For me, the most pythonic* way is the following which pretty much does the same thing but uses the +
operator for concatenating the individual characters in each string:
res = "".join(i + j for i, j in zip(u, l))
print(res)
# 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'
It is also faster than using two join()
calls:
In [5]: l1 = 'A' * 1000000; l2 = 'a' * 1000000
In [6]: %timeit "".join("".join(item) for item in zip(l1, l2))
1 loops, best of 3: 442 ms per loop
In [7]: %timeit "".join(i + j for i, j in zip(l1, l2))
1 loops, best of 3: 360 ms per loop
Faster approaches exist, but they often obfuscate the code.
Note: If the two input strings are not the same length then the longer one will be truncated as zip
stops iterating at the end of the shorter string. In this case instead of zip
one should use zip_longest
(izip_longest
in Python 2) from the itertools
module to ensure that both strings are fully exhausted.
*To take a quote from the Zen of Python: Readability counts.
Pythonic = readability for me; i + j
is just visually parsed more easily, at least for my eyes.
Another way:
res = [''] * len(u) * 2
res[::2] = u
res[1::2] = l
print(''.join(res))
Output:
'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'
Looks like it is faster:
%%timeit
res = [''] * len(u) * 2
res[::2] = u
res[1::2] = l
''.join(res)
100000 loops, best of 3: 4.75 µs per loop
than the fastest solution so far:
%timeit "".join(list(chain.from_iterable(zip(u, l))))
100000 loops, best of 3: 6.52 µs per loop
Also for the larger strings:
l1 = 'A' * 1000000; l2 = 'a' * 1000000
%timeit "".join(list(chain.from_iterable(zip(l1, l2))))
1 loops, best of 3: 151 ms per loop
%%timeit
res = [''] * len(l1) * 2
res[::2] = l1
res[1::2] = l2
''.join(res)
10 loops, best of 3: 92 ms per loop
Python 3.5.1.
u = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
l = 'abcdefghijkl'
zip()
equivalent)min_len = min(len(u), len(l))
res = [''] * min_len * 2
res[::2] = u[:min_len]
res[1::2] = l[:min_len]
print(''.join(res))
Output:
AaBbCcDdEeFfGgHhIiJjKkLl
itertools.zip_longest(fillvalue='')
equivalent)min_len = min(len(u), len(l))
res = [''] * min_len * 2
res[::2] = u[:min_len]
res[1::2] = l[:min_len]
res += u[min_len:] + l[min_len:]
print(''.join(res))
Output:
AaBbCcDdEeFfGgHhIiJjKkLlMNOPQRSTUVWXYZ
With join()
and zip()
.
>>> ''.join(''.join(item) for item in zip(u,l))
'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'
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.
If you want the fastest way, you can combine itertools with operator.add
:
In [36]: from operator import add
In [37]: from itertools import starmap, izip
In [38]: timeit "".join([i + j for i, j in uzip(l1, l2)])
1 loops, best of 3: 142 ms per loop
In [39]: timeit "".join(starmap(add, izip(l1,l2)))
1 loops, best of 3: 117 ms per loop
In [40]: timeit "".join(["".join(item) for item in zip(l1, l2)])
1 loops, best of 3: 196 ms per loop
In [41]: "".join(starmap(add, izip(l1,l2))) == "".join([i + j for i, j in izip(l1, l2)]) == "".join(["".join(item) for item in izip(l1, l2)])
Out[42]: True
But combining izip
and chain.from_iterable
is faster again
In [2]: from itertools import chain, izip
In [3]: timeit "".join(chain.from_iterable(izip(l1, l2)))
10 loops, best of 3: 98.7 ms per loop
There is also a substantial difference between
chain(*
and chain.from_iterable(...
.
In [5]: timeit "".join(chain(*izip(l1, l2)))
1 loops, best of 3: 212 ms per loop
There is no such thing as a generator with join, passing one is always going to be slower as python will first build a list using the content because it does two passes over the data, one to figure out the size needed and one to actually do the join which would not be possible using a generator:
join.h:
/* Here is the general case. Do a pre-pass to figure out the total
* amount of space we'll need (sz), and see whether all arguments are
* bytes-like.
*/
Also if you have different length strings and you don't want to lose data you can use izip_longest :
In [22]: from itertools import izip_longest
In [23]: a,b = "hlo","elworld"
In [24]: "".join(chain.from_iterable(izip_longest(a, b,fillvalue="")))
Out[24]: 'helloworld'
For python 3 it is called zip_longest
But for python2, veedrac's suggestion is by far the fastest:
In [18]: %%timeit
res = bytearray(len(u) * 2)
res[::2] = u
res[1::2] = l
str(res)
....:
100 loops, best of 3: 2.68 ms per loop
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