python - Unicode编码错误: 'ascii' codec can't encode character u'\u2013' in position 7

Question

对于以下代码；

phonenumbers = ['(209) 525-2987', '509-477-4598', None, '229-259–1234']
phoneCheck = re.compile('^[1-9]\d{2}-\d{3}-\d{4}$')

for pn in phonenumbers:
    print pn
    if phoneCheck.match(str(pn)):
        print 'Matched!'
    else:
        print 'Not Matched!'

我在结果中收到此错误，我相信这与电话号码中使用的短划线类型错误有关，我该如何更正此错误以将其标记为“不匹配”？

(209) 576-6546
Not Matched!
509-477-6726
Not Matched!
None
Not Matched!
229-259–9756
Runtime error 
Traceback (most recent call last):
  File "<string>", line 6, in <module>
UnicodeEncodeError: 'ascii' codec can't encode character u'\u2013' in position 7: ordinal not in range(128)

Answer 1

您的诊断是正确的。 (最后一个电话号码中的第二个破折号是某种奇特的破折号，我敢打赌您是从文字处理程序或电子表格中复制并粘贴电话号码的。无论如何......)

这里有一个快速简单的方法:安装 unidecode 包，然后:

import re
import warnings

import unidecode

dash = u'\u2013'
phonenumbers = ['(209) 525-2987', '509-477-4598', None, '229-259' + dash + '1234']
phoneCheck = re.compile('^[1-9]\d{2}-\d{3}-\d{4}$')

# if you pass an ascii string into unidecode, it will complain, but still work.
# Just catch the warnings.
with warnings.catch_warnings():
    warnings.simplefilter("ignore")

    for pn in phonenumbers:
        print pn

        # if pn is None, it's not a phone number (and None will cause unidecode
        # to throw an error)
        if pn and phoneCheck.match(unidecode.unidecode(pn)):
            print 'Matched!'
        else:
            print 'Not Matched!'