How do I interleave strings in Python? [closed]

Answer 1

What about (if the strings are the same length):

s1='abc'
s2='xyz'
s3=''
for x in range(len(s1)):
   s3 += '%s%s'%(s1[x],s2[x])

I'd also like to note that THIS article is now the #1 Google search result for "python interleave strings," which given the above comments I find ironic :-)

Answer 2

A mathematical one, for fun

s1="abc"
s2="xyz"

lgth = len(s1)

ss = s1+s2

print ''.join(ss[i//2 + (i%2)*lgth] for i in xrange(2*lgth))

And another one:

s1="abc"
s2="xyz"

lgth = len(s1)

tu = (s1,s2)

print ''.join(tu[i%2][i//2] for i in xrange(2*lgth))
# or
print ''.join((tu[0] if i%2==0 else tu[1])[i//2] for i in xrange(2*lgth))

Answer 3

Here is one way to do it

>>> s1 = "abc"
>>> s2 = "xyz"
>>> "".join(i for j in zip(s1, s2) for i in j)
'axbycz'

It also works for more than 2 strings

>>> s3 = "123"
>>> "".join(i for j in zip(s1, s2, s3) for i in j)
'ax1by2cz3'

Here is another way

>>> "".join("".join(i) for i in zip(s1,s2,s3))
'ax1by2cz3'

And another

>>> from itertools import chain
>>> "".join(chain(*zip(s1, s2, s3)))
'ax1by2cz3'

And one without zip

>>> b = bytearray(6)
>>> b[::2] = "abc"
>>> b[1::2] = "xyz"
>>> str(b)
'axbycz'

And an inefficient one

>>> ((s1 + " " + s2) * len(s1))[::len(s1) + 1]
'axbycz'