List comprehension, check if item is unique

Question

I am trying to write a list comprehension statement that will only add an item if it's not currently contained in the list. Is there a way to check the current items in the list that is currently being constructed? Here is a brief example:

Input

{
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

Output

["running", "engineering", "dancing", "art", "theatre", "music"]

Code without using a list comprehension

output = []
for name, hobbies in input.items():
    for hobby in hobbies:
        if hobby not in output:
            output.append(hobby)

My Attempt

[hobby for name, hobbies in input.items() for hobby in hobbies if hobby not in ???]

Solution 1

You can use set and set comprehension:

{hobby for name, hobbies in input.items() for hobby in hobbies}

As m.wasowski mentioned, we don't use the name here, so we can use item.values() instead:

{hobby for hobbies in input.values() for hobby in hobbies}

If you really need a list as the result, you can do this (but notice that usually you can work with sets without any problem):

list({hobby for hobbies in input.values() for hobby in hobbies})

Solution 2

As this answer suggests: you can use a uniqueness filter:

def f7(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

and call with:

>>> f7(hobby for name, hobbies in input.items() for hobby in hobbies)
['running', 'engineering', 'dancing', 'art', 'theatre', 'music']

I would implement the uniqueness filter separately since a design rule says "different things should be handled by different classes/methods/components/whatever". Furthermore you can simply reuse this method if necessary.

Another advantage is - as is written at the linked answer - that the order of the items is preserved. For some applications, this might be necessary.

Solution 3

If you really really want a listcomp and only a list-comp, you can do

>>> s = []
>>> [s.append(j)  for i in d.values() for j in i if j not in s]
[None, None, None, None, None, None]
>>> s
['dancing', 'art', 'theatre', 'running', 'engineering', 'music']

Here, s is a result of a side effect and d is your original dictionary. The unique advantage here is that you can preserve the order unlike most other answers here.

Note: This a bad way as it exploits the list-comp and the result is a side effect. Don't do it as a practice, This answer is just to show you that you can achieve it using a list comp alone

Solution 4

sets and dictionaries are your friends here:

from collections import OrderedDict
from itertools import chain # 'flattens' collection of iterables

data = {
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

# using set is the easiest way, but sets are unordered:
print {hobby for hobby in chain.from_iterable(data.values())}
# output:
# set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])


# or use OrderedDict if you care about ordering:
print OrderedDict(
        (hobby, None) for hobby in chain.from_iterable(data.values())
    ).keys()
# output:
# ['dancing', 'art', 'theatre', 'running', 'engineering', 'music']

Solution 5

There's another way of writing this that is a bit more descriptive of what you're actually doing, and doesn't require a nested (double for) comprehension:

output = set.union(*[set(hobbies) for hobbies in input_.values()])

This becomes even nicer when you'd represent the input to be more conceptually sound, i.e. use a set for the hobbies of each person (since there shouldn't be repetitions there either):

input_ = {
    "Stefan" : {"running", "engineering", "dancing"},
    "Bob" : {"dancing", "art", "theatre"}, 
    "Julia" : {"running", "music", "art"}
}

output = set.union(*input_.values())

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