I have a list like this:
l = ['b', '7', 'a', 'e', 'a', '6', 'a', '7', '9', 'c', '7', 'b', '6', '9', '9', 'd', '7', '5', '2', '4', 'c', '7', '8', 'b', '3', 'f', 'f', '7', 'b', '9', '4', '4']
and I want to make a string from it like this:
7bea6a7ac9b796d957427cb8f37f9b44
I did:
l = (zip(l[1:], l)[::2])
s = []
for ll in l:
s += ll
print ''.join(s)
But is there any simpler way? May be, in one line?
You can concatenate each pair of letters, then join the whole result in a generator expression
>>> ''.join(i+j for i,j in zip(l[1::2], l[::2]))
'7bea6a7ac9b796d957427cb8f37f9b44'
You can just use a simple list comprehension to swap (provide you are sure to have an even size) and then join:
''.join([ l[i+1] + l[i] for i in range(0, len(l), 2) ])
Group adjacent list items using zip;
group_adjacent = lambda a, k: zip(*([iter(a)] * k))
Then concatenate them by swapping in the for loop
print (''.join( j+i for i,j in group_adjacent(l,2)) )
