使用列表理解将字典平整为列表的Python方法 [英] Pythonic way to flatten a dictionary into a list using list comprehension

问题描述

我具有以下功能:

def create_list_from_dict1(mydict):
    output = []
    for k, v in openint_dict.items():
        output.append( (k, v['field1'], v['field2'], v['field3']) )

    return output

从本质上讲，它使字典变平，以便我可以对返回列表中元组的字段之一进行排序.

Essentially, it flattens the dictionary, so that I can perform sorting on one of the fields of the tuple in the returned list.

我不喜欢必须硬编码"值字典的字段名称('field1'，...，'fieldN')的事实，而我想要一种更加Python化和优雅的方式这样做是为了使此函数适用于所有包含固定结构(非嵌套)字典作为其值的字典.

I don't like the fact that I am having to 'hard code' the field names of the value dictionary ('field1', ..., 'fieldN'), and I want a more pythonic and elegant way of doing this so that this function works for all dictionaries that contain a fixed structure (non-nested) dictionary as its values.

我想我将不得不使用**kwargs和/或lambda函数，以及列表理解.编写此函数的最pythonic方法是什么?

I imagine that I will have to use **kwargs and/or a lambda function, as well as list comprehension. What would be the most pythonic way to write this function?

推荐答案

您可以这样做:

fields = ("field1", "field2", "field3")

output = [[k] + [mydict[k].get(x) for x in fields] for k in mydict]

在该代码中，我们迭代dict键，并将它们与第二级字典值的选定子集相加.

In that code we iterate dict keys and add them with selected subset of second-level dictionaries values.

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