问题：“ for line in…”导致UnicodeDecodeError：’utf-8’编解码器无法解码字节
这是我的代码，
for line in open('u.item'):
#read each line
每当我运行此代码时，都会出现以下错误：
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xe9 in position 2892: invalid continuation byte
我试图解决这个问题，并在open（）中添加了一个额外的参数，代码看起来像；
for line in open('u.item', encoding='utf-8'):
#read each line
但是，它再次给出相同的错误。那我该怎么办！请帮忙。
Here is my code,
for line in open('u.item'):
#read each line
whenever I run this code it gives the following error:
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xe9 in position 2892: invalid continuation byte
I tried to solve this and add an extra parameter in open(), the code looks like;
for line in open('u.item', encoding='utf-8'):
#read each line
But again it gives the same error.
what should I do then! Please help.
回答 0
正如Mark Ransom所建议的，我找到了解决该问题的正确编码。编码为“ ISO-8859-1”，因此替换open("u.item", encoding="utf-8")为open('u.item', encoding = "ISO-8859-1")可以解决该问题。
As suggested by Mark Ransom, I found the right encoding for that problem. The encoding was "ISO-8859-1", so replacing open("u.item", encoding="utf-8") with open('u.item', encoding = "ISO-8859-1") will solve the problem.
回答 1
同样对我有用，ISO 8859-1将节省很多，哈哈哈，主要是如果使用语音识别API的话
例：
file = open('../Resources/' + filename, 'r', encoding="ISO-8859-1");
Also worked for me, ISO 8859-1 is going to save a lot, hahaha, mainly if using Speech Recognition API’s
Example:
file = open('../Resources/' + filename, 'r', encoding="ISO-8859-1");
回答 2
您的文件实际上并不包含utf-8编码的数据，而是包含其他一些编码。弄清楚编码是什么，并在open调用中使用它。
例如，在Windows-1252编码中，0xe9字符为é。
Your file doesn’t actually contain utf-8 encoded data, it contains some other encoding. Figure out what that encoding is and use it in the open call.
In Windows-1252 encoding for example the 0xe9 would be the character é.
回答 3
尝试使用熊猫阅读
pd.read_csv('u.item', sep='|', names=m_cols , encoding='latin-1')
Try this to read using pandas
pd.read_csv('u.item', sep='|', names=m_cols , encoding='latin-1')
回答 4
如果使用Python 2以下将解决方案：
import io
for line in io.open("u.item", encoding="ISO-8859-1"):
# do something
由于encodingparameter不适用于open()，因此您将收到以下错误：
TypeError：“ encoding”是此函数的无效关键字参数
If you are using Python 2 the following will the solution:
import io
for line in io.open("u.item", encoding="ISO-8859-1"):
# do something
Because encoding parameter doesn’t work with open(), you will be getting the following error:
TypeError: 'encoding' is an invalid keyword argument for this function
回答 5
您可以使用以下方法解决问题：
for line in open(your_file_path, 'rb'):
‘rb’以二进制模式读取文件。在这里阅读更多。希望这会有所帮助！
You could resolve the problem with:
for line in open(your_file_path, 'rb'):
‘rb’ is reading file in binary mode. Read more here.
Hope this will help!
回答 6
这有效：
open('filename', encoding='latin-1')
要么：
open('filename',encoding="IS0-8859-1")
This works:
open('filename', encoding='latin-1')
or:
open('filename',encoding="ISO-8859-1")
回答 7
如果有人在寻找这些，这是在Python 3中转换CSV文件的示例：
try:
inputReader = csv.reader(open(argv[1], encoding='ISO-8859-1'), delimiter=',',quotechar='"')
except IOError:
pass
If someone looking for these, this is an example for converting a CSV file in Python 3:
try:
inputReader = csv.reader(open(argv[1], encoding='ISO-8859-1'), delimiter=',',quotechar='"')
except IOError:
pass
回答 8
有时，当open(filepath)在其中filepath实际上不是一个文件会得到同样的错误，所以，首先要确保你试图打开的文件存在：
import os
assert os.path.isfile(filepath)
希望这会有所帮助。
Sometimes when open(filepath) in which filepath actually is not a file would get the same error, so firstly make sure the file you’re trying to open exists:
import os
assert os.path.isfile(filepath)
hope this will help.
回答 9
您可以这样尝试：
open('u.item', encoding='utf8', errors='ignore')
you can try this way:
open('u.item', encoding='utf8', errors='ignore')
