如何在 python 中的数据框中设置字典中的值

Question

我有一个如下所示的数据框：

创建 df 的代码：

dd = {'name': ["HARDIE'S MOBILE HOME PARK", 'CRESTVIEW RV PARK',
       'HOMESTEAD TRAILER PARK', 'HOUSTON PARK MOBILE HOME PARK',
       'HUDSON MOBILE HOME PARK', 'BEACH DRIVE MOBILE HOME PARK',
       'EVANS TRAILER PARK'],
       'country': ['USA', 'USA', 'USA', 'USA', 'USA', 'USA', 'USA'],
      'coordinates': ['30.44126118, -86.6240656099999',
       '30.7190163500001, -86.5716222299999',
       '30.5115772500001, -86.4628417499999',
       '30.4424195300001, -86.64733076',
       '30.7629176200001, -86.5928893399999', '30.44417349, -86.59951996',
       '30.4427800300001, -86.62941091'],
      'status':['OPEN', 'CLOSED', 'OPEN', 'OPEN', 'OPEN', 'OPEN', 'OPEN']}

df2 = pd.DataFrame(data=dd)

我想做的是创建一个具有以下结构的字典：

{'destination1': 'CRESTVIEW RV PARK; 30.7190163500001, -86.5716222299999',
 'destination2': 'HOMESTEAD TRAILER PARK; 30.5115772500001, -86.4628417499999',
 'destination3': 'HOUSTON PARK MOBILE HOME PARK; 30.4424195300001, -86.64733076',
 'destination4': 'HUDSON MOBILE HOME PARK; 30.7629176200001, -86.5928893399999',
 'destination5': 'BEACH DRIVE MOBILE HOME PARK  ; 30.44417349, -86.59951996'}

如您所见，每个值都必须包含名称；从第二行到最后一行的坐标。我正在使用以下代码来执行此操作：

d1 = {f"destination{k}":v + "; " + i for k in range(1, len(df1)-1) for v,i in zip(df1.name, df1.coordinates)}

但是，这是我得到的输出：

{'destination1': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091',
 'destination2': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091',
 'destination3': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091',
 'destination4': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091',
 'destination5': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091'}

它只读取数据帧的最后一行，每个键都有相同的值，但我想要的是，对于每个键，它的值必须来自数据帧的每一行，从第二行到最后一行.

如果有人知道如何做到这一点，我将非常感谢您的帮助。

Answer 1

你可以这样枚举 zip，

dd = {'name': ["HARDIE'S MOBILE HOME PARK", 'CRESTVIEW RV PARK',
       'HOMESTEAD TRAILER PARK', 'HOUSTON PARK MOBILE HOME PARK',
       'HUDSON MOBILE HOME PARK', 'BEACH DRIVE MOBILE HOME PARK',
       'EVANS TRAILER PARK'],
       'country': ['USA', 'USA', 'USA', 'USA', 'USA', 'USA', 'USA'],
      'coordinates': ['30.44126118, -86.6240656099999',
       '30.7190163500001, -86.5716222299999',
       '30.5115772500001, -86.4628417499999',
       '30.4424195300001, -86.64733076',
       '30.7629176200001, -86.5928893399999', '30.44417349, -86.59951996',
       '30.4427800300001, -86.62941091'],
      'status':['OPEN', 'CLOSED', 'OPEN', 'OPEN', 'OPEN', 'OPEN', 'OPEN']}

df1 = pd.DataFrame(data=dd)

d_out = {
    f"destination{idx+1}":'; '.join(v) for idx, v in enumerate(zip(df1.name[1:], df1.coordinates[1:]))
}

d_out

{'destination1': 'CRESTVIEW RV PARK; 30.7190163500001, -86.5716222299999',
 'destination2': 'HOMESTEAD TRAILER PARK; 30.5115772500001, -86.4628417499999',
 'destination3': 'HOUSTON PARK MOBILE HOME PARK; 30.4424195300001, -86.64733076',
 'destination4': 'HUDSON MOBILE HOME PARK; 30.7629176200001, -86.5928893399999',
 'destination5': 'BEACH DRIVE MOBILE HOME PARK; 30.44417349, -86.59951996',
 'destination6': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091'}

您不必通过字典理解来获得此结果，如果您可以像这样在 pandas 数据框中创建几列，就可以获得此结果。

df1['destination'] = [f"destination{k}" for k in range(len(df1))]
df1['value'] = df1['name'] + "; " + df1['coordinates'] 

df1[['destination', 'value']][1:].set_index("destination").to_dict()['value']

{'destination1': 'CRESTVIEW RV PARK; 30.7190163500001, -86.5716222299999',
 'destination2': 'HOMESTEAD TRAILER PARK; 30.5115772500001, -86.4628417499999',
 'destination3': 'HOUSTON PARK MOBILE HOME PARK; 30.4424195300001, -86.64733076',
 'destination4': 'HUDSON MOBILE HOME PARK; 30.7629176200001, -86.5928893399999',
 'destination5': 'BEACH DRIVE MOBILE HOME PARK; 30.44417349, -86.59951996',
 'destination6': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091'}

Answer 2

您示例中的字典理解有两个 for 循环：

d1 = {
    f"destination{k}":v + "; " + i
    for k in range(1, len(df1)-1)
    for v,i in zip(df1.name, df1.coordinates)
}

在这些循环中，k 独立于 v 和 i 进行迭代。第二个循环有很多问题（要理解它们，只需逐步执行 df1.name、df1.coordinates 和 zip(df1.name, df1.coordinates) 操作，看看这是怎么回事——请注意 df1.name 是一个保留属性，指的是数据框的名称，而不是“名称”列）。

您真正想要的是为每一行循环遍历 df1 中的多个元素。为此，只需使用第一个循环，但在构建值时从 df 访问所需的元素：

d1 = {
    f"destination{k}": (df1.loc[k, 'name'] + "; " + df1.loc[k, 'coordinates'])
    for k in range(1, len(df1)-1)
}

查看有关理解的 this FullStack Python guide's 部分了解更多信息。

或者，（最好）使用 pandas!

d1 = pd.Series(
    df1['name'] + '; ' + df['coordinates'],
    index=('destination' + df.index.astype(str)),
)

如果此时您确实需要字典，可以使用 d1 = d1.to_dict()

将系列转换为字典

如何在 python 中的数据框中设置字典中的值

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