Remove entries in a python set based on condition-CodePudding

I used scipy.spatial.KDTree.query_pairs() which returned a python set of tuples. Let's say, this is the output:

set1 = {(2, 3), (4, 5), (1, 6), (6, 7), (3, 8), (6, 8)}

Next, I want to erase all the tuples in the set which do not fulfill the condition

arr = [6, 7]
tuple[0] in arr or tuple[1] in arr

What would be the most elegant/fast/pythonic way? Should I maybe convert to a numpy array?

CodePudding user response：

You need to iterate over and check every tuple in set1, you can do that using a set comprehension, and any():

>>> set1 = {(2, 3), (4, 5), (1, 6), (6, 7), (3, 8), (6, 8)}
>>> arr = [6, 7]
>>> set2 = set(arr)  # convert to set for O(1) lookup time
>>> set3 = {t for t in set1 if any(x in set2 for x in t)}
>>> set3
{(6, 7), (6, 8), (1, 6)}

CodePudding user response：

One way to approach this would be to iterate through the set and create an array for the elements which fulfill the conditions and then convert the array to a set.

This can be done with a one-line solution such as:

set2 = set([a for a in set1 if a[0] not in key and a[1] not in key])

CodePudding user response：

Hope this is what you are looking for

filtered = [x,y for x,y in set1 if (x,y) == (6,7)]

or try using lambda for complex filters.

CodePudding user response：

Without converting your data to numpy arrays, this is how I would address this problem:

set1 = {(2, 3), (4, 5), (1, 6), (6, 7), (3, 8), (6, 8)}
arr = [6, 7]

filtered = list(filter(lambda xy: bool(set(xy) & set(arr)), set1))

If the length of the elements within set1 grows considerably and you decide to use numpy, then I would just change the sets intersection for numpy's intersect1d

If instead the length of set1 grows, then I suggest you take a look on this other StackOverflow post to see how you can filter a numpy array.