How to delete duplicates in dictionary with list values-CodePudding

I need to delete the elements that are duplicated in a dictionary like this:

{
    1: ['1', '2', '3'],
    2: ['4', '3', '6', '7'],
    3: ['8', '1', '9']
}

as to make the final result like this

{
    1: ['1', '2', '3'],
    2: ['4', '6', '7'],
    3: ['8', '9']
}

Please help me how to do that, I have no idea

CodePudding user response：

Assuming d the input, you can use a set to keep track of the seen values. Here using a dictionary comprehension and "cheating" a bit to add the values:

seen = set()
out = {k: [x for x in v
           if x not in seen and not seen.add(x)]
       for k,v in d.items()}

Output:

{1: ['1', '2', '3'],
 2: ['4', '6', '7'],
 3: ['8', '9']}

Same with a classical loop:

out = {}
seen = set()
for k,v in d.items():
    l = []
    for x in v:
        if x not in seen:
            seen.add(x)
            l.append(x)
    out[k] = l

CodePudding user response：

Rehashing the same old seen-set solution is boring :-P. Let's have some fun with Counters:

from collections import Counter

d = {
    1: ['1', '2', '3'],
    2: ['4', '3', '6', '7'],
    3: ['8', '1', '9']
}

seen = Counter()
for a in d.values():
    uniq = Counter(dict.fromkeys(a, 1))
    a[:] = uniq - seen
    seen |= uniq

print(d)

Each list is first deduplicated on its own by using a dict. Then turned into a Counter so we can conveniently subtract the previously seen values. Write the new ones into the list and add them to the seen ones.

Try it online!

CodePudding user response：

You could do the same with set union and difference operators. As sets are unordered the final list would need to be sorted. Again assuming d is the original dictionary.

s = set()
for k in d:
    z = d[k]
    d[k]= sorted(list(set(z).difference(s)))
    s |= set(z)