I want to swap each pair of characters in a string. \'2143\'
becomes \'1234\'
, \'badcfe\'
becomes \'abcdef\'
.
How
If performance or elegance is not an issue, and you just want clarity and have the job done then simply use this:
def swap(text, ch1, ch2):
text = text.replace(ch2, '!',)
text = text.replace(ch1, ch2)
text = text.replace('!', ch1)
return text
This allows you to swap or simply replace chars or substring. For example, to swap 'ab' <-> 'de' in a text:
_str = "abcdefabcdefabcdef"
print swap(_str, 'ab','de') #decabfdecabfdecabf
Do you want the digits sorted? Or are you swapping odd/even indexed digits? Your example is totally unclear.
Sort:
s = '2143'
p=list(s)
p.sort()
s = "".join(p)
s is now '1234'. The trick is here that list(string) breaks it into characters.
oneliner:
>>> s = 'badcfe'
>>> ''.join([ s[x:x+2][::-1] for x in range(0, len(s), 2) ])
'abcdef'
One more way:
>>> s='123456'
>>> ''.join([''.join(el) for el in zip(s[1::2], s[0::2])])
'214365'
''.join(s[i+1]+s[i] for i in range(0, len(s), 2)) # 10.6 usec per loop
or
''.join(x+y for x, y in zip(s[1::2], s[::2])) # 10.3 usec per loop
or if the string can have an odd length:
''.join(x+y for x, y in itertools.izip_longest(s[1::2], s[::2], fillvalue=''))
Note that this won't work with old versions of Python (if I'm not mistaking older than 2.5).
The benchmark was run on python-2.7-8.fc14.1.x86_64 and a Core 2 Duo 6400 CPU with s='0123456789'*4
.
Loop over length of string by twos and swap:
def oddswap(st):
s = list(st)
for c in range(0,len(s),2):
t=s[c]
s[c]=s[c+1]
s[c+1]=t
return "".join(s)
giving:
>>> s
'foobar'
>>> oddswap(s)
'ofbora'
and fails on odd-length strings with an IndexError exception.