Say I have a list of lists like so:
list = [[1,2,3,4],[4,5,3,2],[7,8,9,2],[5,6,8,9]]
I want to get the indices of the inner lists that contain unique elements. For the example above, the lists at index 2 is the only one that contains 7 and the list at index 3 is the only one that contains 6.
How would one go about implementing this in python?
Here's a solution using Counter. Each inner list is checked for a value that only has a single count, and then the corresponding index is printed (a la enumerate).
from collections import Counter
from itertools import chain
c = Counter(chain.from_iterable(l))
idx = [i for i, x in enumerate(l) if any(c[y] == 1 for y in x)]
print(idx)
[0, 2, 3]
A possible optimisation might include precomputing unique elements in a set to replace the any call with a set.intersection.
c = Counter(chain.from_iterable(l))
u = {k for k in c if c[k] == 1}
idx = [i for i, x in enumerate(l) if u.intersection(x)]
A naive solution:
>>> from collections import Counter
>>> from itertools import chain
>>> my_list = [[1,2,3,4],[4,5,3,2],[7,8,9,2],[5,6,8,9]]
# find out the counts.
>>> counter = Counter(chain(*my_list))
# find the unique numbers
>>> uniques = [element for element,count in counter.items() if count==1]
# find the index of those unique numbers
>>> result = [indx for indx,elements in enumerate(my_list) for e in uniques if e in elements]
>>> result
[0, 2, 3]
using itertools.chain with set.difference(set)
from itertools import chain
l = [[1,2,3,4],[4,5,3,2],[7,8,9,2],[5,6,8,9]]
[i for i in range(len(l)) if set(l[i]).difference(set(chain(*[j for j in l if j!=l[i]])))]
#[0, 2, 3]
