Character encoding in python to replace 'u2019' with '
Question: I have tried numerous ways to encode this to the end result `"BACK RUSHIN'"`
with the most important character being the right apostrophe `'`.

I would like a way of getting to this end result using some of the built in
functions Python has where there is no discrimination between a normal string
and a unicode string.

This was the code I was using to retrieve the string:
`str(unicode(etree.tostring(root.xpath('path')[0],method='text',
encoding='utf-8'),errors='ignore')).strip()`

With the result being: `'BACK RUSHIN'` the thing being the apostrophe `'` is
missing.

Another way was: `root.xpath('path/text()')`

And that result was: `u'BACK RUSHIN\u2019'` in python.

Lastly if I try: `u'BACK RUSHIN\u2019'.encode('ascii', 'replace')`

The result is: `'BACK RUSHIN?'`

Please no replace functions, I would like to make use of pythons codec
libraries. Also no printing the string because it is being held in a variable.

Thanks


Answer: 
    >>> import unidecode
    >>> unidecode.unidecode(u'BACK RUSHIN\u2019')
    "BACK RUSHIN'"
    

[unidecode](https://pypi.python.org/pypi/Unidecode)

