How to split and convert into dictionary and swap the key value

my_list = ["Germany + A", "France + A",  "England + B", "Germany + A"  ]
result = {}
for item in my_list:
    country, key = item.split(' + ')
    if country not in result.setdefault(key, []):
        result[key].append(country)
print(result)

As a side note – use meaningful names, not cryptic one-char names.

As an alternative to using dict.setdefault() one can use collections.defaultdict with default value of list or if the order is not important – set

EDIT: comparison between using dict.setdefault and collections.defaultdict(list)

from collections import defaultdict
from timeit import timeit

my_list = ["Germany + A", "France + A",  "England + B", "Germany + A"  ]

def test1(my_list):
  result = {}
  for item in my_list:
      country, key = item.split(' + ')
      if country not in result.setdefault(key, []):
          result[key].append(country)
  return result

def test2(my_list):
  result = defaultdict(list)
  for item in my_list:
      country, key = item.split(' + ')
      if country not in result[key]:
          result[key].append(country)
  return result

print(timeit('test1(my_list)', setup='from __main__ import test1, my_list', number=100000))
print(timeit('test2(my_list)', setup='from __main__ import test2, my_list', number=100000))

output

0.2819225169987476
0.3298255940026138

at least with small sample data setdefault is a bit faster.

I think opting for a readable solution here is best.

Loop through the list l, and then do the split on ' + '.

Then, append the country names to the appropriate key the first time they are encountered.

Notice the use of collections.defaultdict to initialize the dictionary as a dict of lists.

import collections

l = ["Germany + A", "France + A", "England + B", "Germany + A"]

d = collections.defaultdict(list)

for i in l:
    k, v = i.split(' + ')
    if k not in d[v]:
        d[v].append(k)

print(dict(d))

This gives the output:

{'A': ['Germany', 'France'], 'B': ['England']}

If you want to stick with your original approach, you could again split using ' + ', and put the result into a list using a list comprehension:

m = [i.split(' + ') for i in l]

Then, you would loop through m like this:

for k, v in m:
    if k not in d[v]:
        d[v].append(k)

This is useful if you want the intermediate m list.

l = ["Germany + A", "France + A",  "England + B", "Germany + A"  ]
m = []
dicct = {}
for i in l:
    m.append(i.split('+'))
for k,v in m:
    if v in dicct:
        if k not in dicct[v]:
            dicct[v].append(k)
    else:
        dicct[v] = []
        dicct[v].append(k)

print(dicct)

Another way to achieve this

There’s a great Python lib called pandas which can also do some nice job and give you some flexibility to play with:

# Input
L = ["Germany + A", "France + A",  "England + B", "Germany + A"  ]

# Preprocessing
L = [l.split(' + ') for l in L]

import pandas as pd
df = pd.DataFrame(L, columns=['country','type']) # give the columns some names

See what’s in df:

>>> df
   country type
0  Germany    A
1   France    A
2  England    B
3  Germany    A

# Then, drop duplicate records:
df.drop_duplicates(['country', 'type'], inplace=True)

# Group by type, convert to list for each record and dump to a dict in one shot
grouped = df.groupby('type').apply(lambda x: x['country'].tolist()).to_dict()

And the result:

>>> grouped
{'A': ['Germany', 'France'], 'B': ['England']}

That’s it.

It is not the efficient solution but there are multiple solutions to

multiple questions framed into one

from itertools import groupby

s=["Germany + A", "France + A",  "England + B", "Germany + A"  ]

m=[i.strip(' ').split('+') for i in s]
[['Germany ', ' A'], ['France ', ' A'], ['England ', ' B'], ['Germany ', ' A']]

#Grouping based on alphabets 'A' , B
new=[list(g) for k, g in groupby(sorted(m,reverse=True), lambda x:x[1])]
[[['Germany ', ' A'], ['Germany ', ' A'], ['France ', ' A']],
[['England ', ' B']]]


#swapping alphabet and Countries position
new=[item[::-1] for sublist in new for item in sublist ]
[[' A', 'Germany '], [' A', 'Germany '], [' A', 'France '], [' B', 'England       ']]


dct = dict((key, tuple(v for (k, v) in pairs)) 
           for (key, pairs) in itertools.groupby(new, lambda pair: pair[0]))
{' A': ('Germany ', 'Germany ', 'France '), ' B': ('England ',)}


{k:list(set(v)) for k,v in dct.items()}