我解析了你的嵌套字典并构建了一个字典列表(每行一个).

# some sample input
nested_dict = {
    'string1': {69: [1231,'string2' :{28672: [82,'string3' :{28673: [83,24],22737:[83,94,1103,103],19424: [65,24]}
}

# new list is what we will use to hold each row
new_list = []
for k1 in nested_dict:
    curr_dict = nested_dict[k1]
    for k2 in curr_dict:
        new_dict = {'col1': k1,'col2': k2}
        new_dict.update({'col%d'%(i+3): curr_dict[k2][i] for i in range(len(curr_dict[k2]))})
        new_list.append(new_dict)

# create a DataFrame from new list
df = pd.DataFrame(new_list)

输出：

      col1   col2  col3  col4    col5   col6
0  string2  28672    82    23     NaN    NaN
1  string2  22736    82    93  1102.0  102.0
2  string2  19423    64    23     NaN    NaN
3  string3  19424    65    24     NaN    NaN
4  string3  28673    83    24     NaN    NaN
5  string3  22737    83    94  1103.0  103.0
6  string1     65     1     1     NaN    NaN
7  string1     67   682    12     NaN    NaN
8  string1     69  1231   232     NaN    NaN

假设输入将始终包含足够的数据来创建col1和col2.

我遍历nested_dict.假设nested_dict的每个元素也是字典.我们也循环遍历该字典(curr_dict).键k1和k2用于填充col1和col2.对于其余的键,我们遍历列表内容并为每个元素添加一列.