Checking if List contains all items from another list

Question

What is the easiest/fastest way to check if a list of tuples contains all the the tuples that another list does. For example:

t1 = [ (1,2), (3,4), (5,6), (7,8), (9,10), (11,12) ] t2 = [ (3,4), (11,12) ] 

would be a true case since t1 contains all the tuples in t2. I tried something like:

[i for e in t2 for i in t1 if e in i] sorted(t1) == sorted(t2) 

but this seemed to always return true. Is there a better way?

Answer 1

You can use sets

t1 = [ (1,2), (3,4), (5,6), (7,8), (9,10), (11,12) ] t2 = [ (3,4), (11,12) ] set(t2).issubset(t1) # returns true  # or equivalent use '<=' so set(t2) <= set(t1) # returns true 

Answer 2

For simplicity, you could do this:

print all(x in t1 for x in t2) 

However, that's going to search through t1 for each element in t2. That probably doesn't matter when t1 is small as in this case, but to allow for larger collections I would do this:

s1 = set(t1) print all(x in s1 for x in t2) 

or this:

print set(t1).issuperset(t2) 

This will generally be faster, since in is much faster for sets than for large lists. There's no major performance benefit in converting t2 to a set, regardless of size, so I wouldn't.

As always, it's better if you get your data in the "right" collection to begin with. So if the main purpose of t1 is to look things up in it, use a set in the first place rather than a list.