Most pythonic way to interleave two strings

问题:

What's the most pythonic way to mesh two strings together?

For example:

Input:

u = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
l = 'abcdefghijklmnopqrstuvwxyz'

Output:

'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

回答1:

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.}

回答2:

Faster Alternative

Another way:

res = [''] * len(u) * 2
res[::2] = u
res[1::2] = l
print(''.join(res))

Output:

'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

Speed

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.

Variation for strings with different lengths

u = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
l = 'abcdefghijkl'

Shorter one determines length (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

Longer one determines length (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

回答3:

With join() and zip().

>>> ''.join(''.join(item) for item in zip(u,l))
'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

回答4:

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.

回答5:

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

回答6:

You could also do this using map and operator.add:

from operator import add

u = 'AAAAA'
l = 'aaaaa'

s = "".join(map(add, u, l))

Output:

'AaAaAaAaAa'

What map does is it takes every element from the first iterable u and the first elements from the second iterable l and applies the function supplied as the first argument add. Then join just joins them.

回答7:

Jim's answer is great, but here's my favorite option, if you don't mind a couple of imports:

from functools import reduce
from operator import add

reduce(add, map(add, u, l))

回答8:

A lot of these suggestions assume the strings are of equal length. Maybe that covers all reasonable use cases, but at least to me it seems that you might want to accomodate strings of differing lengths too. Or am I the only one thinking the mesh should work a bit like this:

u = "foobar"
l = "baz"
mesh(u,l) = "fboaozbar"

One way to do this would be the following:

def mesh(a,b):
    minlen = min(len(a),len(b))
    return "".join(["".join(x+y for x,y in zip(a,b)),a[minlen:],b[minlen:]])

回答9:

I like using two fors, the variable names can give a hint/reminder to what is going on:

"".join(char for pair in zip(u,l) for char in pair)

回答10:

Just to add another, more basic approach:

st = ""
for char in u:
    st = "{0}{1}{2}".format( st, char, l[ u.index( char ) ] )

回答11:

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))

回答12:

Feels a bit un-pythonic not to consider the double-list-comprehension answer here, to handle n string with O(1) effort:

"".join(c for cs in itertools.zip_longest(*all_strings) for c in cs)

where all_strings is a list of the strings you want to interleave. In your case, all_strings = [u, l]. A full use example would look like this:

import itertools
a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
b = 'abcdefghijklmnopqrstuvwxyz'
all_strings = [a,b]
interleaved = "".join(c for cs in itertools.zip_longest(*all_strings) for c in cs)
print(interleaved)
# 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

Like many answers, fastest? Probably not, but simple and flexible. Also, without too much added complexity, this is slightly faster than the accepted answer (in general, string addition is a bit slow in python):

In [7]: l1 = 'A' * 1000000; l2 = 'a' * 1000000;

In [8]: %timeit "".join(a + b for i, j in zip(l1, l2))
1 loops, best of 3: 227 ms per loop

In [9]: %timeit "".join(c for cs in zip(*(l1, l2)) for c in cs)
1 loops, best of 3: 198 ms per loop

回答13:

You could use iteration_utilities.roundrobin¹

u = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
l = 'abcdefghijklmnopqrstuvwxyz'

from iteration_utilities import roundrobin
''.join(roundrobin(u, l))
# returns 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

or the ManyIterables class from the same package:

from iteration_utilities import ManyIterables
ManyIterables(u, l).roundrobin().as_string()
# returns 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'

---

^{1 This is from a third-party library I have written: iteration_utilities.}

回答14:

I would use zip() to get a readable and easy way:

result = ''
for cha, chb in zip(u, l):
    result += '%s%s' % (cha, chb)

print result
# 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz'