python将json嵌套到具有指定标题的csv/xlsx

Question

使用下面的json，它是最外层的对象数组，带有对象的进一步嵌套数组。

data = [{"a": [{"a1": [{"id0": [{"aa": [{"aaa": 97}, {"aab": "one"}], "ab": [{"aba": 97}, {"abb": ["one", "two"]}]}]}, {"id1": [{"aa": [{"aaa": 23}]}]}]}, {"a2": []}]}, {"b": [{"b1": [{"Common": [{"bb": [{"value": 4}]}]}]}]}]

我需要将其写入 csv（或 .xlsx 文件）

到目前为止我尝试了什么？

data_file = open('data_file.csv', 'w')
csv_writer = csv.writer(data_file)
for row in data:
  csv_writer.writerow(row)
data_file.close() 

这会给出一个空文件“data_file.csv”。

另外，我如何将标题添加到 CSV。我将标题存储在如下列表中

hdrs = ['Section', 'Subsection', 'pId', 'Group', 'Parameter', 'Value'] 

- 这对应于五个级别的键

预期的 CSV 输出

+---------+------------+--------+-------+-----------+----------+
| Section | Subsection |  pId   | Group | Parameter |  Value   |
+---------+------------+--------+-------+-----------+----------+
| a       | a1         | id0    | aa    | aaa       | 97       |
| a       | a1         | id0    | aa    | aab       | one      |
| a       | a1         | id0    | ab    | aba       | 97       |
| a       | a1         | id0    | ab    | abb       | one, two |
| a       | a1         | id1    | aa    | aaa       | 23       |
| a       | a2         |        |       |           |          |
| b       | b1         | Common | bb    | value     | 4        |
+---------+------------+--------+-------+-----------+----------+

预期的 XLSX 输出

Answer 1

以下代码能够按照预期格式解析提供的数据。

from typing import List

def parse_recursive(dat)->List[List]:
    ret=[]
    if type(dat) is list:
        for item in dat:
            if type(item)==dict:
                for k in item:
                    #print(k, item[k], sep=" # ")#debug print
                    if item[k]==[]: #empty list
                        ret.append([k])
                    else:
                        for l in parse_recursive(item[k]):
                            #print(k,l,sep=" : ") #debug print
                            ret.append([k]+l) #always returns List of List
            else: #Right now only possibility is string eg. "one", "two"
                return [[",".join(dat)]]
    else: #can be int or string eg. 97, "23"
        return [[dat]]

    return ret


def write_to_csv(file_name:str, fields:List, row_data:List[List]):
    import csv
    with open(file_name, 'w') as csvfile:  
        # creating a csv writer object  
        csvwriter = csv.writer(csvfile)  
        # writing the fields  
        csvwriter.writerow(fields)  
        # writing the data rows  
        csvwriter.writerows(row_data)


if __name__=="__main__":
    org_data = [{"a": [
        {"a1": [
            {"id0": [
                {
                    "aa": [
                        {"aaa": 97},
                        {"aab": "one"}],
                    "ab": [
                        {"aba": 97},
                        {"abb": ["one", "two"]}
                        ]
                }
            ]
            },
            {"id1": [
                {"aa": [
                    {"aaa": 23}]}]}
            ]
        },
        {"a2": []}
        ]},
        {"b": [{"b1": [{"Common": [{"bb": [{"value": 4}]}]}]}]}]
    print(parse_recursive(org_data)) #Debug

    file_name="data_file.csv"
    fields=['Section', 'Subsection', 'pId', 'Group', 'Parameter', 'Value']
    write_to_csv(file_name, fields, parse_recursive(org_data))

parse_recursive 尝试按照我尝试从您的输入和输出格式推断的规则解析任意深度字典。

以下是您提供的输入的parse_recursive 的输出 -

mahorir@mahorir-Vostro-3446:~/Desktop$ python3 so.py 
[['a', 'a1', 'id0', 'aa', 'aaa', 97], ['a', 'a1', 'id0', 'aa', 'aab', 'one'], ['a', 'a1', 'id0', 'ab', 'aba', 97], ['a', 'a1', 'id0', 'ab', 'abb', 'one,two'], ['a', 'a1', 'id1', 'aa', 'aaa', 23], ['a', 'a2'], ['b', 'b1', 'Common', 'bb', 'value', 4]]

write_to_csv 是一个写入 csv 文件的简单函数。

Answer 2

这是一个有趣的问题... 这里的格式确实有两个问题：

1. 数据是字典列表，他们真的只是想要字典。例如他们想要{"foo": 1, "bar": 2}，但将其格式化为[{"foo": 1}, {"bar": 2}]。

   一个。我不在这里评判。他们这样做可能是有原因的。只是让我们解析有点烦。

2. 数据有时会被截断；如果通常有 5 个级别的深度，有时如果他们没有超过一个点的数据，他们就会忽略它。例如'a2' 在您的示例中。

所以我将展示解决这些问题的两种可能方法。

熊猫之路

此解决方案与此处提到的其他解决方案有些不同。让我知道你的想法：

import pandas as pd
from copy import deepcopy

hdrs = ['Section', 'Subsection', 'pId', 'Group', 'Parameter', 'Value']

js = [{"a": [{"a1": [{"id0": [{"aa": [{"aaa": 97}, {"aab": "one"}],
                               "ab": [{"aba": 98}, {"abb": ["one", "two"]}]}]},
                     {"id1": [{"aa": [{"aaa": 23}]}]}
                    ]},
             {"a2": []}
            ]},
      {"b": [{"b1": [{"Common": [{"bb": [{"value": 4}]}]}]}]}]

def list_to_dict(lst):
    """convert a list of dicts as you have to a single dict

    The idea here is that you have a bunch of structures that look
    like [{x: ...}, {y: ...}] that should probably have been stored as
    {x:..., y:...}. So this function does that (but just one level in).
    
    Note:
    If there is a duplicate key in one of your dicts (meaning you have
    something like [{x:...},...,{x:...}]), then this function will overwrite
    it without warning!
    """
    d = {}
    for new_d in lst:
        d.update(new_d)
    return d

def recursive_parse(lst, levels):
    "Parse the nested json into a single pandas dataframe"
    name = levels.pop(0)  # I should have used a counter instead
    d = list_to_dict(lst)  # get a sensible dict instead of the list of dicts
    if len(levels) <= 1: # meaning there are no more levels to be parsed.
        if len(d) == 0:
            d = {'': ''} # to handle the uneven depths (e.g. think 'a2')
        return pd.Series(d, name=levels[-1])
    if len(d) == 0: # again to handle the uneven depths of json
        d = {'': []}
    # below is a list-comprehension to recursively parse the thing.
    d = {k: recursive_parse(v, deepcopy(levels)) for k, v in d.items()}
    return pd.concat(d)

def json_to_df(js, headers):
    "calls recursive_parse, and then adds the column names and whatnot"
    df = recursive_parse(js, deepcopy(headers))
    df.index.names = headers[:-1]
    df = df.reset_index()
    return df
df = json_to_df(js, hdrs)
display(df)

输出正是您想要的数据框（但带有您可能不想要的索引列）。如果您之后将其写入 csv，请这样做：

df.to_csv('path/to/desired/file.csv', index=False)

这有意义吗？

极简方式

更好的版本（不使用 pandas）...

import csv

hdrs = ['Section', 'Subsection', 'pId', 'Group', 'Parameter', 'Value']

js = [{"a": [{"a1": [{"id0": [{"aa": [{"aaa": 97}, {"aab": "one"}],
                               "ab": [{"aba": 98}, {"abb": ["one", "two"]}]}]},
                     {"id1": [{"aa": [{"aaa": 23}]}]}
                    ]},
             {"a2": []}
            ]},
      {"b": [{"b1": [{"Common": [{"bb": [{"value": 4}]}]}]}]}]

def list_of_dicts_to_lists(lst, n_levels=len(hdrs)):
    if n_levels == 1:
        if isinstance(lst, list):
            if len(lst) == 0: # we fill the shorter ones with empty lists
                lst = None # replacing them back to None
            else: # [1, 2] => "1,2"
                lst = ','.join(str(x) for x in lst if x is not None)
        return [[lst]] # the later ones are going to be lists of lists so let's start out that way to keep everything consistent.
    if len(lst) == 0:
        lst = [{None: []}] # filling with an empty list
    output = []
    for d in lst:
        for k, v in d.items():
            tmp = list_of_dicts_to_lists(v, n_levels - 1)
            for x in tmp:
                output.append([k] + x)
    return output

def to_csv(values, header, outfile):
    with open(outfile, 'w', newline='') as csv_file:
        # pretty much straight from the docs @
        # https://docs.python.org/3.7/library/csv.html
        csv_writer = csv.writer(csv_file, quoting=csv.QUOTE_MINIMAL)
        csv_writer.writerow(header)
        for line in values:
            csv_writer.writerow(line)
    return True

rows = list_of_dicts_to_lists(js)
to_csv(rows, hdrs, 'tmp.csv')

我现在看到这个解决方案与这里的其他答案非常相似......我的错。