展平python中每个键的值

我有一个这样的字典：

migration_dict = {'30005': ['key42750','key43119','key44103',['key333'],['key444'],['keyxx']],'30003': ['key43220','key42244','key42230',['keyzz'],['kehh']]}

我怎样才能压平每个键的值以便得到类似的东西：

migration_dict = {'30005': ['key42750','key333','key444','keyxx'],'keyzz','kehh']}

解决方法

您可以编写递归函数来展平值列表,并在字典理解中使用它来构建新字典：

def flatten(lst):
   for x in lst:
      if isinstance(x,list):
         for y in flatten(x): # yield from flatten(...) in Python 3
            yield y           #
      else:
         yield x

migration_dict = {k: list(flatten(v)) for k,v in dct.items()}
print(migration_dict)
# {'30005': ['key42750','kehh']}

它处理dict值列表中的任何嵌套深度.