结合字典Python中多个键的值

Question

In Python, I have the following dictionary of sets:

{
    1: {'Hello', 'Bye'},
    2: {'Bye', 'Do', 'Action'},
    3: {'Not', 'But', 'No'},
    4: {'No', 'Yes'}
}

My goal is combine the keys which contain match values (like in this example, "Bye" and "No"), so the result will look like this:

{
    1: {'Hello', 'Bye', 'Do', 'Action'},
    3: {'Not', 'But', 'No', 'Yes'}
}

Is there a way to do this?

Answer 1

If there are overlapping matches and you want the longest matches:

from collections import defaultdict

d = {
    1: {'Hello', 'Bye'},
    2: {'Bye', 'Do', 'Action'},
    3: {'Not', 'But', 'No'},
    4: {'No', 'Yes'}
}
grp = defaultdict(list)

# first group all keys with common words
for k, v in d.items():
    for val in v:
        grp[val].append(k)


# sort the values by lengths to find longest matches.    
for v in sorted(grp.values(), key=len, reverse=True):
    for val in v[1:]:
       if val not in d:
           continue
           # use first ele as the key and union to existing values
       d[v[0]] |= d[val]
       del d[val]


print(d)

if you don't have overlaps you can just:

grp = defaultdict(list)

for k, v in d.items():
    for val in v:
        grp[val].append(k)

for v in grp.values():
    for val in v[1:]:
        d[v[0]] |= d[val]
        del d[val]

Or if you want a new dict:

new_d = {}
for v in grp.values():
    if len(v) > 1:
        k = v[0]
        new_d[k] = d[k]
        for val in v[1:]:
            new_d[k] |= d[val]

All three give you the following but key order could be different:

{1: set(['Action', 'Do', 'Bye', 'Hello']), 3: set(['Not', 'Yes', 'But', 'No'])}

Answer 2

If there is no overlapping matches:

a = {1: {'Hello', 'Bye'}, 2: {'Bye', 'Do', 'Action'}, 3: {'Not', 'But', 'No'}, 4: {'No', 'Yes'}}
output = {}
for k, v in a.items():
    if output:
        for k_o, v_o in output.items():
            if v_o.intersection(v):
                output[k_o].update(v)
                break
        else:
            output[k] = v
    else:
        output[k] = v
print(output)

Output:

{1: {'Action', 'Bye', 'Do', 'Hello'}, 3: {'But', 'No', 'Not', 'Yes'}}