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 xmigration_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.
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