简而言之，为了确保在Python 2中正确处理unicode:

使用io。打开文件读写 使用from __future__ import unicode_literals 配置其他数据输入/输出(例如，数据库，网络)使用unicode 如果不能将输出配置为utf-8，请将输出转换为print(text。编码(“ascii”、“替换”).decode ())

有关解释，请参阅@Alastair McCormack的详细回答。

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

"UnicodeDecodeError: 'ascii' codec can't decode byte"

错误原因:input_string必须是unicode，但给出了str

"TypeError: Decoding Unicode is not supported"

此错误的原因:试图将unicode input_string转换为unicode

因此，首先检查你的input_string是否为str，并在必要时转换为unicode:

if isinstance(input_string, str):
   input_string = unicode(input_string, 'utf-8')

其次，上面只是改变了类型，但没有删除非ascii字符。如果你想删除非ascii字符:

if isinstance(input_string, str):
   input_string = input_string.decode('ascii', 'ignore').encode('ascii') #note: this removes the character and encodes back to string.

elif isinstance(input_string, unicode):
   input_string = input_string.encode('ascii', 'ignore')

我得到了字符串“PastelerÃ-a Mallorca”同样的问题，我用:

unicode("Pastelería Mallorca", 'latin-1')

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。

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')