Python嵌套json解析和拆分值
我试图拆分url并将其分配给变量,我得到了期望的结果,但我知道有一种方法可以临时编写当前代码 JSON文件Python嵌套json解析和拆分值,python,json,Python,Json,我试图拆分url并将其分配给变量,我得到了期望的结果,但我知道有一种方法可以临时编写当前代码 JSON文件 { "Result": [ "Url::Link::Url1", "Url::Link::Url2", "Url::Link::Url3", "Url::Link::Url4", "Url::Link::Url5", "Url::Link::Url6", "Url::Link::Url7" ], "Record": [ "Re
{
"Result": [
"Url::Link::Url1",
"Url::Link::Url2",
"Url::Link::Url3",
"Url::Link::Url4",
"Url::Link::Url5",
"Url::Link::Url6",
"Url::Link::Url7"
],
"Record": [
"Record::Label::Music1",
"Record::Label::Music2",
"Record::Label::Music3"
],
}
输出url1=url1
url2=url2
…我可能会使用列表理解:
>>> urls = [i.split('::')[2]
... for _ in range(len(result['Result']))
... for i in result['Result'][_:_+1]]
>>> for url in urls:
... print(url)
...
Url1
Url2
Url3
Url4
Url5
Url6
Url7
您可以使用:
urls = [e.split('::')[-1] for e in result['Result']]
print(*urls)
输出:
Url1 Url2 Url3 Url4 Url5 Url6 Url7
如果只想将前n个URL分配给某些n个变量,可以使用:
url1, url2, url3 = [e.split('::')[-1] for e in result['Result'][:3]]
您的代码在哪里?最好在中发布您的问题,不要忘记添加您自己的代码。粘贴代码时遇到问题code@rusu_ro1添加了每个的代码应该有url1、url2、url3?
url1, url2, url3 = [e.split('::')[-1] for e in result['Result'][:3]]