0.81016074  0.18582826 -1.05729503  1.87637834]\n",
      " [ 0.81016074  0.18582826  0.85855728  1.87637834]\n",
      " [ 0.81016074  0.18582826  0.85855728 -0.53294156]\n",
      " ...\n",
      " [-1.23432296  0.18582826  0.85855728  1.87637834]\n",
      " [-1.23432296  1.45111372 -1.05729503  1.87637834]\n",
      " [-1.23432296  0.18582826 -1.05729503  1.87637834]]\n"
     ]
    }
   ],
   "source": [
    "from sklearn.preprocessing import StandardScaler\n",
    "\n",
    "scaler = StandardScaler()\n",
    "\n",
    "standardized_data = scaler.fit_transform(df1)\n",
    "\n",
    "print(standardized_data)\n",
    "\n",
    "\n",
    "#put your code here"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "             X2        X3        X4         Y\n",
      "0      0.810161  0.185828 -1.057295  1.876378\n",
      "1      0.810161  0.185828  0.858557  1.876378\n",
      "2      0.810161  0.185828  0.858557 -0.532942\n",
      "3      0.810161  0.185828 -1.057295 -0.532942\n",
      "4     -1.234323  0.185828 -1.057295 -0.532942\n",
      "...         ...       ...       ...       ...\n",
      "29995 -1.234323  1.451114 -1.057295 -0.532942\n",
      "29996 -1.234323  1.451114  0.858557 -0.532942\n",
      "29997 -1.234323  0.185828  0.858557  1.876378\n",
      "29998 -1.234323  1.451114 -1.057295  1.876378\n",
      "29999 -1.234323  0.185828 -1.057295  1.876378\n",
      "\n",
      "[30000 rows x 4 columns]\n"
     ]
    }
   ],
   "source": [
    "standardized_df = pd.DataFrame(standardized_data, columns=df1.columns)\n",
    "\n",
    "print(standardized_df)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "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": 12,
   "metadata": {},
   "outputs": [],
   "source": [
    "kf=KFold(n_splits=5, shuffle= True)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
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       "    }\n",
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       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>X2</th>\n",
       "      <th>X3</th>\n",
       "      <th>X4</th>\n",
       "      <th>Y</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>0.810161</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>1.876378</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>0.810161</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>0.858557</td>\n",
       "      <td>1.876378</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>0.810161</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>0.858557</td>\n",
       "      <td>-0.532942</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>0.810161</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>-0.532942</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>-0.532942</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>...</th>\n",
       "      <td>...</td>\n",
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       "      <th>29995</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>1.451114</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>-0.532942</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>29996</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>1.451114</td>\n",
       "      <td>0.858557</td>\n",
       "      <td>-0.532942</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>29997</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>0.858557</td>\n",
       "      <td>1.876378</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>29998</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>1.451114</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>1.876378</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>29999</th>\n",
       "      <td>-1.234323</td>\n",
       "      <td>0.185828</td>\n",
       "      <td>-1.057295</td>\n",
       "      <td>1.876378</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "<p>30000 rows × 4 columns</p>\n",
       "</div>"
      ],
      "text/plain": [
       "             X2        X3        X4         Y\n",
       "0      0.810161  0.185828 -1.057295  1.876378\n",
       "1      0.810161  0.185828  0.858557  1.876378\n",
       "2      0.810161  0.185828  0.858557 -0.532942\n",
       "3      0.810161  0.185828 -1.057295 -0.532942\n",
       "4     -1.234323  0.185828 -1.057295 -0.532942\n",
       "...         ...       ...       ...       ...\n",
       "29995 -1.234323  1.451114 -1.057295 -0.532942\n",
       "29996 -1.234323  1.451114  0.858557 -0.532942\n",
       "29997 -1.234323  0.185828  0.858557  1.876378\n",
       "29998 -1.234323  1.451114 -1.057295  1.876378\n",
       "29999 -1.234323  0.185828 -1.057295  1.876378\n",
       "\n",
       "[30000 rows x 4 columns]"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "standardized_df"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [],
   "source": [
    "X2 = standardized_df[[\"X2\", \"X3\", \"X4\"]]\n",
    "y2 = df[\"Y\"]\n",
    "\n",
    "X2_train, X2_test, y2_train, y2_test = model_selection.train_test_split(X2, y2, test_size=0.2, random_state=100) \n",
    "#split into train and test"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "KNeighborsClassifier(n_neighbors=7)"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "knn = KNeighborsClassifier(n_neighbors=7)\n",
    "  \n",
    "knn.fit(X2_train, y2_train)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "              precision    recall  f1-score   support\n",
      "\n",
      "           0       0.77      1.00      0.87      4625\n",
      "           1       0.00      0.00      0.00      1375\n",
      "\n",
      "    accuracy                           0.77      6000\n",
      "   macro avg       0.39      0.50      0.44      6000\n",
      "weighted avg       0.59      0.77      0.67      6000\n",
      "\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "C:\\Users\\KC\\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\\KC\\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\\KC\\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": [
    "print(classification_report(y2_test, logistic_model.predict(X2_test)))"
   ]
  },
  {
   "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": "markdown",
   "metadata": {},
   "source": [
    "### even with the standardized data the knn models and the logistic model seems to have the same results so I would say choosing one over the another would not make a difference in this case in particular when there is almost not correlation between X and Y\n",
    "\n",
    "\n",
    "#put your code here"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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