从元组更改为列表，反之亦然

Question

how could i change [('a', 1), ('c', 3), ('b', 2)] to ['a',1,'c',3,'b',2] and vice versa? Thanks

Answer 1

Going in the first direction from [('a', 1), ('c', 3), ('b', 2)] to ['a',1,'c',3,'b',2] is flattening a list . Taking the accepted answer from there and modifying for this example:

>>> L = [('a', 1), ('c', 3), ('b', 2)]
>>> list(itertools.chain(*L))
['a', 1, 'c', 3, 'b', 2]

This uses itertools.chain which is a nice tool from itertools that flattens sequences quite nicely.

Going the opposite way is zipping:

>>> L = ['a', 1, 'c', 3, 'b', 2]
>>> zip(L[0::2], L[1::2])
[('a', 1), ('c', 3), ('b', 2)]

zip takes two lists and combines list 1's first element with list 2's first element in a tuple, and so on down the lengths of the list. So in this one, we basically take the even-indexed elements as our first list ( L[0::2] ), and then the odd-indexed elements as our second ( L[1::2] )

Answer 2

Here it's my take on it (assuming a contains first list and b second):

b = []
for i in a:
    b.extend(i)

And reverse:

c = []
for i in range(0, len(b) - 1, 2):
   c.append((b[i], b[i+1]))

Answer 3

L = [('a', 1), ('c', 3), ('b', 2)]
flat = [x for p in L for x in p]
assert flat == ['a', 1, 'c', 3, 'b', 2]

it = iter(flat)
L2 = zip(it, it)
assert L2 == L

print "Success"

Answer 4

There are already plenty of correct answers here, so this is just a reminder not to use sum() to flatten lists as although it looks like a neat solution unfortunately the performance is quadratic

In [1]: L=[('a',x) for x in range(10)]

In [2]: timeit sum(L,())
100000 loops, best of 3: 2.78 us per loop

In [3]: L=[('a',x) for x in range(100)]

In [4]: timeit sum(L,())
10000 loops, best of 3: 108 us per loop

In [5]: L=[('a',x) for x in range(1000)]

In [6]: timeit sum(L,())
100 loops, best of 3: 8.02 ms per loop