如果list不是None，则获取list的第一个元素：Python

Question

Here is my issue:

I am doing an LDAP search in Python. The search will return a dictionary object:

{'mail':['user@mail.com'],'mobile':['07852242242'], 'telephoneNumber':['01112512152']}

As you can see, the returned dictionary contains list values.

Sometimes, no result will be found:

{'mail':None, 'mobile':None, 'telephoneNumber':['01112512152']}

To extract the required values, I am using get() as to avoid exceptions if the dictionary item does not exist.

return {"uname":x.get('mail')[0], "telephone":x.get('telephoneNumber')[0], "mobile":x.get('mobile')[0]}

I want to return my own dictionary as above, with just the string values, but Im struggling to find an efficient way to check that the lists are not None and keep running into index errors or type errors:

(<type 'exceptions.TypeError'>, TypeError("'NoneType' object is unsubscriptable",)

Is there a way to use a get() method on a list, so that if the list is None it wont throw an exception???

{"uname":x.get('mail').get(0)}

What is the most efficient way of getting the first value of a list or returning None without using:

if isinstance(x.get('mail'),list):

or

if x.get('mail') is not None:

Answer 1

If you want to flatten your dictionary, you can just do:

>>> d = {'mail':None, 'mobile':None, 'telephoneNumber':['01112512152']}
>>> 
>>> dict((k,v and v[0] or v) for k,v in d.items())
{'mail': None, 'mobile': None, 'telephoneNumber': '01112512152'}

If you'd also like to filter, cutting off the None values, then you could do:

>>> dict((k,v[0]) for k,v in d.items() if v)
{'telephoneNumber': '01112512152'}

Answer 2

Try the next:

return {"uname":(x.get('mail') or [None])[0], ...

It is a bit unreadable, so you probably want to wrap it into some helper function.

Answer 3

You could do something like this:

input_dict = {'mail':None, 'mobile':None, 'telephoneNumber':['01112512152']}

input_key_map = {
    'mail': 'uname',
    'telephoneNumber': 'telephone',
    'mobile': 'mobile',
}

dict((new_name, input_dict[old_name][0]) 
      for old_name, new_name in input_key_map.items() if input_dict.get(old_name))

# would print:
{'telephone': '01112512152'}

Answer 4

I'm not sure if there's a straightforward way to do this, but you can try:

 default_value = [None]
 new_x = dict((k, default_value if not v  else v) for k, v in x.iteritems())

And use new_x instead of x .