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 ???]

Asked By: Stefan Bossbaly

||

Source

Answer 1

Use a set:

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

myset = set()
for _, value in dict.items():
    for item in value:
        myset.add(item)

print(myset)

Answered By: nullptr

Answer 2

How about this:

set(dict['Bob']+dict['Stefan']+dict['Julia'])
>>> set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])

Or more nicely:

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

list_ = []
for y in dict.keys():
    list_ = list_ + dict[y]
list_ = set(list_)
>>> list_
set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])

you can apply the list function to list_ like list(list_) to return a list rather than a set.

Answered By: Plug4

Answer 3

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})

Answered By: geckon

Answer 4

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.

Answered By: Willem Van Onsem

Answer 5

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']

Answered By: m.wasowski

Answer 6

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

Answered By: Bhargav Rao

Answer 7

A list comprehension is not well-suited for this problem. I think a set comprehension would be better, but since that was already shown in another answer, I’ll show a way of solving this problem with a compact one-liner:

list(set(sum(hobbies_dict.values(), [])))

Another interesting solution using bitwise or operator which serves as a union operator for sets:

from operator import or_
from functools import reduce # Allowed, but unnecessary in Python 2.x
list(reduce(or_, map(set, hobbies_dict.values())))

Or (unintentional pun, I swear), instead of using bitwise or operator, just use set.union and pass it the unpacked set-mapping of your values. No need to import or_ and reduce! This idea is inspired by Thijs van Dien’s answer.

list(set.union(*map(set, hobbies_dict.values())))

Answered By: Shashank

Answer 8

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())

Answered By: Thijs van Dien

List comprehension, check if item is unique

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