python - 在 Python 中将 CSV 转换为 UTF-8

Question

我正在尝试创建一个没有标题的重复 CSV。当我尝试这样做时，我收到以下错误:

UnicodeDecodeError: 'utf-8' codec can't decode byte 0xa0 in position 1895: invalid start byte.

我已经阅读了关于 CSV 和 Unicode 编码的 python UTF-8 documentation 并实现了它。
但是，生成的输出文件中没有数据。不确定我在这里做错了什么。

import csv

path =  '/Users/johndoe/file.csv'

with open(path, 'r') as infile, open(path + 'final.csv', 'w') as outfile:

    def unicode_csv(infile, outfile):
        inputs = csv.reader(utf_8_encoder(infile))
        output = csv.writer(outfile)

        for index, row in enumerate(inputs):
            yield [unicode(cell, 'utf-8') for cell in row]
            if index == 0:
                 continue
        output.writerow(row)

    def utf_8_encoder(infile):
        for line in infile:
            yield line.encode('utf-8')

unicode_csv(infile, outfile)

Answer 1

解决方案是简单地将两个附加参数添加到

with open(path, 'r') as infile:

这两个参数是 encoding='UTF-8' 和 errors='ignore'。这使我可以创建没有标题和 UnicodeDecodeError 的原始 CSV 的副本。下面是完成的代码。

import csv

path =  '/Users/johndoe/file.csv'

with open(path, 'r', encoding='utf-8', errors='ignore') as infile, open(path + 'final.csv', 'w') as outfile:
     inputs = csv.reader(infile)
     output = csv.writer(outfile)

     for index, row in enumerate(inputs):
         # Create file with no header
         if index == 0:
             continue
         output.writerow(row)