I have a list like

myl = ['A','B','C','D','E','F'] #length always even

Now my desired output is 'AB','CD','EF'

I tried

>>> myl = ['A','B','C','D','E','F']
>>> even_pos = myl[::2]
>>> odd_pos = myl[::-2]
>>> odd_pos.reverse()
>>> newlist = zip(even_pos,odd_pos)
>>> for x in newlist:
...     print "".join(list(x))
...
...
AB
CD
EF
>>>

I don't like this way because I think this is too much.

So, is there any better way to achieve my output.

• python

You can do this concisely using a list comprehension or generator expression:

>>> myl = ['A','B','C','D','E','F']
>>> [''.join(myl[i:i+2]) for i in range(0, len(myl), 2)]
['AB', 'CD', 'EF']
>>> print '\n'.join(''.join(myl[i:i+2]) for i in range(0, len(myl), 2))
AB
CD
EF

You could replace ''.join(myl[i:i+2]) with myl[i] + myl[i+1] for this particular case, but using the ''.join() method is easier for when you want to do groups of three or more.

Or an alternative that comes from the documentation for zip():

>>> map(''.join, zip(*[iter(myl)]*2))
['AB', 'CD', 'EF']

Why is your method so complicated? You could do basically what you did, but in one line, like so:

[ "".join(t) for t in zip(myl[::2], myl[1::2]) ]

F.J's answer is more efficient though.

How about this?

>>> ["%s%s" % (myl[c], myl[c+1]) for c in range(0, 6, 2)]
['AB', 'CD', 'EF']

I'd probably write:

[myl[i] + myl[i + 1] for i in xrange(len(myl), step=2)]