Matching element from value list in a dictionary to another element in the same list in python

Question

I'm using the following: http://deron.meranda.us/data/nicknames.txt which has the nicknames for most of the names. I'm using it as follows:

def load_names():
    with open('file.txt', 'r') as infile:
        outdict = collections.defaultdict(list)
        for i, line in enumerate(infile.readlines()):
            tmp = line.strip().split('\t')
            outdict[tmp[1]].append((tmp[0]))

    return dict(outdict)

This returns a dictionary where the actual names as keys and all their nicknames as a list of values for that key.

Now, when I have a

namelist = ['KEN', 'KENNY', 'KENNETH', 'MITCH', 'MITCHELL', 'LIZ', 'LIZZIE', 'ELIZABETH']

then I want to output only the different names ie, ['KENNETH', 'MITCHELL', 'ELIZABETH'] , which I'm able to do by:

l = load_names()
temp =[]
for i in namelist:
    v = l.get(i)
    if v is not None:
        l3 = [x for x in namelist if x not in v]
        temp.append(l3)

print(temp)   
s = set.intersection(*map(set,temp))     
s = list(s)
print(s)

However, I also want it to deal the case like ['KEN', 'KENNY', 'MITCH', 'MITCHELL', 'LIZ', 'LIZZIE', 'ELIZABETH'] , this should output ['KEN', 'MITCHELL', 'ELIZABETH'] ie, the two nicknames 'KEN' and 'KENNY' belong to the same key so, I want to consider them as one and have only one of them in the final list. Also, if I have a namelist as ['KENNETH, 'ZHAO', 'MITCH', 'MITCHELL'] then it should output: ['KENNETH', 'ZHAO', 'MITCHELL'] ie, for such names that do not occur in the dictionary at all (in either key or value), then output list should still have that name. How do I get that?

UPATE:

outdict = {'KENNETH': ['KEN', 'KENNY'], 'MITCHELL': ['MITCH'], 'ELIZABETH' : ['LIZ', 'LIZZIE'], .....}

namelist1 = ['KEN', 'KENNY', 'KENNETH', 'MITCH', 'MITCHELL', 'LIZ', 'LIZZIE', 'ELIZABETH']

output1 = `['KENNETH', 'MITCHELL', 'ELIZABETH']`

I'm getting the above output from the code I have put.

However, I want to be able to get the following outputs also when the namelists are as follows:

namelist2 = ['KEN', 'KENNY', 'MITCH', 'MITCHELL', 'LIZ', 'LIZZIE', 'ELIZABETH']`

output2 = `['KEN', 'MITCHELL', 'ELIZABETH']`

namelist3 = ['KENNETH, 'ZHAO', 'MITCH', 'MITCHELL']`
output3 =  `['KENNETH', 'ZHAO', 'MITCHELL']`

Got the answer myself:

l = load_names()
    temp =[]
    e = {}
    for n in namelist:
        if n in l.keys():
            temp.append(n)


    for ix in namelist: 
        for key, i in l.items():
            if key in namelist:
                continue
            else:
                b=0
                if ix in i:
                    b=1
                    f=0
                    for x in temp:
                        if x in i:
                            f=1
                            break
                    if f == 0:
                        temp.append(ix)
                        break
                b=0
                for k2, loc in l.items():
                    if ix in loc:
                        b=1
                        break
                    elif ix == k2:
                        b=1
                        break
                if b == 0:
                    temp.append(ix)   
                    break
    print(temp)

Answer 1

try this,

def getOddout(st):
    out = []
    for s in st:
        if not any([s in r for r in st if s != r]):
            out.append(s)
    return out
namelist = ['KEN', 'KENNY', 'KENNETH', 'MITCH', 'MITCHELL', 'LIZ', 'LIZZIE', 'ELIZABETH']
print getOddout(namelist)

if its not put your dict object

Answer 2

Adding the lists to a dict and then returning the keys works, but there has to be a better way.

l = load_names()
e = {}
for k, v in l.items():
    for n in namelist:
        if n in v and k not in e:
            e[k] = v
print('dict', e.keys())

Outputs

['ELIZABETH', 'MITCHELL', 'KENNETH']