Question

实际值包含我需要分组的完整列表。需要拆分为动态组。

它应该拆分为对象数组。

c = {
      'Date#1': '07/03/2018', 
      'Item#1': '789807', 
      'Description#1': 'Wooden Blocks', 
      'Qty#1': '4', 
      'Unit_Price#1': '$10.00', 
      'Discount#1': '$2.00', 
      'Total#1': '$38.00', 
      'Date#2': '07/03/2018', 
      'Item#2': '789808', 
      'Description#2': 'Magnetic Alphabets', 
      'Qty#2': '5', 
      'Unit_Price#2': '$10.00', 
      'Discount#2': '$2.00', 
      'Total#2': '$48.00', 
      'Date#3': '07/03/2018', 
      'Item#3': '769804', 
      'Description#3': 'Building Blocks Flat', 
      'Qty#3': '3', 
      'Unit_Price#3': '$23.00', 
      'Discount#3': '$2.00', 
      'Total#3': '$67.00'
 }

实际值包含我需要分组的完整列表。

需要拆分为动态组。它应该拆分为对象数组。我需要的是

 c = [{
        'Date#1': '07/03/2018', 
        'Item#1': '789807', 
        'Description#1': 'Wooden Blocks', 
        'Qty#1': '4', 
        'Unit_Price#1': '$10.00', 
        'Discount#1': '$2.00', 
        'Total#1': '$38.00' 
       },
       {
        'Date#2': '07/03/2018', 
        'Item#2': '789808', 
        'Description#2': 'Magnetic Alphabets', 
        'Qty#2': '5', 
        'Unit_Price#2': '$10.00', 
        'Discount#2': '$2.00', 
        'Total#2': '$48.00'
       },
       {
        'Date#3': '07/03/2018', 
        'Item#3': '769804', 
        'Description#3': 'Building Blocks Flat', 
        'Qty#3': '3', 
        'Unit_Price#3': '$23.00', 
        'Discount#3': '$2.00', 
        'Total#3': '$67.00'
       }]

Answer 1

直到最近，在Python中，dict个对象都是无序的。要订购字典类型，您需要使用OrderedDict模块（docs here）中的collections：

from collections import OrderedDict
from pprint import pprint

c = OrderedDict((
      ('Date#1', '07/03/2018'),
      ('Item#1', '789807'),
      ('Description#1', 'Wooden Blocks'),
      ('Qty#1', '4'),
      ('Unit_Price#1', '$10.00'),
      ('Discount#1', '$2.00'),
      ('Total#1', '$38.00'),
      ('Date#2', '07/03/2018'),
      ('Item#2', '789808'),
      ('Description#2', 'Magnetic Alphabets'),
      ('Qty#2', '5'),
      ('Unit_Price#2', '$10.00'),
      ('Discount#2', '$2.00'),
      ('Total#2', '$48.00'),
      ('Date#3', '07/03/2018'),
      ('Item#3', '769804'),
      ('Description#3', 'Building Blocks Flat'),
      ('Qty#3', '3'),
      ('Unit_Price#3', '$23.00'),
      ('Discount#3', '$2.00'),
      ('Total#3', '$67.00')
 ))

groups, actual_group = [], OrderedDict()

for k, v in c.items():
    if k.startswith('Date'):
        if actual_group:
            groups.append(actual_group)
        actual_group = OrderedDict()
    actual_group[k] = v

if actual_group:
    groups.append(actual_group)

pprint(groups)

此代码将产生3组OrderedDict：

[OrderedDict([('Date#1', '07/03/2018'),
              ('Item#1', '789807'),
              ('Description#1', 'Wooden Blocks'),
              ('Qty#1', '4'),
              ('Unit_Price#1', '$10.00'),
              ('Discount#1', '$2.00'),
              ('Total#1', '$38.00')]),
 OrderedDict([('Date#2', '07/03/2018'),
              ('Item#2', '789808'),
              ('Description#2', 'Magnetic Alphabets'),
              ('Qty#2', '5'),
              ('Unit_Price#2', '$10.00'),
              ('Discount#2', '$2.00'),
              ('Total#2', '$48.00')]),
 OrderedDict([('Date#3', '07/03/2018'),
              ('Item#3', '769804'),
              ('Description#3', 'Building Blocks Flat'),
              ('Qty#3', '3'),
              ('Unit_Price#3', '$23.00'),
              ('Discount#3', '$2.00'),
              ('Total#3', '$67.00')])]

python 3 json分成几组

1 个答案: