字符编码在python中的处理_字符编码在python中用'u2019'替换'

最新推荐文章于 2023-11-15 14:21:42 发布

weixin_39610785

最新推荐文章于 2023-11-15 14:21:42 发布

阅读量508

点赞数

文章标签：字符编码在python中的处理

本文探讨了在Python中使用内置函数处理字符串时，如何确保'字符在'BACKRUSHIN'这样的字符串中正确显示，重点介绍了unidecode库的应用。作者分享了尝试过的多种方法，并最终通过unidecode.unidecode()函数解决了问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

解决方案>>> import unidecode

>>> unidecode.unidecode(u'BACK RUSHIN\u2019')

"BACK RUSHIN'"