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创建一个新的 pandas 列,它是列表的列表

[英]Create a new pandas column that is a list of lists

I have two tables that look something like this:我有两个看起来像这样的表:

data = [['tom', [3,5]], ['nick', [3,8]], ['juli', [3]]] 
dfA = pd.DataFrame(data, columns = ['Name', 'job_id'])


data1 = [['coder', 3], ['cook', 5], ['cop', 8]]
df_B = pd.DataFrame(data1, columns = ['job', 'job_id']) 

And I'd like to add a column to the first table such that it looks like this:我想在第一个表中添加一列,使其看起来像这样:

data_comb = [['tom', ['coder','cook']], ['nick', ['coder','cop']], ['juli', ['coder']]] 
df_comb = pd.DataFrame(data_comb, columns = ['Name', 'jobs_done'])

I am getting an unhashable list error likely due to the list within the column.由于列中的列表,我收到了一个不可散列的列表错误。 Pointers on how to solve this will be appreciated.关于如何解决这个问题的指针将不胜感激。

You could use a dictionary to map the lists:您可以使用字典 map 列表:

lookup = {key : value for value, key in data1}
dfA['job_id'] = dfA.job_id.apply(lambda x : [lookup[v] for v in x])
print(dfA)

Output Output

   Name         job_id
0   tom  [coder, cook]
1  nick   [coder, cop]
2  juli        [coder]
mapper = df_B.set_index('job_id').to_dict()['job']
dfA['job_id'] = dfA['job_id'].apply(lambda lst: [mapper.get(x) for x in lst])

output: output:

>>>dfA

    Name    job_id
0   tom     ['coder', 'cook']
1   nick    ['coder', 'cop']
2   juli    ['coder']

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