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Accepted
Fernando Javier León
Asked: 2022-06-11 00:03:35 +0800 CST 2022-06-11 00:03:35 +0800 CST

How to build a dictionary with keys and values ​​as dictionary lists in Python

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I have this data called RemainingBlocks and RemainingInfoBlocks :

bloquesRestantes = ['Articulos', 'Proyectos']
bloquesInfoRestante = {
   "Articulos":[
      {
         "pais":"Peru",
         "indice":"Artículos Publicados de Divulgación Local",
         "fecha_publicacion":"2005-02-04",
         "mapeo":[
            "Pais",
            "Indice",
            "Fecha Publicacion"
         ]
      },
      {
         "pais":"Turquia",
         "indice":"Artículos",
         "fecha_publicacion":"1995-02-04",
         "mapeo":[
            "Pais",
            "Indice",
            "Fecha Publicacion"
         ]
      }
   ],
   "Proyectos":[
      {
         "fecha_inicio":"2005-02-01",
         "codigo_proyecto":"",
         "descripcion":"Asadsadsadsadsa sadsadsadsa",
         "mapeo":[
            "Fecha Inicio",
            "Codigo Proyecto",
            "Descripcion",  
         ]
      },
      {
         "fecha_inicio":"2008-08-08",
         "codigo_proyecto":"008",
         "descripcion":"Descripcion test",
         "mapeo":[
            "Fecha Inicio",
            "Codigo Proyecto",
            "Descripcion",  
         ]
      }
   ]
}

I use that information and through the zip() function I map the information with the data element mapeoin bloquesInfoRestantethe following way:

listaData = dict()
for i in bloquesRestantes:
  for bloqueInformacion in bloquesInfoRestante[i]:
    resultados = dict(zip(bloqueInformacion['mapeo'], bloqueInformacion.values()))
    listaData[i] = [{k:v for k,v in resultados.items()}]
print(listaData)

So far so good, now in the line where it is listaData[i]what I try is that for each dictionary key bloquesInfoRestantethat are 'Articles', 'Projects' . to be able to obtain the information of the list of each one.

But the way I'm doing it, I'm only getting the last element of each key in bloquesInfoRestantethe following way:

listaData = {
   "Articulos":[
      {
         "Pais":"Turquia",
         "Indice":"Artículos",
         "Fecha Publicacion":"1995-02-04"
      }
   ],
   "Proyectos":[
      {
         "Fecha Inicio":"2008-08-08",
         "Codigo Proyecto":"008",
         "Descripcion":"Descripcion test",
      }
   ],
}

When my intention is to get the data like this:

listaData = {
   "Articulos":[
      {
         "Pais":"Peru",
         "Indice":"Artículos Publicados de Divulgación Local",
         "Fecha Publicacion":"2005-02-04"
      },
      {
         "Pais":"Turquia",
         "Indice":"Artículos",
         "Fecha Publicacion":"1995-02-04"
      }

   ],
   "Proyectos":[
      {
         "Fecha Inicio":"2005-02-01",
         "Codigo Proyecto":"",
         "Descripcion":"Asadsadsadsadsa sadsadsadsa",
      }, 
      {
         "Fecha Inicio":"2008-08-08",
         "Codigo Proyecto":"008",
         "Descripcion":"Descripcion test",
      }
   ],
}

My question is how can I get the information in this way. I hope you can help me and thanks in advance.

python

1 Answers

  1. Best Answer

    Since we want to remove the key "mapeo"inside each contained element from the dictionary "Articulos"and "Proyectos"( bloquesRestantes), the dictionaries in Pythonhave a method pop.

    For this, it is necessary to iterate the elements contained in both "Articulos"and "Proyectos"( bloquesRestantes), taking into account that, when iterating the elements in a dictionary, they do not always maintain the same order.

    In this case, as the structure shown always has the key "mapeo", an index can be created enumeratefor this purpose.

    If in case the key "mapeo"does not exist... a conditional is included to verify that the internal dictionary in each element of "Articulos"and "Proyectos"(they are lists) said key exists to apply the methodpop

    Additionally, it is required to replace the keys "codigo_proyecto"a ""Codigo Proyecto""for them use the method replaceand title.

    With regard to replacing the keys, I am going to opt for the method pop... this would imply using a loop whilebecause the length of the dictionary would change as it iterates, in this sense a loop forwould throw an error.

    data_a_limpiar = 'mapeo'
data_lista = bloquesInfoRestante.copy() # para preservar nuestra data cruda

for bloquesRestantes, data_cruda in data_lista.items():
    for id, _ in enumerate(data_cruda):
        # verificamos si exite la clave "mapeo" para eliminarla
        if data_a_limpiar in data_lista[bloquesRestantes][id].keys():
            data_lista[bloquesRestantes][id].pop(data_a_limpiar, None)
        
        # Se emplea un bucle while porque la longitud del diccionario cambia 
        index = 0
        claves = list(data_lista[bloquesRestantes][id].keys())
        while index < len(claves):
            clave = claves[index]
            nueva_clave = clave.replace('_',' ').title()
            data_lista[bloquesRestantes][id][nueva_clave] = data_lista[bloquesRestantes][id].pop(clave)
            index += 1
            

# Resultado
print(data_lista)
"""
{
   "Articulos": [
      {
         "Indice": "Artículos Publicados de Divulgación Local",
         "Fecha Publicacion": "2005-02-04",
         "Pais": "Peru"
      },
      {
         "Indice": "Artículos",
         "Pais": "Turquia",
         "Fecha Publicacion": "1995-02-04"
      }
   ],
   "Proyectos": [
      {
         "Codigo Proyecto": "",
         "Descripcion": "Asadsadsadsadsa sadsadsadsa",
         "Fecha Inicio": "2005-02-01"
      },
      {
         "Descripcion": "Descripcion test",
         "Codigo Proyecto": "008",
         "Fecha Inicio": "2008-08-08"
      }
   ]
}
"""

    By having nested data up to the third level, the exposed solution is complex O(n3)in that sense, it is not recommended for large volumes of data... it can be better worked with clusters or dividing the data to facilitate its processing through distributed computing.

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