将字典转换为列表列表

Question

I've got a defaultdict that looks like...

defaultdict(int,
              {" u'CAMILLE'": 10,
               " u'SAHARA'": 1,
               " u'JEREMIAH'": 114,
               " u'EDISON'": 9,
               ...}

I need something like...

[[u'CAMILLE', 10],
 [u'SAHARA', 1],
 [u'JEREMIAH',114],
 [u'EDISON', 9],
 ...]

both

firstnames = [lambda x,y:list(x,y) for k,v in firstnames.items()]

and

firstnames = [lambda x,y:[x,y] for k,v in firstnames.items()]

produce

[<function __main__.<lambda>>,
 <function __main__.<lambda>>,
 <function __main__.<lambda>>,
 <function __main__.<lambda>>,
 ...]

which is obviously not what I'm intending. How can I correct this code?

Answer 1

There is no need to use lambda s:

firstnames = [[k,v] for k,v in firstnames.items()]

or even shorter:

firstnames = [list(t) for t in firstnames.items()]

lambda creates an anonymous function. This means you have generated a list of functions that take as input two parameters (here x and y ) and return a list [x,y] . The k and v are not even taken into account in your approach.

Answer 2

Using map will also work:

firstnames = map(list, firstnames.items())

Note that in python3 this will build a generator (so, as Willem comment says, its better to inmediatly use list() around the map call so the values are taken at the call moment).

firstnames = list(map(list, firstnames.items()))