Up2143, todos OK, 1 falla, error en test

Author: jcordovaj

Workbench/Taller/HC280123.ipynb

@@ -440,15 +440,23 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 76,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "ename": "TypeError",
+     "evalue": "list indices must be integers or slices, not str",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
+      "\u001b[1;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
+      "Cell \u001b[1;32mIn[76], line 74\u001b[0m\n\u001b[0;32m     71\u001b[0m             \u001b[39mreturn\u001b[39;00m diccionario_par\n\u001b[0;32m     72\u001b[0m     \u001b[39mreturn\u001b[39;00m res\n\u001b[1;32m---> 74\u001b[0m OrdenarDiccionario(dicc, \u001b[39m'\u001b[39;49m\u001b[39mclave1\u001b[39;49m\u001b[39m'\u001b[39;49m) \n",
+      "Cell \u001b[1;32mIn[76], line 68\u001b[0m, in \u001b[0;36mOrdenarDiccionario\u001b[1;34m(diccionario_par, clave, descendente)\u001b[0m\n\u001b[0;32m     66\u001b[0m         \u001b[39mfor\u001b[39;00m i \u001b[39min\u001b[39;00m \u001b[39mrange\u001b[39m(\u001b[39m0\u001b[39m, cantidadValores):\n\u001b[0;32m     67\u001b[0m             newArr2\u001b[39m.\u001b[39mappend(newArrSorted[i][key])\n\u001b[1;32m---> 68\u001b[0m         res[key] \u001b[39m=\u001b[39m newArr2\n\u001b[0;32m     69\u001b[0m         \u001b[39m#Si no tienen valores devolvemos el mismo diccionario:\u001b[39;00m\n\u001b[0;32m     70\u001b[0m \u001b[39melse\u001b[39;00m:\n\u001b[0;32m     71\u001b[0m         \u001b[39mreturn\u001b[39;00m diccionario_par\n",
+      "\u001b[1;31mTypeError\u001b[0m: list indices must be integers or slices, not str"
+     ]
+    }
+   ],
    "source": [
-    "dicc = {\n",
-    "        'clave1':['c','a','b'],\n",
-    "        'clave2':['casa','auto','barco'],\n",
-    "        'clave3':[1,2,3]\n",
-    "        }\n",
     "def OrdenarDiccionario(diccionario_par, clave, descendente=True):\n",
     "    '''\n",
     "    Esta función recibe como parámetro un diccionario, cuyas listas de valores tienen el mismo\n",
@@ -493,7 +501,7 @@
     "\n",
     "    #print (len(diccionario_par[list(diccionario_par.keys())[0]]))\n",
     "    # chequeamos qué valor estamos reciendo\n",
-    "    print()\n",
+    "    #print()\n",
     "    if len(diccionario_par[list(diccionario_par.keys())[0]])>0: \n",
     "        cantidadValores=len(diccionario_par[list(diccionario_par.keys())[0]])\n",
     "        # convertimos el diccionario de listas, a una lista de diccionarios:\t\n",
@@ -502,21 +510,26 @@
     "            for key in diccionario_par:\n",
     "                newObj[key]=diccionario_par[key][i]\n",
     "            newArr.append(newObj)\n",
-    "    #print(newArr) Aquí se generó una lista de diccionarios\n",
-    "    # Ahora ordenemos\n",
-    "    newArrSorted = sorted(newArr, key=lambda d: d[clave], reverse=not (descendente))\n",
-    "    print (newArrSorted) #verificar contenido del nuevo arreglo\n",
-    "    # Convertimos la lista de diccionarios, en un diccionario de listas:\n",
-    "    for key in diccionario_par:\n",
-    "        newArr2 = []\n",
-    "        for i in range(0, cantidadValores):\n",
-    "            newArr2.append(newArrSorted[i][key])\n",
-    "        res[key] = newArr2\n",
+    "            #print(newArr) Aquí se generó una lista de diccionarios\n",
+    "        # Ahora ordenemos\n",
+    "        newArrSorted = sorted(newArr, key=lambda d: d[clave], reverse=not (descendente))\n",
+    "        # print (newArrSorted) #verificar contenido del nuevo arreglo\n",
+    "        # Convertimos la lista de diccionarios, en un diccionario de listas:\n",
+    "        for key in diccionario_par:\n",
+    "            newArr2 = []\n",
+    "            for i in range(0, cantidadValores):\n",
+    "                newArr2.append(newArrSorted[i][key])\n",
+    "            res[key] = newArr2\n",
     "            #Si no tienen valores devolvemos el mismo diccionario:\n",
     "    else:\n",
-    "        return diccionario_par\n",
+    "            return diccionario_par\n",
+    "    \n",
     "    return res\n",
-    "\n",
+    "dicc = {\n",
+    "        'clave1':['c','a','b'],\n",
+    "        'clave2':['casa','auto','barco'],\n",
+    "        'clave3':[1,2,3]\n",
+    "        }\n",
     "OrdenarDiccionario(dicc, 'clave1') "
    ]
   }

Workbench/Taller/OrdenarDiccionario.py

@@ -0,0 +1,34 @@
+def OrdenarDiccionario(diccionario_par, clave, descendente=True):
+    cantidadValores=0 
+    newArr=[] 		  
+    newArrSorted=[]   
+    res ={}
+    if not (type(diccionario_par) is dict):  
+        return None
+    if not (clave in diccionario_par.keys()): 
+        return None                                 
+    if len(diccionario_par[list(diccionario_par.keys())[0]])>0: 
+        cantidadValores=len(diccionario_par[list(diccionario_par.keys())[0]])
+        for i in range(0, cantidadValores):
+            newObj={}
+            for key in diccionario_par:
+                newObj[key]=diccionario_par[key][i]
+            newArr.append(newObj)
+        newArrSorted = sorted(newArr, key=lambda d: d[clave], reverse=not (descendente))
+        for key in diccionario_par:
+            newArr2 = []
+            for i in range(0, cantidadValores):
+                newArr2.append(newArrSorted[i][key])
+            res[key] = newArr2
+    else:
+            return diccionario_par
+    
+    return res
+
+dicc = {
+        'clave1':['c','a','b'],
+        'clave2':['casa','auto','barco'],
+        'clave3':[1,2,3]
+        }
+
+print (OrdenarDiccionario(dicc,"clave1"))