我有一个简单的Json文件
input.json
[
{
"title": "Person",
"type": "object",
"required": "firstName",
"min_max": "200/600"
},
{
"title": "Person1",
"type": "object2",
"required": "firstName1",
"min_max": "230/630"
},
{
"title": "Person2",
"type": "object2",
"required": "firstName2",
"min_max": "201/601"
},
{
"title": "Person3",
"type": "object3",
"required": "firstName3",
"min_max": "2000/6000"
},
{
"title": "Person4",
"type": "object4",
"required": "firstName4",
"min_max": "null"
},
{
"title": "Person4",
"type": "object4",
"required": "firstName4",
"min_max": "1024 / 256"
},
{
"title": "Person4",
"type": "object4",
"required": "firstName4",
"min_max": "0"
}
]
我正在尝试使用新数据创建一个新的json文件。我想拆分" min_max"进入两个不同的领域,即。,min和max。下面是用python编写的代码。
import json
input=open('input.json', 'r')
output=open('test.json', 'w')
json_decode=json.load(input)
result = []
for item in json_decode:
my_dict={}
my_dict['title']=item.get('title')
my_dict['min']=item.get('min_max')
my_dict['max']=item.get('min_max')
result.append(my_dict)
data=json.dumps(result, output)
output.write(data)
output.close()
如何将字符串拆分为两个不同的值。此外,是否有可能按顺序打印json输出。
答案 0 :(得分:0)
表+ Python == Pandas
import pandas as pd
# Read old json to a dataframe
df = pd.read_json("input.json")
# Create two new columns based on min_max
# Removes empty spaces with strip()
# Returns [None,None] if length of split is not equal to 2
df['min'], df['max'] = (zip(*df['min_max'].apply
(lambda x: [i.strip() for i in x.split("/")]
if len(x.split("/"))== 2 else [None,None])))
# 'delete' (drop) min_max column
df.drop('min_max', axis=1, inplace=True)
# output to json again
df.to_json("test.json",orient='records')
结果:
[{'max': '600',
'min': '200',
'required': 'firstName',
'title': 'Person',
'type': 'object'},
{'max': '630',
'min': '230',
'required': 'firstName1',
'title': 'Person1',
'type': 'object2'},
{'max': '601',
'min': '201',
'required': 'firstName2',
'title': 'Person2',
'type': 'object2'},
{'max': '6000',
'min': '2000',
'required': 'firstName3',
'title': 'Person3',
'type': 'object3'},
{'max': None,
'min': None,
...