问题描述
我和一个朋友有一个项目的数据:
arr = [{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'Thirdkey 1', 'Value':100},
{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':130},
{'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 1', 'Value':230},
{'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':300},
{'Key':'Key 2', 'SecondKey':'SecondKey 4', 'ThirdKey':'ThirdKey 1', 'Value':111},
{'Key':'Key 2', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':400},
{'Key':'Key 2', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':230}
]
我试过这样的功能:
def array_to_dict(arr, classification):
dct = tree()
for d in arr:
dct[d["Key"]]=(d)
pprint.pprint(dct)
def tree():
return defaultdict(tree)
def array_to_dict_recursive(arr, classification):
result = defaultdict(arr)
for v, k in arr:
result[k].append(v)
final_result = [{'type': k, 'items': v} for k, v in result.items()]
print (str(final_result))
def array_cool(arr):
#pprint.pprint(arr)
arr.sort(key=operator.itemgetter('Key'))
pprint.pprint(arr)
list1= []
print("")
for key, items in itertools.groupby(arr, operator.itemgetter('Key')):
list1.append(list(items))
pprint.pprint(list1)
我希望它以这种方式显示为 JSON 文件:
{
"Key 1": {
"SecondKey 2": {
"ThirdKey 2": [
{
"Key": "Key 1",
"Value": 300,
"SecondKey": "SecondKey 2",
"ThirdKey": "ThirdKey 2"
}
],
"ThirdKey 1": [
{
"Key": "Key 1",
"Value": 230,
"SecondKey": "SecondKey 2",
"ThirdKey": "ThirdKey 1"
}
]
},
"SecondKey 1": {
"ThirdKey 2": [
{
"Key": "Key 1",
"Value": 130,
"SecondKey": "SecondKey 1",
"ThirdKey": "ThirdKey 2"
}
],
"ThirdKey 1": [
{
"Key": "Key 1",
"Value": 100,
"SecondKey": "SecondKey 1",
"ThirdKey": "ThirdKey 1"
}
]
}
},
"Key 2": {
"SecondKey 4": {
"ThirdKey 1": [
{
"Key": "Key 2",
"Value": 111,
"SecondKey": "SecondKey 4",
"ThirdKey": "ThirdKey 1"
}
]
},
"SecondKey 2": {
"ThirdKey 2": [
{
"Key": "Key 2",
"Value": 400,
"SecondKey": "SecondKey 2",
"ThirdKey": "ThirdKey 2"
}
]
},
"SecondKey 1": {
"ThirdKey 2": [
{
"Key": "Key 2",
"Value": 230,
"SecondKey": "SecondKey 1",
"ThirdKey": "ThirdKey 2"
}
]
}
}
}
我尝试过排序,但是排序将 Key 放在第二个位置,并干扰了下一个排序过程,它不起作用。
1楼
尝试
print json.dumps(arr, indent=4, sort_keys=True)
有关更多信息,请参阅文档:
json.dumps(obj, skipkeys=False, ensure_ascii=True, check_circular=True, allow_nan=True, cls=None, indent=None, separators=None, encoding="utf-8", default=None, sort_keys=False, **千瓦)
2楼
Python dicts 是未排序的。 你有几个选择。
使用将排序应用于标准字典的工具,例如:
import json
print json.dumps(thedict, indent=4, sort_keys=True)
使用保持 dict 条目创建顺序的类型:
from collections import OrderedDict
d = OrderedDict()
d[KEY] = VALUE
d[KEY][SUBKEY] = VALUE
etc.
或者最简单的,只需按照您想要的顺序将它们添加到数组/列表中。
3楼
这是你想要的第一次尝试。 它会让你开始,然后你可以简化它并让它变得漂亮。 关键点:
-
json.dumps()
按照看到的顺序输出字典键,因此明确排序的 OrderedDict 可能是您获得它的唯一方法。 - 您的结果与输入的数据结构不同,因此您必须构建它
- Python 和 JSON 中的字典没有有序键。 因此,当您将结果导入其他内容时,订单可能会丢失。
- 这段代码非常低效,并且可能在大型数据集上表现不佳。
您的数据结构:
-
每个
Key
字典,按顺序排序-
每个
SecondKey
字典,反向排序-
每个
ThirdKey
dict ,反向排序-
一个包含每个值的字典(没有对值进行排序)的列表,其中包含这些已排序的键和值
-
Key1:外循环的
Key
- 值:当前项目的值
-
SecondKey:来自外循环的
SecondKey
值 -
第三键:来自外循环的
ThirdKey
键值
-
Key1:外循环的
-
一个包含每个值的字典(没有对值进行排序)的列表,其中包含这些已排序的键和值
-
每个
-
每个
.
from collections import OrderedDict
import json
arr = [{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 1', 'Value':100},
{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':130},
{'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 1', 'Value':230},
{'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':300},
{'Key':'Key 2', 'SecondKey':'SecondKey 4', 'ThirdKey':'ThirdKey 1', 'Value':111},
{'Key':'Key 2', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':400},
{'Key':'Key 2', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':230}
]
def arr_to_ordereddict():
d = OrderedDict()
keys = set(a['Key'] for a in arr)
#print(sorted(keys))
for k1 in sorted(keys):
k1_dict = OrderedDict()
d[k1] = k1_dict
arr2 = [a for a in arr if a['Key'] == k1]
keys2 = set(a['SecondKey'] for a in arr2)
#print('\t', k1, sorted(keys2, reverse=True))
for k2 in sorted(keys2, reverse=True):
k2_dict = OrderedDict()
k1_dict[k2] = k2_dict
arr3 = [a for a in arr if a['SecondKey'] == k2]
keys3 = set(a['ThirdKey'] for a in arr3)
#print('\t\t', k1, k2, sorted(keys3, reverse=True))
for k3 in sorted(keys3, reverse=True):
k3_list = []
k2_dict[k3] = k3_list
for item in (a for a in arr3 if a['ThirdKey'] == k3):
value_dict = OrderedDict()
value_dict['Key'] = k1
value_dict['Value'] = item['Value']
value_dict['SecondKey'] = k2
value_dict['ThirdKey'] = k3
k3_list.append(value_dict)
return d
print(json.dumps(arr_to_ordereddict(), indent=4))