在某些情况下，当你检查你的默认编码(打印sys.getdefaultencoding())时，它会返回你使用的是ASCII。如果您更改为UTF-8，它将不起作用，这取决于变量的内容。 我找到了另一种方法:

import sys
reload(sys)  
sys.setdefaultencoding('Cp1252')

Got a same error and this solved my error. Thanks! python 2 and python 3 differing in unicode handling is making pickled files quite incompatible to load. So Use python pickle's encoding argument. Link below helped me solve the similar problem when I was trying to open pickled data from my python 3.7, while my file was saved originally in python 2.x version. https://blog.modest-destiny.com/posts/python-2-and-3-compatible-pickle-save-and-load/ I copy the load_pickle function in my script and called the load_pickle(pickle_file) while loading my input_data like this:

input_data = load_pickle("my_dataset.pkl")

load_pickle函数在这里:

def load_pickle(pickle_file):
    try:
        with open(pickle_file, 'rb') as f:
            pickle_data = pickle.load(f)
    except UnicodeDecodeError as e:
        with open(pickle_file, 'rb') as f:
            pickle_data = pickle.load(f, encoding='latin1')
    except Exception as e:
        print('Unable to load data ', pickle_file, ':', e)
        raise
    return pickle_data

我发现最好的方法是始终转换为unicode -但这很难实现，因为在实践中，您必须检查并将每个参数转换为您编写的包含某种形式的字符串处理的每个函数和方法。

因此，我提出了以下方法，以保证从任何一个输入中获得unicode或字节字符串。简而言之，包括并使用以下lambdas:

# guarantee unicode string
_u = lambda t: t.decode('UTF-8', 'replace') if isinstance(t, str) else t
_uu = lambda *tt: tuple(_u(t) for t in tt) 
# guarantee byte string in UTF8 encoding
_u8 = lambda t: t.encode('UTF-8', 'replace') if isinstance(t, unicode) else t
_uu8 = lambda *tt: tuple(_u8(t) for t in tt)

例子:

text='Some string with codes > 127, like Zürich'
utext=u'Some string with codes > 127, like Zürich'
print "==> with _u, _uu"
print _u(text), type(_u(text))
print _u(utext), type(_u(utext))
print _uu(text, utext), type(_uu(text, utext))
print "==> with u8, uu8"
print _u8(text), type(_u8(text))
print _u8(utext), type(_u8(utext))
print _uu8(text, utext), type(_uu8(text, utext))
# with % formatting, always use _u() and _uu()
print "Some unknown input %s" % _u(text)
print "Multiple inputs %s, %s" % _uu(text, text)
# but with string.format be sure to always work with unicode strings
print u"Also works with formats: {}".format(_u(text))
print u"Also works with formats: {},{}".format(*_uu(text, text))
# ... or use _u8 and _uu8, because string.format expects byte strings
print "Also works with formats: {}".format(_u8(text))
print "Also works with formats: {},{}".format(*_uu8(text, text))

这里有更多关于这个的推理。

我有同样的错误，url包含非ascii字符(值> 128的字节)，我的解决方案:

url = url.decode('utf8').encode('utf-8')

注意:utf-8, utf8只是别名。只使用'utf8'或'utf-8'应该以同样的方式工作

在我的情况下，为我工作，在Python 2.7中，我认为这个赋值改变了str内部表示中的“某些东西”——即。，它强制正确解码url中支持的字节序列，并最终将字符串放入utf-8 STR中，所有的魔法都在正确的地方。 Python中的Unicode对我来说是一种黑魔法。 希望有用

我遇到了同样的问题，但它不适用于Python 3。我遵循了这个方法，解决了我的问题:

enc = sys.getdefaultencoding()
file = open(menu, "r", encoding = enc)

在读取/写入文件时，必须设置编码。

最后我明白了:

as3:/usr/local/lib/python2.7/site-packages# cat sitecustomize.py
# encoding=utf8  
import sys  

reload(sys)  
sys.setdefaultencoding('utf8')

让我查一下:

as3:~/ngokevin-site# python
Python 2.7.6 (default, Dec  6 2013, 14:49:02)
[GCC 4.4.5] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import sys
>>> reload(sys)
<module 'sys' (built-in)>
>>> sys.getdefaultencoding()
'utf8'
>>>

上面显示了python的默认编码是utf8。那么错误就不再存在了。