将2个列表的字符串表示转换为列表

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我想用字符串创建一个列表 . 我的字符串:

[[1544434140,asd,asd,asd]]

你可以看到我的String包含一个列表在另一个列表中,如果我想这样做,这绝对是必要的:

[[1544434140,asd,asd,asd][1544434140,asd,asd,asd]]

我已经尝试过:

literal_eval(mystring)

eval(mystring)

但两者都从ast.literal_eval抛出这样的异常:

Traceback (most recent call last):
  File "/usr/local/lib/python3.6/socketserver.py", line 317, in _handle_request_noblock
    self.process_request(request, client_address)
  File "/usr/local/lib/python3.6/socketserver.py", line 348, in process_request
    self.finish_request(request, client_address)
  File "/usr/local/lib/python3.6/socketserver.py", line 361, in finish_request
    self.RequestHandlerClass(request, client_address, self)
  File "/usr/local/lib/python3.6/socketserver.py", line 696, in __init__
    self.handle()
  File "/usr/local/lib/python3.6/http/server.py", line 418, in handle
    self.handle_one_request()
  File "/usr/local/lib/python3.6/http/server.py", line 406, in handle_one_request
    method()
  File "/home/dev/requestHandler.py", line 223, in do_GET
    addrequestResult = rh.addRequestToQue(self.path)
  File "/home/dev/requestHandler.py", line 86, in addRequestToQue
    resultOK = self.addToList(request)
  File "/home/dev/requestHandler.py", line 181, in addToList
    doAdd(request, userque, userhistory)
  File "/home/dev/requestHandler.py", line 143, in doAdd
    userque = literal_eval(userque)
  File "/usr/local/lib/python3.6/ast.py", line 85, in literal_eval
    return _convert(node_or_string)
  File "/usr/local/lib/python3.6/ast.py", line 61, in _convert
    return list(map(_convert, node.elts))
  File "/usr/local/lib/python3.6/ast.py", line 61, in _convert
    return list(map(_convert, node.elts))
  File "/usr/local/lib/python3.6/ast.py", line 84, in _convert
    raise ValueError('malformed node or string: ' + repr(node))
ValueError: malformed node or string: <_ast.Name object at 0x7fa6751838d0>

任何想法我如何将我的字符串与2封装列表转换为一个列表?

2回答

  • 0

    所以@Narendra Lucky的答案确实有效,但还有第二个解决方案:

    Mylist = json.loads(MyString)
    

  • 0

    根据我的假设,如果asd不是变量,需要','在flattern lists.If任何让我知道

    Try this Format string eval() will work

    mystring = "[[1544434140,'asd','asd','asd'],[1544434140,'asd','asd','asd']]"
    print eval(mystring)
    
    Result:[[1544434140, 'asd', 'asd', 'asd'], [1544434140, 'asd', 'asd', 'asd']]
    
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