英文:
Summing values in json objects in Python
问题
这是两个JSON对象。我想要将它们合并，但在键相同的地方，字段"obj_count"应该被累加。在Python中是否有解决方法？
以下是一个示例：
这是第一个JSON对象：
[{"text": "pen and ink and watercolour", "id": "x32505 ", "obj_count": 1855},{"text": "watercolour", "id": "x33202 ", "obj_count": 674},{"text": "pencil", "id": "AAT16013 ", "obj_count": 297}]
这是第二个JSON对象：
[{"text": "pen and ink and watercolour", "id": "x32505 ", "obj_count": 807},{"text": "watercolour", "id": "x33202 ", "obj_count": 97},{"text": "ink", "id": "AAT15012 ", "obj_count": 297}]
我想要的结果如下：
[{"text":"pen and ink and watercolour","id":"x32505 ","obj_count": 2662 #累加},{"text":"watercolour","id":"x33202 ","obj_count": 771 #累加},{"text":"ink","id":"AAT15012 ","obj_count":297},{"text":"pencil","id":"AAT16013 ","obj_count":297}]
英文:
I have two JSON objects. I want to merge them but wherever the keys are the same the field obj_count should be summed. Is there any way around it in python?
Here is an example of it:
This is the 1st JSON object
[{"text": " pen and ink and watercolour", "id": "x32505 ", "obj_count": 1855},{"text": " watercolour", "id": "x33202 ", "obj_count": 674},{"text": "pencil", "id": "AAT16013 ", "obj_count": 297}]
And here is the second json object
[{"text": " pen and ink and watercolour", "id": "x32505 ", "obj_count": 807},{"text": " watercolour", "id": "x33202 ", "obj_count": 97},{"text": " ink", "id": "AAT15012 ", "obj_count": 297}]
What I want is something like this:
[{"text":" pen and ink and watercolour","id":"x32505 ","obj_count": 2662 #summed},{"text":" watercolour","id":"x33202 ","obj_count": 771 #summed},{"text":" ink","id":"AAT15012 ","obj_count":297},{"text":"pencil","id":"AAT16013 ","obj_count":297}]
答案1
得分: 3
使用一个dict来存储是否已经见过一个id
- 如果已经见过，将它们的
obj_count相加 - 如果没有见过，只需保存该项
values_a = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 1855},{"text": "watercolour", "id": "x33202", "obj_count": 674},{"text": "pencil", "id": "AAT16013", "obj_count": 297}]values_b = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 807},{"text": "watercolour", "id": "x33202", "obj_count": 97},{"text": "ink", "id": "AAT15012", "obj_count": 297}]result = {}for item in [*values_a, *values_b]:if item['id'] in result:result[item['id']]['obj_count'] += item['obj_count']else:result[item['id']] = item# 转回项目列表result = list(result.values())
英文:
Use a dict to store whether you have seen an id or not
- if you have, sum their
obj_count - if you haven't, just save the item
<!-- -->
values_a = [{"text": " pen and ink and watercolour", "id": "x32505 ", "obj_count": 1855},{"text": " watercolour", "id": "x33202 ", "obj_count": 674},{"text": "pencil", "id": "AAT16013 ", "obj_count": 297}]values_b = [{"text": " pen and ink and watercolour", "id": "x32505 ", "obj_count": 807},{"text": " watercolour", "id": "x33202 ", "obj_count": 97},{"text": " ink", "id": "AAT15012 ", "obj_count": 297}]result = {}for item in [*values_a, *values_b]:if item['id'] in result:result[item['id']]['obj_count'] += item['obj_count']else:result[item['id']] = item# back to list of itemsresult = list(result.values())
答案2
得分: 1
是的
可以使用json模块来进行加载和保存（以下代码未使用）
def sum_list_of_dict(source, add):for add_elem in add:found = Falsefor source_elem in source:if add_elem["id"] == source_elem["id"]:source_elem["obj_count"] += add_elem["obj_count"]found = Truebreak # 不应该存在重复项if not found:source.append(add_elem)return sourcedata1 = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 1855},{"text": "watercolour", "id": "x33202", "obj_count": 674},{"text": "pencil", "id": "AAT16013", "obj_count": 297},]data2 = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 807},{"text": "watercolour", "id": "x33202", "obj_count": 97},{"text": "ink", "id": "AAT15012", "obj_count": 297},]data3 = sum_list_of_dict(data1, data2)# 仅用于美观打印from pprint import pprintpprint(data3)
输出
[{'id': 'x32505', 'obj_count': 2662, 'text': 'pen and ink and watercolour'},{'id': 'x33202', 'obj_count': 771, 'text': 'watercolour'},{'id': 'AAT16013', 'obj_count': 297, 'text': 'pencil'},{'id': 'AAT15012', 'obj_count': 297, 'text': 'ink'}]
英文:
Yes
Any loading/saving can be done with the json module (not used below though)
def sum_list_of_dict(source, add):for add_elem in add:found = Falsefor source_elem in source:if add_elem["id"] == source_elem["id"]:source_elem["obj_count"] += add_elem["obj_count"]found = Truebreak # dupes should not be presentif not found:source.append(add_elem)return sourcedata1 = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 1855},{"text": "watercolour", "id": "x33202", "obj_count": 674},{"text": "pencil", "id": "AAT16013", "obj_count": 297},]data2 = [{"text": "pen and ink and watercolour", "id": "x32505", "obj_count": 807},{"text": "watercolour", "id": "x33202", "obj_count": 97},{"text": "ink", "id": "AAT15012", "obj_count": 297},]data3 = sum_list_of_dict(data1, data2)# just for pretty printingfrom pprint import pprintpprint(data3)
output
[{'id': 'x32505', 'obj_count': 2662, 'text': 'pen and ink and watercolour'},{'id': 'x33202', 'obj_count': 771, 'text': 'watercolour'},{'id': 'AAT16013', 'obj_count': 297, 'text': 'pencil'},{'id': 'AAT15012', 'obj_count': 297, 'text': 'ink'}]
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