如何将此列表转换为数据框架？我想知道如何在蟒蛇中做到这一点？

Question

我在python中有一个列表，如下所示。我想把它变成一个数据框架。我尝试过这样做：pd.DataFrame(myList)，然而‘’列存储了一个列表，但是我想将原点和quantityLeads键存储在相同的数据格式中

myList = [
   {
      "id":3105052,
      "title":"Ebook Relat�rios Gerenciais",
      "offering":"Institucional",
      "created_date":"2022-06-28"
      "inserted_date":"2022-06-28",
      "channel":"Social",
      "start_date":"2022-06-28",
      "end_date":"2022-06-28",
      "origins":[
         {
            "origin":"LinkedIn",
            "quantityLeads":"1"
         },
         {
            "origin":"Facebook",
            "quantityLeads":"1"
         }
      ]
   },
   {
      "id":3105052,
      "title":"Ebook Relat�rios Gerenciais",
      "offering":"Institucional",
      "inserted_date":"2022-06-28",
      "created_date":"2022-06-28",
      "channel":"Direct",
      "start_date":"2022-06-28",
      "end_date":"2022-06-28",
      "origins":[
         {
            "origin":"Desconhecida",
            "quantityLeads":"2"
         }
      ]
   },
   {
      "id":2918513,
      "title":"Ebook Direct To Consumer",
      "offering":"Supply Chain",
      "created_date":"2022-06-28",
      "inserted_date":"2022-06-28",
      "channel":"Social",
      "start_date":"2022-06-28",
      "end_date":"2022-06-28",
      "origins":[
         {
            "origin":"LinkedIn",
            "quantityLeads":"1"
         }
      ]
   }
]

Answer 1

在追求简单的过程中，您可以使用如下内容来简化字典结构：

for row in myList:
   row["origin"] = row["origins"][0]["origin"]
   row["quantityLeads"] = row["origins"][0]["quantityLeads"]
   del row["origins"]

df = pd.DataFrame(myList)
print(df)

输出：

        id                        title       offering created_date inserted_date channel  start_date    end_date        origin quantityLeads
0  3105052  Ebook Relat�rios Gerenciais  Institucional   2022-06-28    2022-06-28  Social  2022-06-28  2022-06-28      LinkedIn             1
1  3105052  Ebook Relat�rios Gerenciais  Institucional   2022-06-28    2022-06-28  Direct  2022-06-28  2022-06-28  Desconhecida             2
2  2918513     Ebook Direct To Consumer   Supply Chain    2022-06-28    2022-06-28  Social  2022-06-28  2022-06-28      LinkedIn             1

顺便提一下，对于上面的myList示例，在第一个条目的created_date之后缺少一个逗号，这会导致一个错误。

编辑:如果源列表中有可变数量的项目，但是每个项目都有相同的键，那么我们也可以迭代这些项。

for row in myList:
   origins_list = row["origins"]
   counter = 0
   for item in origins_list:
      row["origin_" + str(counter)] = item["origin"]
      row["quantityLeads_" + str(counter)] = item["quantityLeads"]
      counter += 1

   del row["origins"]

df = pd.DataFrame(myList)
print(df)

Answer 2

如果“源”中有多个元素，您可能会首先爆炸，创建“原产地”、"quantityLeads“，然后决定如何处理其余的数据。

df = pd.DataFrame(myList)
df = df.explode('origins')
df[['origin', 'quantityLeads']] = pd.DataFrame(df['origins'].tolist())
df.drop('origins', axis=1, inplace=True)

打印(Df)：

        id                        title       offering created_date  \
0  3105052  Ebook Relat�rios Gerenciais  Institucional   2022-06-28   
1  3105052  Ebook Relat�rios Gerenciais  Institucional   2022-06-28   
2  2918513     Ebook Direct To Consumer   Supply Chain   2022-06-28   

  inserted_date channel  start_date    end_date        origin quantityLeads  
0    2022-06-28  Social  2022-06-28  2022-06-28      LinkedIn             1  
1    2022-06-28  Direct  2022-06-28  2022-06-28  Desconhecida             2  
2    2022-06-28  Social  2022-06-28  2022-06-28      LinkedIn             1  

Answer 3

对我来说很管用。

df = pd.DataFrame(myList)
df = df.explode('origins')
df['origin'] = df.origins.str.get('origin')
df['quantityLeads'] = df.origins.str.get('quantityLeads')
df.drop('origins', axis=1, inplace=True)