Python: How can I replace full-width characters with half-width characters?

Question

If this was PHP, I would probably do something like this:

function no_more_half_widths($string){
  $foo = array(\'１\',\'２\',\'３\',\'４\',\'５\',\'６\',\'７\',\'８         


        

Answer 1

I don't think there's a built-in function to do multiple replacements in one pass, so you'll have to do it yourself.

One way to do it:

>>> src = (u'１',u'２',u'３',u'４',u'５',u'６',u'７',u'８',u'９',u'１０')
>>> dst = ('1','2','3','4','5','6','7','8','9','0')
>>> string = u'a１２３'
>>> for i, j in zip(src, dst):
...     string = string.replace(i, j)
... 
>>> string
u'a123'

Or using a dictionary:

>>> trans = {u'１': '1', u'２': '2', u'３': '3', u'４': '4', u'５': '5', u'６': '6', u'７': '7', u'８': '8', u'９': '9', u'０': '0'}
>>> string = u'a１２３'
>>> for i, j in trans.iteritems():
...     string = string.replace(i, j)
...     
>>> string
u'a123'

Or finally, using regex (and this might actually be the fastest):

>>> import re
>>> trans = {u'１': '1', u'２': '2', u'３': '3', u'４': '4', u'５': '5', u'６': '6', u'７': '7', u'８': '8', u'９': '9', u'０': '0'}
>>> lookup = re.compile(u'|'.join(trans.keys()), re.UNICODE)
>>> string = u'a１２３'
>>> lookup.sub(lambda x: trans[x.group()], string)
u'a123'