Question

Json对象（输出）： [424783，[198184]，[605]，[644]，[296]，[2048]，424694，[369192]，[10139]， [152532]，[397538]，[1420]]

<<<已删除代码>>>

所需的输出：

424783,198184
424783,605
424783,644
424783,296
424783,2048
424694,369192
424694,10139
424694,152532
424694,397538
424694,1420

Answer 1

从您的数据看来，应该将非方括号的项目视为第一列的值（即 key ），将方括号的项目视为第二列的值，使用< em> key 在它们之前。您可以纯粹以程序方式进行此操作：

getFilteredList

哪个应该产生包含以下内容的import csv import json src = '''[424783, [198184], [605], [644], [296], [2048], 424694, [369192], [10139], [152532], [397538], [1420]]''' with open('output.csv', 'w', newline='') as f: # Python 2.x: open('output.csv', 'wb') writer = csv.writer(f) # create a simple CSV writer current_key = None # a container for the last seen / cached 'key' for element in json.loads(src): # parse the structure and iterate over it if isinstance(element, list): # if the element is a 'list' writer.writerow((current_key, element[0])) # write to csv w/ cached key else: current_key = element # cache the element as the key for following entries：

424783,198184
424783,605
424783,644
424783,296
424783,2048
424694,369192
424694,10139
424694,152532
424694,397538
424694,1420

Answer 2

itertools.groupby对于Python初学者来说有点挑战，但是在浏览一系列项目并分组处理时非常方便。在这种情况下，我们按不是Python列表的项目进行分组。

从每组嵌套的int中，我们将在累加器列表中创建一个或多个条目。

一旦加载了累加器列表，下面的代码将打印出结果，轻松转换为写入文件。

import ast
from itertools import groupby
from collections import namedtuple


# this may be JSON, but it's also an ordinary Python nested list of ints, so safely parseable using
# ast.literal_eval()
text = "[424783, [198184], [605], [644], [296], [2048], 424694, [369192], [10139], [152532], [397538], [1420]]"
items = ast.literal_eval(text)

# a namedtuple to hold each record, and a list to accumulate them
DataRow = namedtuple("DataRow", "old_id new_id")
accumulator = []

# use groupby to process the entries in groups, depending on whether the items are lists or not
key = None
for is_data, values in groupby(items, key=lambda x: isinstance(x, list)):
    if not is_data:
        # the sole value the next record key
        key = list(values)[0]
    else:
        # the values are the collection of lists until the next key
        accumulator.extend(DataRow(key, v[0]) for v in values)

# dump out as csv
for item in accumulator:
    print("{old_id},{new_id}".format_map(item._asdict()))

打印：

424783,198184
424783,605
424783,644
424783,296
424783,2048
424694,369192
424694,10139
424694,152532
424694,397538
424694,1420

Answer 3

我认为使用itertools.groupby()是一个很好的方法，因为对项目进行分组是完成所需内容的主要工作。

这是一种相当简单的使用方法：

import csv
from itertools import groupby
import json

json_src = '''[424783, [198184], [605], [644], [296], [2048],
               424694, [369192], [10139], [152532], [397538], [1420]]'''

def xyz():
    return json.loads(json_src)

def abc():
    json_processed = xyz()
    output_filename = 'y.csv'

    with open(output_filename, 'w', newline='') as csvfile:
        writer = csv.writer(csvfile)
        for is_list, items in groupby(json_processed, key=lambda v: isinstance(v, list)):
            if is_list:
                new_ids = [item[0] for item in items]
            else:
                old_id = next(items)
                continue

            for new_id in new_ids:
                writer.writerow([old_id, new_id])

abc()

产生的csv文件内容：

424783,198184
424783,605
424783,644
424783,296
424783,2048
424694,369192
424694,10139
424694,152532
424694,397538
424694,1420

尝试将以下json输出为csv

3 个答案: