What\'s the most pythonic way to mesh two strings together?
For example:
Input:
u = \'ABCDEFGHIJKLMNOPQRSTUVWXYZ\'
l = \'abcdefghijklmnopqrst
Potentially faster and shorter than the current leading solution:
from itertools import chain
u = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
l = 'abcdefghijklmnopqrstuvwxyz'
res = "".join(chain(*zip(u, l)))
Strategy speed-wise is to do as much at the C-level as possible. Same zip_longest() fix for uneven strings and it would be coming out of the same module as chain() so can't ding me too many points there!
Other solutions I came up with along the way:
res = "".join(u[x] + l[x] for x in range(len(u)))
res = "".join(k + l[i] for i, k in enumerate(u))
On Python 2, by far the faster way to do things, at ~3x the speed of list slicing for small strings and ~30x for long ones, is
res = bytearray(len(u) * 2)
res[::2] = u
res[1::2] = l
str(res)
This wouldn't work on Python 3, though. You could implement something like
res = bytearray(len(u) * 2)
res[::2] = u.encode("ascii")
res[1::2] = l.encode("ascii")
res.decode("ascii")
but by then you've already lost the gains over list slicing for small strings (it's still 20x the speed for long strings) and this doesn't even work for non-ASCII characters yet.
FWIW, if you are doing this on massive strings and need every cycle, and for some reason have to use Python strings... here's how to do it:
res = bytearray(len(u) * 4 * 2)
u_utf32 = u.encode("utf_32_be")
res[0::8] = u_utf32[0::4]
res[1::8] = u_utf32[1::4]
res[2::8] = u_utf32[2::4]
res[3::8] = u_utf32[3::4]
l_utf32 = l.encode("utf_32_be")
res[4::8] = l_utf32[0::4]
res[5::8] = l_utf32[1::4]
res[6::8] = l_utf32[2::4]
res[7::8] = l_utf32[3::4]
res.decode("utf_32_be")
Special-casing the common case of smaller types will help too. FWIW, this is only 3x the speed of list slicing for long strings and a factor of 4 to 5 slower for small strings.
Either way I prefer the join
solutions, but since timings were mentioned elsewhere I thought I might as well join in.