Question

所以我试图打开并读取没有字段名称的csv文件。根据我所做的研究，我很确定它是用UTF-8编码的。我的csv有这种格式：

1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

我使用以下内容打开并阅读它：

def parseCSVCounter(csv_file):

with codecs.open(csv_file, "r", "utf-8-sig","strict", -1) as f:
    f = str(f)
    relayreader = csv.reader(f, delimiter=',')
    for row in relayreader:
        print(row)

        try:
            #row[0] = unicode(row[0], 'latin-1')
            counter(row)
            print('starting row..')

        except UnicodeDecodeError, e:
            print('something went wrong1')
            print e

        except Exception, e:
            print('something went wrong')
            print e

这会生成

Starting Command..
['<']
something went wrong
invalid literal for int() with base 10: '<'
['o']
something went wrong
invalid literal for int() with base 10: 'o'
........
starting row..
['9']
starting row..
['3']
starting row..
['8']
starting row..
['2']
starting row..
['8']
starting row..
['>']
something went wrong
invalid literal for int() with base 10: '>'`

我减少了这一点以证明我的观点。它似乎会自动为我生成字段名称。使用csv.DictReader(fieldnames = 'foo')，我可以在序列中指定字段名称。如何让csv.reader()忽略缺少字段名称？

Answer 1

你不需要致电str(f);直接使用文件对象：

with codecs.open(csv_file, "r", "utf-8-sig", "strict") as f:
    relayreader = csv.reader(f, delimiter=',')

您正在尝试将str(f)的输出读作CSV文件，而这是一个以下形式的字符串：

<open file '/path/to/file', mode 'rb' at 0x105f10d20>

您可以从错误输出中看到;它拼写出<，o等，一直到内存地址数字和结束>。

请注意，utf-8-sig编解码器可以处理文件开头存在的UTF-8编码BOM，但除非预期该BOM存在，否则正常{{1}编解码器就可以了。

没有字段名称CSV Reader Python UTF-8

1 个答案: