78-English-Alphabet-classification-UCI-dataset.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h1><center><div style=\"direction:rtl;font-family:B Lotus, B Nazanin, Tahoma\">طبقه بندی حروف انگلیسی</div></center></h1>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">لود کتابخانه‌های مورد استفاده</div>\n"
   ]
  },
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   "execution_count": 1,
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   "source": [
    "import numpy as np\n",
    "import cv2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">لود مجموعه داده (dataset)</div>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "[Dataset from UCI Machine Learning Repository](https://archive.ics.uci.edu/ml/datasets/Letter+Recognition)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(20000, 17)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "# Load the data and convert the letters to numbers\n",
    "data= np.loadtxt('dataset/letter-recognition.data', dtype= 'float32', delimiter = ',',\n",
    "                    converters= {0: lambda ch: ord(ch)-ord('A')})\n",
    "data.shape"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">تقسیم داده برای آموزش و آزمون و تبدیل به فرمت ورودی مدل</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(10000, 17)\n",
      "(10000, 17)\n"
     ]
    }
   ],
   "source": [
    "# Split the dataset in two, with 10000 samples each for training and test sets\n",
    "train, test = np.vsplit(data,2)\n",
    "print(train.shape)\n",
    "print(test.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">جداکردن کردن ویژگی‌ها از لیبل در داده</div>\n",
    "<div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">می‌دانیم ستون اول لیبل و مابقی ستون‌ها ویژگی ها هستند.</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(10000, 16)\n",
      "(10000, 1)\n"
     ]
    }
   ],
   "source": [
    "y_train = train[:,:1]\n",
    "X_train = train[:,1:]\n",
    "print(X_train.shape)\n",
    "print(y_train.shape)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(10000, 16)\n",
      "(10000, 1)\n"
     ]
    }
   ],
   "source": [
    "y_test = test[:,:1]\n",
    "X_test = test[:,1:]\n",
    "print(X_test.shape)\n",
    "print(y_test.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">دو سلول بالا دقیقا معادل سلول زیر هستند</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(10000, 16)\n",
      "(10000, 1)\n",
      "(10000, 16)\n",
      "(10000, 1)\n"
     ]
    }
   ],
   "source": [
    "# Split trainData and testData into features and responses\n",
    "y_train, X_train = np.hsplit(train,[1])\n",
    "y_test, X_test = np.hsplit(test,[1])\n",
    "\n",
    "print(X_train.shape)\n",
    "print(y_train.shape)\n",
    "print(X_test.shape)\n",
    "print(y_test.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">آموزش مدل</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
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     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "# Initiate the kNN, classify, measure accuracy\n",
    "knn = cv2.ml.KNearest_create()\n",
    "knn.train(X_train, cv2.ml.ROW_SAMPLE, y_train)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## <div style=\"direction:rtl;text-align:right;font-family:B Lotus, B Nazanin, Tahoma\">بررسی نتیجه و دقت مدل</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "93.06\n"
     ]
    }
   ],
   "source": [
    "ret, result, neighbours, dist = knn.findNearest(X_test, k=5)\n",
    "# Now we check the accuracy of classification\n",
    "matches = result==y_test\n",
    "correct = np.count_nonzero(matches)\n",
    "accuracy = correct*100.0/result.size\n",
    "print( accuracy )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here we used k=5. What happens if you try other values of k? Can you find a value that maximizes accuracy (minimizes the number of errors)?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "https://docs.opencv.org/3.4/d8/d4b/tutorial_py_knn_opencv.html"
   ]
  }
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