numpy.unique 给出了错误的 output 的集合列表

Question

I have a list of sets given by,

sets1 = [{1},{2},{1}]

When I find the unique elements in this list using numpy's unique , I get

np.unique(sets1)
Out[18]: array([{1}, {2}, {1}], dtype=object)

As can be seen seen, the result is wrong as {1} is repeated in the output.

When I change the order in the input by making similar elements adjacent, this doesn't happen.

sets2 = [{1},{1},{2}]

np.unique(sets2)
Out[21]: array([{1}, {2}], dtype=object)

Why does this occur? Or is there something wrong in the way I have done?

Answer 1

What happens here is that the np.unique function is based on the np._unique1d function from NumPy (see the code here ), which itself uses the .sort() method.

Now, sorting a list of sets that contain only one integer in each set will not result in a list with each set ordered by the value of the integer present in the set. So we will have (and that is not what we want):

sets = [{1},{2},{1}]
sets.sort()
print(sets)

# > [{1},{2},{1}]
# ie. the list has not been "sorted" like we want it to

Now, as you have pointed out, if the list of sets is already ordered in the way you want, np.unique will work (since you would have sorted the list beforehand).

One specific solution (though, please be aware that it will only work for a list of sets that each contain a single integer) would then be:

np.unique(sorted(sets, key=lambda x: next(iter(x))))

Answer 2

That is because set is unhashable type

{1} is {1} # will give False

you can use python collections.Counter if you can can convert the set to tuple like below

from collections import Counter
sets1 = [{1},{2},{1}]
Counter([tuple(a) for a in sets1])