{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Diccionarios"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Los diccionarios tambien son un tipo primitivo, se ponen los elementos entre llaves, la clave y el valor se separan con ':' ({} = diccionario vacio) . La clave no tiene porque ser un string, solo hace falta que se le pueda aplicar una funcion de hash (objeto no mutable)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{}"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = {}\n",
    "a"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'pepe': 1, 'juan': 3.14, 'maria': [1, 2, 3]}"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = {'pepe': 1, 'juan': 3.14, 'maria': [1, 2, 3]}\n",
    "a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos acceder por clave"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a['pepe']"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos consultar si una clave existe en el diccionario"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "'pepe' in a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Y usar cada elemento dependiendo de su tipo"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a['maria'][1]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos cambiar el valor de la clave"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'pepe': 1, 'juan': (1, 2), 'maria': [1, 2, 3]}"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a['juan'] = (1,2)\n",
    "a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Y tenemos las operaciones esperadas sobre diccionarios, como obtener todas las claves o todos sus valores en una lista "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "dict_keys(['pepe', 'juan', 'maria'])"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.keys()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "dict_values([1, (1, 2), [1, 2, 3]])"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.values()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos fusionar diccionarios"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'pepe': 1, 'juan': (1, 2), 'maria': [1, 2, 3], 'luis': 3, 'jose': 324}"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.update({'luis':3, 'jose': 324})\n",
    "a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos borrar claves"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'juan': (1, 2), 'maria': [1, 2, 3], 'luis': 3, 'jose': 324}"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.pop('pepe')\n",
    "a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos tener como valores otros diccionarios e indexarlos a diferentes niveles"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'pepe': {'perez': 33, 'garcia': 27}}"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b = {'pepe': {'perez': 33, 'garcia': 27} }\n",
    "b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "33"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b['pepe']['perez']"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Si damos un valor a una clave que no esta en el diccionario la añadimos"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'pepe': {'perez': 33, 'garcia': 27, 'ruiz': 22}}"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b['pepe']['ruiz'] = 22\n",
    "b"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podemos obtener el diccionario como una lista de tuplas (clave, valor)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "dict_items([('perez', 33), ('garcia', 27), ('ruiz', 22)])"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b['pepe'].items()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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