Python “for line in...”导致 UnicodeDecodeError: 'utf-8' codec can't decode byte
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"for line in..." results in UnicodeDecodeError: 'utf-8' codec can't decode byte
提问by SujitS
Here is my code,
这是我的代码，
for line in open('u.item'):
#read each line
whenever I run this code it gives the following error:
每当我运行此代码时，它都会出现以下错误：
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;
我试图解决这个问题并在 open() 中添加一个额外的参数，代码看起来像；
for line in open('u.item', encoding='utf-8'):
#read each line
But again it gives the same error. what should I do then! Please help.
但它再次给出了同样的错误。那我该怎么办！请帮忙。
采纳答案by SujitS
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.
正如 Mark Ransom 所建议的，我找到了解决该问题的正确编码。编码是“ISO-8859-1”，所以替换open("u.item", encoding="utf-8")为open('u.item', encoding = "ISO-8859-1")可以解决问题。
回答by Mark Ransom
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 opencall.
您的文件实际上并不包含 utf-8 编码数据，它包含一些其他编码。弄清楚该编码是什么并在open调用中使用它。
In Windows-1252 encoding for example the 0xe9would be the character é.
例如，在 Windows-1252 编码中，0xe9将是字符é.
回答by user6832484
If someone looking for these, this is an example for converting a CSV file in Python 3:
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
回答by Shashank
Try this to read using pandas
试试这个使用熊猫阅读
pd.read_csv('u.item', sep='|', names=m_cols , encoding='latin-1')
回答by Jeril
If you are using Python 2the following will the solution:
如果您正在使用Python 2以下解决方案：
import io
for line in io.open("u.item", encoding="ISO-8859-1"):
# do something
Because encodingparameter doesn't work with open(), you will be getting the following error:
由于encoding参数不适用于open()，您将收到以下错误：
TypeError: 'encoding' is an invalid keyword argument for this function
回答by Ryoji Kuwae Neto
Also worked for me, ISO 8859-1 is going to save a lot, hahaha, mainly if using Speech Recognition API's
也对我有用，ISO 8859-1 会节省很多，哈哈哈，主要是如果使用语音识别 API 的
Example:
例子：
file = open('../Resources/' + filename, 'r', encoding="ISO-8859-1");
回答by xtluo
Sometimeswhen open(filepath)in which filepathactually is not a file would get the same error, so firstly make sure the file you're trying to open exists:
有时当open(filepath)其中filepath实际上不是文件时会出现相同的错误，因此首先确保您尝试打开的文件存在：
import os
assert os.path.isfile(filepath)
hope this will help.
希望这会有所帮助。
回答by Ozcar Nguyen
回答by Ayesha Siddiqa
This works:
这有效：
open('filename', encoding='latin-1')
or:
或者：
open('filename',encoding="IS0-8859-1")
回答by FaridLU
you can try this way:
你可以试试这种方式：
open('u.item', encoding='utf8', errors='ignore')
