Comparing elements between elements in two lists of tuples

Question

Here is what I am looking to do.I have two list of tuples. Build a list of elements such that the first element in a tuple in list1 matches the first element in a tuple in list 2

list1 = [('a', 2), ('b', 3), ('z', 5)]

list2 = [('a', 1), ('b', 2), ('c', 3)]

list3 = ['a','b']

Note: There can be no duplicate first elements

After looking at python list comprehensions, this is what I have done

[x[0] for x in list1 if (x[0] in [y[0] for y in list2])]

My questions is would this be how an experienced python programmer would code this up? Having coded this up myself I still find this fairly hard to read. If not how else would you do it

Answer 1

I'd use zip():

In [25]: l1 = [('a', 2), ('b', 3), ('z', 5)]

In [26]: l2 = [('a', 1), ('b', 2), ('c', 3)]

In [27]: [x[0] for x,y in zip(l1,l2) if x[0]==y[0]]
Out[27]: ['a', 'b']

EDIT: After reading your comment above it looks like you're looking for something like this:

In [36]: [x[0] for x in l1 if any(x[0]==y[0] for y in l2)]
Out[36]: ['a', 'b']

or using sets:

In [43]: from operator import itemgetter

In [44]: set(map(itemgetter(0),l1)) & set(map(itemgetter(0),l2))
Out[44]: set(['a', 'b'])

Answer 2

I think you want to use sets here:

set(x[0] for x in list1).intersection(y[0] for y in list2)

or using syntactic sugar:

{x[0] for x in list1} & {y[0] for y in list2}

Both of which result in:

set(['a', 'b'])