如何将pandas列转换为逐行列表，并以逐行列为基础创建字典

Question

我有一个像下面这样的数据框架。

id1            ids1                         Name1        Name2      ids2                     ID     col1  Goal     col2    col3       
0   85643        234,34,11223,345,345_2         aasd1        vaasd1    2234,354,223,35,3435     G-0001     1   NaN       3       1      
1   85644        2343,355,121,34                aasd2                                           G-0001     2   56.0000   4       22     
2   8564312      24 , 23 ,244 ,2421 ,567 ,789   aabsd1                                          G-0002     3   NaN       32      33     
3   8564314      87 ,35 ,67_1                   aabsd2       averabsd   387 ,355 ,667_1         G-00002    4   89.0000   43      44 

我想基于name1、ids1和id1列以及name2、ids2和id1列创建列表。在列ids1和ids2中有一些额外的空间，需要在传输到列表时删除。例如，最终结果如下所示:

aasd1 = ['85643','234','34','11223','345','345_2']
vaasd1 = ['85643','2234','354','223','35','3435']
aasd2 = ['85644','2343','355','121','34']
aabsd1= ['8564312','24','23','244','2421','567','789'] 
...........

另外，我想基于name1和name2列创建一个字典。就像下面

dic = {"aasd1":aasd1, "vaasd1":vaasd1, "aasd2":aasd2, "vaasd1":vaasd1, "aabsd1":aabsd1, "aabsd2":aabsd2, "aabsd2":aabsd2}

任何帮助都会很感激。提前感谢!!

Answer 1

使用说明:

df['ids1'] = df['ids1'].str.split(',').apply(lambda x: [val.strip() for val in x]).tolist()
data = df.set_index('Name1').apply(lambda x: [x['id1']] + x['ids1'], axis=1).to_dict()

OUTPUT:

for key in data.keys():
print(key, ': ', data[key])
#OUTPUT
aasd1 :  [85643, '234', '34', '11223', '345', '345_2']
aasd2 :  [85644, '2343', '355', '121', '34']
aabsd1 :  [8564312, '24', '23', '244', '2421', '567', '789']
averabsd :  [8564314, '87', '35', '67_1']

For Second Part use:

data = df.assign(d=df['Name1']).set_index('d')['Name1'].to_dict()

OUTPUT:

{'aasd1': 'aasd1', 'aasd2': 'aasd2', 'aabsd1': 'aabsd1', 'averabsd': 'averabsd'}