我在Python2.7中遇到了这个错误。我在尝试运行许多python程序时遇到了这种情况，但我设法用这个简单的脚本重现了它:

#!/usr/bin/env python

import subprocess
import sys

result = subprocess.Popen([u'svn', u'info'])
if not callable(getattr(result, "__enter__", None)) and not callable(getattr(result, "__exit__", None)):
    print("foo")
print("bar")

在成功的情况下，它应该打印出'foo'和'bar'，如果你不在svn文件夹中，可能会有一个错误消息。

在失败时，它应该打印'UnicodeDecodeError: 'ascii' codec不能解码字节0xc4在位置39:序号不在范围(128)'。

在尝试重新生成区域设置和这个问题中发布的许多其他解决方案后，我了解到发生了错误，因为我的PATH环境变量中编码了一个特殊字符(ĺ)。在` ~/中固定PATH后。Bashrc '，然后退出我的会话并再次进入，(显然是在查找'~/。Bashrc’没有起作用)，问题就消失了。

这招对我很管用:

    file = open('docs/my_messy_doc.pdf', 'rb')

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

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

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

Encode将unicode对象转换为字符串对象。我认为你正在尝试编码一个字符串对象。首先将结果转换为unicode对象，然后将该unicode对象编码为'utf-8'。 例如

    result = yourFunction()
    result.decode().encode('utf-8')

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。