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Home/ Questions/Q 7636951
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T07:49:02+00:00 2026-05-31T07:49:02+00:00

problem is self explanatory: import simplejson as json a = u[(datetime.datetime(2012, 3, 13, 14,

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problem is self explanatory:

import simplejson as json
a = u"[(datetime.datetime(2012, 3, 13, 14, 50, 13, 996833), 'ACTIVE', [u'my.test.service', '{}'])]"
json.loads(a)

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/simplejson/__init__.py", line 384, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/dist-packages/simplejson/decoder.py", line 402, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python2.7/dist-packages/simplejson/decoder.py", line 418, in raw_decode
    obj, end = self.scan_once(s, idx)
simplejson.decoder.JSONDecodeError: Expecting object: line 1 column 1 (char 1)

Question is obvious: How can I convert my string to a list?

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  1. Editorial Team
    Editorial Team
    2026-05-31T07:49:04+00:00Added an answer on May 31, 2026 at 7:49 am

    You can’t.
    (datetime.datetime(2012, 3, 13, 14, 50, 13, 996833) has no meaning in JSON.
    Conversely, if you try to do the opposite, you’ll notice the problem:

    json.dumps([(datetime.datetime(2012, 3, 13, 14, 50, 13, 996833), 'ACTIVE' [u'my.test.service', '{}'])])
    
    TypeError: datetime.datetime(2012, 3, 13, 14, 50, 13, 996833) is not JSON serializable
    

    Edit:

    Actually, reading your question again made me notice you did not specify you wanted to use JSON, just “convert the string to a list”. Not sure what your use case is, but this will work in your case:

    In [23]: a = "[(datetime.datetime(2012, 3, 13, 14, 50, 13, 996833), 'ACTIVE', [u'my.test.service', '{}'])]"
    In [24]: eval(a)
    Out[24]: [(datetime.datetime(2012, 3, 13, 14, 50, 13, 996833), 'ACTIVE', [u'my.test.service', '{}'])]
    
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