Check item membership in set in Python

Question

Hello I've been coding for a couple of months now and know the basics, but I'm having a set membership problem for which I can't find a solution. I have a list of lists of pairs of integers, and I want to remove the list that have the "a" integer in them. I thought using sets was the easiest way. Bellow is the code: ``` ## This is the item to test against. a = set([3]) ## This is the list to test. groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]] ## This is a list that will contain the lists present ## in groups which do not contain "a" groups_no_a = [] for group in groups: group = set(group) if a in group: groups_no_a.append(group) ## I thought the problem had something to do with ## clearing the variable so I put this in, ## but to no remedy. group.clear() print groups_no_a ``` I had also tried using `s.issubset(t)` until I realized that this tested if every element in `s` in `t`. Thank you!

Accepted Answer

You want to test if there is no intersection:

```
if not a & group:
```

or

```
if not a.intersection(group):
```

or, inversely, that the sets are disjoint:

```
if a.isdisjoint(group):
```

The method forms take any iterable, you don't even have to turn `group` into a set for that. The following one-liner would work too:

```
groups_no_a = [group for group in groups if a.isdisjoint(group)]
```

Demo:

```
>>> a = set([3]) 
>>> groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]]     
>>> [group for group in groups if a.isdisjoint(group)]
[[1, 2], [5, 4]]
```

If all you are testing for is one element, then it could be that creating sets is going to cost more in performance than what you gain in testing for membership, and just doing:

```
3 not in group
```

where `group` is a short list.

You can use the `timeit` module to compare pieces of Python code to see what works best for your specific typical list sizes.

Check item membership in set in Python

Solution

Problem