I have written a function which will return several dictionaries. For example:
def func()
return c # <---- nested dictionary
if __name__ == "__main__":
ans = func()
print ans
If I print the ans:
{u'ok': 1.0, u'result': [{u'price': 129.7, u'_id': datetime.datetime(2015, 2, 23, 9, 32)}, {u'price': 129.78, u'_id': datetime.datetime(2015, 2, 23, 9, 33)},
print ans.get('_id')
If I print this, the result is None.
How can I get _id?
You could use a list comprehension.
In [19]: ans = {u'ok': 1.0, u'result': [{u'price': 129.7, u'_id': datetime.datetime(2015, 2, 23, 9, 32)}, {u'price': 129.78, u'_id': datetime.datetime(2015, 2, 23, 9, 33)}]}
In [24]: [i['_id'] for i in ans['result']]
Out[24]: [datetime.datetime(2015, 2, 23, 9, 32), datetime.datetime(2015, 2, 23, 9, 33)]
In [25]: [i.get('_id') for i in ans['result']]
Out[25]: [datetime.datetime(2015, 2, 23, 9, 32), datetime.datetime(2015, 2, 23, 9, 33)]
for i in ans['result']:
print i['_id']
_id - the OP is interested in getting all of the _id values.From looking at your trace it seems that c is a dictionary containing various other dictionary.
print ans["result"][0]["_id"]
Should return the value you want.
func() is actually returning a nested dictionary. Look closely at what your print is telling you.
So ans = func() is a nested dictionary:
{u'ok': 1.0, u'result': [{u'price': 129.7, u'_id': datetime.datetime(2015, 2, 23, 9, 32)}, {u'price': 129.78, u'_id': datetime.datetime(2015, 2, 23, 9, 33)},
Hence ans['result'] is itself another dict, or apparently a list containing a dict.
[i['_id'] for i in ans['result']]