CODEWITHSUNDEEP
pythonlistdictionary
João Nunes
João Nunes

Reputation: 1

Dictionary value replace

Imagine I have a dictionary like this:

d = {'1':['a'], '2':['b', 'c', 'd'], '3':['e', 'f'], '4':['g']}

Each key of the dictionary represents a unique person of a certain class.
Each key must have only one value.
Key's with one value represent the correct reassignment.
Key's with more than one value represent the possibilities. One of those values is the most correct for that key.

I have a processed list with the most correct value.

LIST = ['c', 'e']

I must now iterate values of LIST through values of dictionary when len(values) > 1 and replace them to look like this:

d = {'1':['a'], '2':['c'], '3':['e'], '4':['g']}

Upvotes: 0

Views: 58

Answers (2)

Chiheb Nexus
Chiheb Nexus

Reputation: 9257

Another approach in one statement using dict comprehension:

d = {'1':['a'], '2':['b', 'c', 'd'], '3':['e', 'f'], '4':['g']}
a = ['c', 'e']

output = {k: v if not any(j in set(a) for j in v) else list(set(v) & set(a)) if v and isinstance(v, (list, tuple)) else [] for k, v in d.items()}

# More readeable like this:
# {
#    k: v if not any(j in set(a) for j in v) else list(set(v) & set(a))
#       if v and isinstance(v, (list, tuple))
#           else [] for k, v in d.items()
# }

print(output)

Output:

{'1': ['a'], '2': ['c'], '3': ['e'], '4': ['g']}

Upvotes: 0

cs95
cs95

Reputation: 402333

Initialise your correct values inside a set.

correct = {'c', 'e'}
# correct = set(LIST)

Now, assuming list values with more than one element can have only a single correct element, you can build a dictionary using a conditional comprehension:

d2 = {k : list(correct.intersection(v)) if len(v) > 1 else v for k, v in d.items()}
print(d2)
# {'1': ['a'], '2': ['c'], '3': ['e'], '4': ['g']}

If there can be more than one possible correct value, you can take just the first one.

d2 = {}
for k, v in d.items():
    if len(v) > 1:
        c = list(correct.intersection(v))
        v = c[:1]
    d2[k] = v

print(d2)
# {'1': ['a'], '2': ['c'], '3': ['e'], '4': ['g']}

If you meant to mutate d in-place (because making a full copy can be expensive), then the above solution simplifies to

for k, v in d.items():
    if len(v) > 1:
        c = list(correct.intersection(v))
        d[k] = c[:1]

print(d)
# {'1': ['a'], '2': ['c'], '3': ['e'], '4': ['g']}

Upvotes: 2

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