solve coflict

.ipynb_checkpoints/Encoding_decoding-checkpoint.ipynb

@@ -213,8 +213,8 @@
     "    diff       (510, 638): difference of min and max of the 4 neighbors\n",
     "    boundary   (2300,): the boundary values after subtracting the very first pixel value\n",
     "    predict    (325380,): the list of predicted values\n",
-    "    bins       (num_bins)\n",
-    "    A       (3 X 3): system of equation\n",
+    "    bins       (num_bins - 1,): a list of threshold to cut the bins\n",
+    "    A          (3 X 3): system of equation\n",
     "    \n",
     "    \"\"\"\n",
     "    # get the prediction error and difference\n",
@@ -306,6 +306,9 @@
    "outputs": [],
    "source": [
     "def encoder(error, list_dic, diff, bound, bins):\n",
+    "    \"\"\"\n",
+    "    This function en\n",
+    "    \"\"\"\n",
     "    # copy the error matrix (including the boundary)\n",
     "    encoded = np.copy(error).astype(int).astype(str).astype(object)\n",
     "    #diff = np.reshape(diff,(510,638))\n",
@@ -490,57 +493,7 @@
    "id": "a282f9e6",
    "metadata": {},
    "outputs": [],
-   "source": [
-    "def predict_pix_lstsq(tiff_list):\n",
-    "    \"\"\"\n",
-    "    Predict the next pixel using a fit hyperplane of the four closest pixels.\n",
-    "    The gradient measure in this function is the summed distance to the fitted hyperplane\n",
-    "    of each of the four points, aka the residual from the least squares function. The previous\n",
-    "    predict_pix function uses the difference between the minimal and maximal pixels of the surrounding\n",
-    "    four.\n",
-    "    \"\"\"\n",
-    "\n",
-    "    image = tiff_list\n",
-    "    image = Image.open(image)    #Open the image and read it as an Image object\n",
-    "    image = np.array(image)[1:,:]    #Convert to an array, leaving out the first row because the first row is just housekeeping data\n",
-    "    image = image.astype(int)\n",
-    "    A = np.array([[3,0,-1],[0,3,3],[1,-3,-4]]) # the matrix for system of equation\n",
-    "    z0 = image[0:-2,0:-2]   # get all the first pixel for the entire image\n",
-    "    z1 = image[0:-2,1:-1]   # get all the second pixel for the entire image\n",
-    "    z2 = image[0:-2,2::]    # get all the third pixel for the entire image\n",
-    "    z3 = image[1:-1,0:-2]   # get all the forth pixel for the entire image\n",
-    "    # calculate the out put of the system of equation\n",
-    "    y0 = np.ravel(-z0+z2-z3)\n",
-    "    y1 = np.ravel(z0+z1+z2)\n",
-    "    y2 = np.ravel(-z0-z1-z2-z3)\n",
-    "    y = np.vstack((y0,y1,y2))\n",
-    "    \n",
-    "    # use numpy solver to solve the system of equations all at once\n",
-    "    predict = np.round(np.round((np.linalg.solve(A,y)[-1]),1)) #round the solution to the nearest integer so that encoding/decoding is easier\n",
-    "    \n",
-    "    points = np.array([[-1,-1,1], [-1,0,1], [-1,1,1], [0,-1,1]]) #Matrix system of points that will be used to solve the least squares fitting hyperplane\n",
-    "\n",
-    "    \n",
-    "    \n",
-    "    # flatten the neighbor pixels and stack them together\n",
-    "    z0 = np.ravel(z0)\n",
-    "    z1 = np.ravel(z1)\n",
-    "    z2 = np.ravel(z2)\n",
-    "    z3 = np.ravel(z3)\n",
-    "    neighbor = np.vstack((z0,z1,z2,z3)).T\n",
-    "    \n",
-    "    f, res, rank, s = la.lstsq(points, neighbor.T, rcond=None) \n",
-    "    \n",
-    "    \n",
-    "    # calculate the difference\n",
-    "    diff = np.max(neighbor,axis = 1) - np.min(neighbor, axis=1)\n",
-    "    \n",
-    "    # flatten the image to a vector\n",
-    "    image = np.ravel(image[1:-1,1:-1])\n",
-    "    error = image-predict\n",
-    "    \n",
-    "    return image, predict, res, error, A, diff"
-   ]
+   "source": []
   }
  ],
  "metadata": {
@@ -559,7 +512,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.8.11"
+   "version": "3.9.1"
   }
  },
  "nbformat": 4,

Encoding_decoding.ipynb

@@ -307,7 +307,7 @@
    "source": [
     "def encoder(error, list_dic, diff, bound, bins):\n",
     "    \"\"\"\n",
-    "    This function \n",
+    "    This function en\n",
     "    \"\"\"\n",
     "    # copy the error matrix (including the boundary)\n",
     "    encoded = np.copy(error).astype(int).astype(str).astype(object)\n",
@@ -493,57 +493,7 @@
    "id": "a282f9e6",
    "metadata": {},
    "outputs": [],
-   "source": [
-    "def predict_pix_lstsq(tiff_list):\n",
-    "    \"\"\"\n",
-    "    Predict the next pixel using a fit hyperplane of the four closest pixels.\n",
-    "    The gradient measure in this function is the summed distance to the fitted hyperplane\n",
-    "    of each of the four points, aka the residual from the least squares function. The previous\n",
-    "    predict_pix function uses the difference between the minimal and maximal pixels of the surrounding\n",
-    "    four.\n",
-    "    \"\"\"\n",
-    "\n",
-    "    image = tiff_list\n",
-    "    image = Image.open(image)    #Open the image and read it as an Image object\n",
-    "    image = np.array(image)[1:,:]    #Convert to an array, leaving out the first row because the first row is just housekeeping data\n",
-    "    image = image.astype(int)\n",
-    "    A = np.array([[3,0,-1],[0,3,3],[1,-3,-4]]) # the matrix for system of equation\n",
-    "    z0 = image[0:-2,0:-2]   # get all the first pixel for the entire image\n",
-    "    z1 = image[0:-2,1:-1]   # get all the second pixel for the entire image\n",
-    "    z2 = image[0:-2,2::]    # get all the third pixel for the entire image\n",
-    "    z3 = image[1:-1,0:-2]   # get all the forth pixel for the entire image\n",
-    "    # calculate the out put of the system of equation\n",
-    "    y0 = np.ravel(-z0+z2-z3)\n",
-    "    y1 = np.ravel(z0+z1+z2)\n",
-    "    y2 = np.ravel(-z0-z1-z2-z3)\n",
-    "    y = np.vstack((y0,y1,y2))\n",
-    "    \n",
-    "    # use numpy solver to solve the system of equations all at once\n",
-    "    predict = np.round(np.round((np.linalg.solve(A,y)[-1]),1)) #round the solution to the nearest integer so that encoding/decoding is easier\n",
-    "    \n",
-    "    points = np.array([[-1,-1,1], [-1,0,1], [-1,1,1], [0,-1,1]]) #Matrix system of points that will be used to solve the least squares fitting hyperplane\n",
-    "\n",
-    "    \n",
-    "    \n",
-    "    # flatten the neighbor pixels and stack them together\n",
-    "    z0 = np.ravel(z0)\n",
-    "    z1 = np.ravel(z1)\n",
-    "    z2 = np.ravel(z2)\n",
-    "    z3 = np.ravel(z3)\n",
-    "    neighbor = np.vstack((z0,z1,z2,z3)).T\n",
-    "    \n",
-    "    f, res, rank, s = la.lstsq(points, neighbor.T, rcond=None) \n",
-    "    \n",
-    "    \n",
-    "    # calculate the difference\n",
-    "    diff = np.max(neighbor,axis = 1) - np.min(neighbor, axis=1)\n",
-    "    \n",
-    "    # flatten the image to a vector\n",
-    "    image = np.ravel(image[1:-1,1:-1])\n",
-    "    error = image-predict\n",
-    "    \n",
-    "    return image, predict, res, error, A, diff"
-   ]
+   "source": []
   }
  ],
  "metadata": {
@@ -562,7 +512,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.8.11"
+   "version": "3.9.1"
   }
  },
  "nbformat": 4,