我在一个文件中有这个JSON:

{
    "maps": [
        {
            "id": "blabla",
            "iscategorical": "0"
        },
        {
            "id": "blabla",
            "iscategorical": "0"
        }
    ],
    "masks": [
        "id": "valore"
    ],
    "om_points": "value",
    "parameters": [
        "id": "valore"
    ]
}

我编写了这个脚本来打印所有JSON数据:

import json
from pprint import pprint

with open('data.json') as f:
    data = json.load(f)

pprint(data)

该程序引发了一个例外:

Traceback (most recent call last):
  File "<pyshell#1>", line 5, in <module>
    data = json.load(f)
  File "/usr/lib/python3.5/json/__init__.py", line 319, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python3.5/json/decoder.py", line 339, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python3.5/json/decoder.py", line 355, in raw_decode
    obj, end = self.scan_once(s, idx)
json.decoder.JSONDecodeError: Expecting ',' delimiter: line 13 column 13 (char 213)

如何解析JSON并提取其值?


当前回答

你的data.json应该是这样的:

{
 "maps":[
         {"id":"blabla","iscategorical":"0"},
         {"id":"blabla","iscategorical":"0"}
        ],
"masks":
         {"id":"valore"},
"om_points":"value",
"parameters":
         {"id":"valore"}
}

您的代码应为:

import json
from pprint import pprint

with open('data.json') as data_file:    
    data = json.load(data_file)
pprint(data)

注意,这只适用于Python 2.6及更高版本,因为它取决于with语句。在Python 2.5中,使用from __future_importwith_statement,在Python<=2.4中,请参见Justin Peel的答案,该答案基于此。

现在,您还可以访问如下单个值:

data["maps"][0]["id"]  # will return 'blabla'
data["masks"]["id"]    # will return 'valore'
data["om_points"]      # will return 'value'

其他回答

你的data.json应该是这样的:

{
 "maps":[
         {"id":"blabla","iscategorical":"0"},
         {"id":"blabla","iscategorical":"0"}
        ],
"masks":
         {"id":"valore"},
"om_points":"value",
"parameters":
         {"id":"valore"}
}

您的代码应为:

import json
from pprint import pprint

with open('data.json') as data_file:    
    data = json.load(data_file)
pprint(data)

注意,这只适用于Python 2.6及更高版本,因为它取决于with语句。在Python 2.5中,使用from __future_importwith_statement,在Python<=2.4中,请参见Justin Peel的答案,该答案基于此。

现在,您还可以访问如下单个值:

data["maps"][0]["id"]  # will return 'blabla'
data["masks"]["id"]    # will return 'valore'
data["om_points"]      # will return 'value'

这里是修改后的data.json文件:

{
    "maps": [
        {
            "id": "blabla",
            "iscategorical": "0"
        },
        {
            "id": "blabla",
            "iscategorical": "0"
        }
    ],
    "masks": [{
        "id": "valore"
    }],
    "om_points": "value",
    "parameters": [{
        "id": "valore"
    }]
}

您可以使用以下行在控制台上调用或打印数据:

import json
from pprint import pprint
with open('data.json') as data_file:
    data_item = json.load(data_file)
pprint(data_item)

打印的预期输出(data_item['parameters'][0]['id']):

{'maps': [{'id': 'blabla', 'iscategorical': '0'},
          {'id': 'blabla', 'iscategorical': '0'}],
 'masks': [{'id': 'valore'}],
 'om_points': 'value',
 'parameters': [{'id': 'valore'}]}

打印的预期输出(data_item['parameters'][0]['id']):

valore

您的数据不是有效的JSON格式。当“masks”和“parameters”元素应该具有{}时,您具有[]:

[]表示JSON数组,在Python中称为列表{}用于JSON对象,在Python中称为dict

以下是JSON文件的外观:

{
    "maps": [
        {
            "id": "blabla",
            "iscategorical": "0"
        },
        {
            "id": "blabla",
            "iscategorical": "0"
        }
    ],
    "masks": {
        "id": "valore"
    },
    "om_points": "value",
    "parameters": {
        "id": "valore"
    }
}

然后您可以使用代码:

import json
from pprint import pprint

with open('data.json') as f:
    data = json.load(f)

pprint(data)

使用数据,您现在还可以找到如下值:

data["maps"][0]["id"]
data["masks"]["id"]
data["om_points"]

试试这些,看看它是否开始有意义。