group and classify words as well as characters

Given I understand it correctly, this is more a matter of constructing a datastructure that, for a given tag, constructs the correct list. We can do this by constructing a dictionary that only takes into account singular tags. Later, when a person queries on multiple tags, we calculate the intersection. This thus makes it compact to represent, as well as easy to extract for example all the elements with tag AC and this will list elements with tag ABCD, ACD, ZABC, etc.

We can thus construct a parser:

from collections import defaultdict

class Hunspell(object):

    def __init__(self, data):
        self.data = data

    def __getitem__(self, tags):
        if not tags:
            return self.data.get(None, [])

        elements = [self.data.get(tag ,()) for tag in tags]
        data = set.intersection(*map(set, elements))
        return [e for e in self.data.get(tags[0], ()) if e in data]

    @staticmethod
    def load(f):
       data = defaultdict(list)
       for line in f:
           try:
               element, tags = line.rstrip().split('/', 1)
               for tag in tags:
                   data[tag].append(element)
               data[None].append(element)
           except ValueError:
               pass  # element with no tags
       return Hunspell(dict(data))

The list processing at the end of __getitem__ is done to retrieve the elements in the correct order.

We can then load the file into memory with:

>>> with open('test.txt') as f:
...     h = Hunspell.load(f)

and query it for arbitrary keys:

>>> h['SE']
['girl', 'house', 'wind']
>>> h['ES']
['girl', 'house', 'wind']
>>> h['1']
['house']
>>> h['']
['test', 'girl', 'home', 'house', 'man', 'wind']
>>> h['S3']
['house']
>>> h['S2']
['house']
>>> h['SE2']
['house']
>>> h[None]
['test', 'girl', 'home', 'house', 'man', 'wind']
>>> h['4']
[]

querying for non-existing tags will result in an empty list. Here we thus postpone the "intersection" process upon calling. We can in fact already generate all possible intersections, but this will result in a large datastructure, and perhaps a large amount of work

Try this:

myl=dict()
with open('test.txt') as f :
    for l in f:
        l = l.rstrip()
        try:
            tags = l.split('/')[1]
            myl.setdefault(tags,[])
            myl[tags].append(l.split('/')[0])
            for t in tags:
                myl.setdefault(t,[])
                myl[t].append( l.split('/')[0])
        except:
            pass
keys=myl.keys()
for k1 in keys:
    for k2 in keys:
        if len(set(k1).intersection(k2))==len(set(k1)) and k1!=k2:
            myl[k1].extend([myk2v for myk2v in myl[k2] if myk2v not in myl[k1]]) 
print(myl)

Output

{'S': ['test', 'test', 'girl', 'house', 'wind'], 'SE': ['girl', 'house', 'wind'], 'E': ['girl', 'house', 'man', 'man', 'wind'], '': ['home', 'test', 'test', 'girl', 'house', 'wind', 'man', 'man'], 'SE123': ['house'], '1': ['house'], '2': ['house'], '3': ['house'], 'ES': ['wind', 'girl', 'house']}

In the last two for loops of the program, the set of k1 and k2 are taken first, and then, the intersection of the two sets is compared. If the length of the intersection is equal to the length of set k1, the value of the key k2 should be in the key k1, so the value of the key k2 is added to the key k1.

A much simpler Version of what willem van onsem already answered

data = defaultdict(list)
with open('text.txt') as f:
    for line in f.readlines():
        try:
            element,tags = line.rstrip().split('/', 1)
            print(element)
            for tag in tags:
                data[tag].append(element)
                data[None].append(element)
        except ValueError:
            pass

def parse(data,tag):
    if(tag==None or tag==''):
        return set(data[None])
    elements = [set(data[tag_i]) for tag_i in tag]
    return set.intersection(*map(set, elements))



>>> parse(data,'ES') 
>>> {'girl', 'house', 'wind'}
>>> parse(data,None)
>>> {'girl', 'house', 'man', 'wind'}
>>> parse(data,'')
>>> {'girl', 'house', 'man', 'wind'}