Finding intersection/difference between python lists

Question

I have two python lists:

a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]

b = ['the', 'when', 'send', 'we', 'us']

I need to filter out all the elements from a that are similar to those in b. Like in this case, I should get:

c = [('why', 4), ('throw', 9), ('you', 1)]

What should be the most effective way?

Answer 1

A list comprehension should work:

c = [item for item in a if item[0] not in b]

Or with a dictionary comprehension:

d = dict(a)
c = {key: value for key in d.iteritems() if key not in b}

Answer 2

A list comprehension will work.

a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]
b = ['the', 'when', 'send', 'we', 'us']
filtered = [i for i in a if not i[0] in b]

>>>print(filtered)
[('why', 4), ('throw', 9), ('you', 1)]

Answer 3

in is nice, but you should use sets at least for b. If you have numpy, you could also try np.in1d of course, but if it is faster or not, you should probably try.

# ruthless copy, but use the set...
b = set(b)
filtered = [i for i in a if not i[0] in b]

# with numpy (note if you create the array like this, you must already put
# the maximum string length, here 10), otherwise, just use an object array.
# its slower (likely not worth it), but safe.
a = np.array(a, dtype=[('key', 's10'), ('val', int)])
b = np.asarray(b)

mask = ~np.in1d(a['key'], b)
filtered = a[mask]

Sets also have have the methods difference, etc. which probably are not to useful here, but in general probably are.