从字典列表中的字典值中查找唯一字符串对的优雅方式

Question

I have a list of dictionaries:

    some_list = [
                  {
                     item1:'a', 
                     item2:['1', '2']
                  }, 
                  {
                     item1:'b', 
                     item2:['1', '2']
                  }, 
                  {
                     item1:'a', 
                     item2:['1']
                  }
                ]

I would like to get:

['a 1', 'a 2', 'b 1', 'b 2'] where each value from item 1 is paired with value from item 2 for each dictionary, and then only the unique strings are left.

I can think of an obvious way to do it, namely:

1. iterate through the some_list;
2. for each dictionary, get a each['item_1'] and each['item_2']
3. for member of each['item_2'] do each['item_1'] + ' ' + member
4. Now, make the list into a set and I have my unique values

I am wondering if these is a more elegant way to do it using list comprehension.

Answer 1

If order doesn't matter, then you can easily translate your logic into a set-comprehnsion (and convert to list if you'd like):

In [1]: some_list = [
   ...:                   {
   ...:                      'item1':'a',
   ...:                      'item2':['1', '2']
   ...:                   },
   ...:                   {
   ...:                      'item1':'b',
   ...:                      'item2':['1', '2']
   ...:                   },
   ...:                   {
   ...:                      'item1':'a',
   ...:                      'item2':['1']
   ...:                   }
   ...:                 ]

In [2]: {f"{d['item1']} {v}" for d in some_list for v in d['item2']}
Out[2]: {'a 1', 'a 2', 'b 1', 'b 2'}

Answer 2

def fun(item):
     return [item[item1]+' '+k for k in item[item2]]
res = []
[res.append(fun(i)) for i in some_list if(fun(i)) not in res]
print res

this should work

Answer 3

another formatting option, runs back to 2.7

sorted(map(' '.join,
           {(d['item1'], v)
            for d in some_list
                for v in d['item2']}))

still a 'one-liner' but with decorative line breaks and indentation

inner list comp same as other ans, arrived at independently without seeing it 1st