Python - 如何将嵌套字典加载到 Pandas 数据框中？

Question

我有一个长嵌套字典，结构如下，我该如何将其加载到 Pandas 数据框中？ Feed、Spindle Speed 和 Tool 的子键始终保持不变，但上面的两个级别(Heading) >、N1 等以及 4001、4002 等在整个词典中都是唯一的，或者至少在顺序上是唯一的。

我知道这样的事情:

pd.DataFrame.from_dict({(i,j): dictionary[i][j] 
                           for i in dictionary.keys() 
                           for j in dictionary[i].keys()},
                       orient='index')

但这看起来像一个数据透视表，我更喜欢使用带有冗余信息的数据框(例如4001)来运行整个列。

{
    "4001": {
        "Heading": {
            "Feed": [],
            "Spindle Speed": [],
            "Tool": []
        },
        "N1": {
            "Feed": [],
            "Spindle Speed": [],
            "Tool": [
                "0800"
            ]
        },
        "N10 ": {
            "Feed": [
                0.01,
                0.0006,
                0.0001,
                0.0006,
                0.0001,
                0.0006,
                0.0002,
                0.02,
                0.0004
            ],
            "Spindle Speed": [
                "M3S2630"
            ],
            "Tool": [
                "1616"
            ]
        }
    },
    "4002": {
        "Heading": {
            "Feed": [],
            "Spindle Speed": [],
            "Tool": []
        },
        "N1": {
            "Feed": [],
            "Spindle Speed": [],
            "Tool": [
                "9900"
            ]
        },
        "N10": {
            "Feed": [
                0.01,
                0.001,
                0.0004,
                0.001,
                0.005
            ],
            "Spindle Speed": [],
            "Tool": [
                "3838"
            ]
        }
    },     
    "4003": {...
             ...
             ...

理想情况下，数据框看起来像这样:

Program Operation Number    Feed        Tool      Spindle Speed
4001    Heading             []          []        []
4001    N1                  []          ['0800']  []
4001    N10                 [0.01, ...] ['1616']  ['M3S2630']

Answer 1

你就快到了。您只需重置 multi_index 并给出正确的列名称:

pd.DataFrame.from_dict({(i,j): dictionary[i][j] 
                           for i in dictionary.keys() 
                           for j in dictionary[i].keys()},
                       orient='index').reset_index().rename(
    {'level_0': 'Program', 'level_1': 'Operation Number'}, axis=1)