返回两个不同长度的字符串中的交替字母

Question

I currently have a code written that recursively takes the letters within two strings and returns the new word with alternating letters. I would like to optimize this code so that if the first or second word was longer it will still return the remaining letters in the longer string.

def alt(s,t):
    if len(s) != len(t):
        return 
    elif s == '' and t == '':
        return ''
    else:
        return s[0] + t[0] + alt(s[1:], t[1:])

Desired Output:

>>> alt('hello','bye')
'hbeylelo'

Answer 1

Just test for s and t are empty and return the other value if one of them is:

def alt(s, t):
    if not s:
        return t
    elif not t:
        return s
    else:
        return s[0] + t[0] + alt(s[1:], t[1:])

Even if both s and t are empty, the empty string is returned, which is a perfectly valid end-state.

You could shorten this to:

def alt(s, t):
    if not (s and t):
        return s + t
    return s[0] + t[0] + alt(s[1:], t[1:])

so the end-state is reached whenever either s or t is empty (or they both are).

This produces your desired output:

>>> alt('hello', 'bye')
'hbeylelo'

An iterative version would be:

from itertools import chain
try:
    # Python 2
    from itertools import izip_longest as zip_longest
except ImportError:
    # Python 3
    from itertools import zip_longest

def alt_iterative(s, t):
    return ''.join(chain.from_iterable(zip_longest(s, t, fillvalue='')))

This uses the itertools.zip_longest() function to do most of the work.

Answer 2

The problem itself is not inherently recursive. An iterative solution may be simpler in this case.

Python 2:

from itertools import izip_longest
def alt(s, t):
    return ''.join(a + b for a, b in izip_longest(s, t, fillvalue=''))

Python 3:

from itertools import zip_longest
def alt(s, t):
    return ''.join(a + b for a, b in zip_longest(s, t, fillvalue=''))

Answer 3

If you like a one-liner using itertools.chain and zip (or itertools.zip_longest / izip_longest ):

# Python 2
return ''.join(chain(*zip(s, t))) + s[len(t):] + t[len(s):]
# or
return ''.join(chain(*izip_longest(s, t, fillvalue='')))

# Python 3
return ''.join(chain(*zip_longest(s, t, fillvalue='')))