review tarea gaussj

parent 4f069cc7
......@@ -817,10 +817,124 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": []
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Inversa(C)=\n",
" [[-2. -2.07692308 -0.38461538]\n",
" [ 1. 1.07692308 0.38461538]\n",
" [-0. -0.23076923 -0.15384615]]\n",
"\n",
"np.linalg.inv(C)=\n",
" [[-2. -2.07692308 -0.38461538]\n",
" [ 1. 1.07692308 0.38461538]\n",
" [-0. -0.23076923 -0.15384615]]\n",
"Trying:\n",
" A = np.array([[2, 4, 3], [0, 1, -1], [3, 5, 7]])\n",
"Expecting nothing\n",
"ok\n",
"Trying:\n",
" Inversa(A)\n",
"Expecting:\n",
" array([[ 4. , -4.33333333, -2.33333333],\n",
" [-1. , 1.66666667, 0.66666667],\n",
" [-1. , 0.66666667, 0.66666667]])\n",
"ok\n",
"Trying:\n",
" C = np.array([[1, 3, 5], [-2, -4, -5], [3, 6, 1]])\n",
"Expecting nothing\n",
"ok\n",
"Trying:\n",
" Inversa(C)\n",
"Expecting:\n",
" array([[-2. , -2.07692308, -0.38461538],\n",
" [ 1. , 1.07692308, 0.38461538],\n",
" [-0. , -0.23076923, -0.15384615]])\n",
"ok\n",
"1 items had no tests:\n",
" __main__\n",
"1 items passed all tests:\n",
" 4 tests in __main__.Inversa\n",
"4 tests in 2 items.\n",
"4 passed and 0 failed.\n",
"Test passed.\n"
]
},
{
"data": {
"text/plain": [
"TestResults(failed=0, attempted=4)"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"\n",
"import numpy as np\n",
"import sys\n",
"\n",
"def Inversa(A):\n",
" '''Función que calcula la inversa (si existe) de una matriz de n x n mediante el metodo de eliminación\n",
" de Gauss-Jordan.\n",
" \n",
" Args: \n",
" A (np.array): Lista de Arreglos que representa la matriz con valores tipo float.\n",
" Se dice que la matriz inversa A-1 tiene inversa, si A-1*A = I, donde I es la matriz identidad \n",
" \n",
" Returns:\n",
" np.array: La matriz inversa de A\n",
" \n",
" Ejemplos:\n",
" >>> A = np.array([[2, 4, 3], [0, 1, -1], [3, 5, 7]])\n",
" >>> Inversa(A)\n",
" array([[ 4. , -4.33333333, -2.33333333],\n",
" [-1. , 1.66666667, 0.66666667],\n",
" [-1. , 0.66666667, 0.66666667]])\n",
" \n",
" >>> C = np.array([[1, 3, 5], [-2, -4, -5], [3, 6, 1]])\n",
" >>> Inversa(C)\n",
" array([[-2. , -2.07692308, -0.38461538],\n",
" [ 1. , 1.07692308, 0.38461538],\n",
" [-0. , -0.23076923, -0.15384615]])\n",
" '''\n",
" if len(A) != len(A[0]):\n",
" print('La matriz no es cuadrada, por lo tanto NO INVERTIBLE estrictamente')\n",
" else:\n",
" A = np.array(A, dtype=np.float64)\n",
" n = len(A)\n",
" Inver = np.eye(n, dtype=np.float64)\n",
" #Método L por eliminación de Gauss, p es el pivote\n",
" for p in range(n-1):\n",
" if A[p,p] == 0: #Verifica que el pivote no sea cero\n",
" sys.exit('La matriz es singular, por lo tanto NO INVERTIBLE')\n",
" A[p,:], Inver[p,:] = A[p,:] / A[p,p], Inver[p,:] / A[p,p]\n",
" for k in range(p+1,n):\n",
" A[k,:], Inver[k,:] = A[p,:] * A[k,p] - A[k,:], Inver[p,:] * A[k,p] - Inver[k,:]\n",
" #Método U por eliminación de Gauss\n",
" for p in reversed(range(1,n)):\n",
" if A[p,p] == 0: #Verifica que el pivote no sea cero\n",
" sys.exit('La matriz es singular, por lo tanto NO INVERTIBLE')\n",
" A[p,:], Inver[p,:] = A[p,:] / A[p,p], Inver[p,:] / A[p,p]\n",
" for k in reversed(range(p)):\n",
" A[k,:], Inver[k,:] = A[p,:] * A[k,p] - A[k,:], Inver[p,:] * A[k,p] - Inver[k,:]\n",
" return Inver\n",
"\n",
"C = np.array([[1, 3, 5],\n",
" [-2, -4, -5],\n",
" [3, 6, 1]], dtype=float)\n",
"\n",
"print('Inversa(C)=\\n',Inversa(C))\n",
"print('\\nnp.linalg.inv(C)=\\n', np.linalg.inv(C))\n",
"import doctest\n",
"doctest.testmod(verbose=True)\n"
]
},
{
"cell_type": "markdown",
......@@ -1994,7 +2108,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.5.3"
"version": "3.7.1"
}
},
"nbformat": 4,
......
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