如何在python中压缩两个元组列表

Question

I have two lists of tuples, for example:

a = [(1,2,3),(4,5,6),(7,8,9)]
b = [(1,'a'),(4,'b'),(7,'c')]

The first element of each tuple in a and b are matched, I want to get a list like this:

merged = [(1,2,3,'a'),(4,5,6,'b'),(7,8,9,'c')]

Perhaps I will have another list like:

c = [(1,'xx'),(4,'yy'),(7,'zz')]

and merge it to "merged" list later, I tried "zip" and "map" which are not right for this case.

Answer 1

>>> a = [(1,2,3),(4,5,6),(7,8,9)]
>>> b = [(1,'a'),(4,'b'),(7,'c')]
>>> 
>>> [x + (z,) for x, (y, z) in zip(a, b)]
[(1, 2, 3, 'a'), (4, 5, 6, 'b'), (7, 8, 9, 'c')]

to check if first elements actually match,

>>> [x + y[1:] for x, y in zip(a, b) if x[0] == y[0]]

Answer 2

def merge(a,b):
    for ax, (first, bx) in zip(a,b):
        if ax[0] != first:
            raise ValueError("Items don't match")
        yield ax + (bx,)

print list(merge(a,b))
print list(merge(merge(a,b),c))

Answer 3

>>> [a[i]+(k,) for i,(j, k) in enumerate(b)]
[(1, 2, 3, 'a'), (4, 5, 6, 'b'), (7, 8, 9, 'c')]

Using timeit this is the fastest of the posted solutions to return a merged list.

Answer 4

[ (x,y,z,b[i][1]) for i,(x,y,z) in enumerate(a) if x == b[i][0] ]

这可确保匹配然后合并值。