Question

以下格式的数据帧必须像“ op_df”那样转换，

ip_df=pd.DataFrame({'class':['I','II','III'],'details':[[{'sec':'A','assigned_to':'tom'},{'sec':'B','assigned_to':'sam'}],[{'sec':'B','assigned_to':'joe'}],[]]})
ip_df:
     class    details
0    I        [{'sec':'A','assigned_to':'tom'},{'sec':'B','assigned_to':'sam'}]
1    II       [{'sec':'B','assigned_to':'joe'}]
2    III      []

假定所需的输出数据帧为

 op_df:
      class sec assigned_to 
 0    I     A   tom    
 1    I     B   sam
 2    II    B   joe
 3    III   NaN NaN

如何将“详细信息”列的每个字典更改为新行，并以字典的键作为列名和字典的值作为其各自的列值？

我尝试过，

ip_df.join(ip_df['details'].apply(pd.Series))

而我无法像“ op_df”那样构图。

Answer 1

我确信有更好的方法，但是我必须解构您的详细信息列表，并按如下所示创建数据框：

dict_values = {'class':['I','II','III'],'details':[[{'sec':'A','assigned_to':'tom'},{'sec':'B','assigned_to':'sam'}],[{'sec':'B','assigned_to':'joe'}],[]]}

all_values = []

for cl, detail in zip(dict_values['class'], dict_values['details']):

    if len(detail) > 0:
        for innerdict in detail:
            row = {'class': cl}
            for innerkey in innerdict.keys():
                row[innerkey] = innerdict[innerkey]
            all_values.append(row)
    else:
        row = {'class': cl}
        all_values.append(row)

op_df = pd.DataFrame(all_values)

熊猫：如何将一列中的字典列表映射为新行

1 个答案: