将 s3 的内容写入 CSV

Question

我正在创建一个脚本，将我的 s3 数据抓取到我的本地计算机。通常，我收到的数据是配置单元分区的数据。即使文件确实存在，我也会收到 No such file or directory 错误。有人可以解释我做错了什么以及我应该如何以不同的方式处理这个问题？这是错误引用的一段代码：

bucket = conn.get_bucket(bucket_name)
for sub in bucket.list(prefix = 'some_prefix'):
        matched = re.search(re.compile(read_key_pattern), sub.name)
        if matched:
            with open(sub.name, 'rb') as fin:
                reader = csv.reader(fin, delimiter = '\x01')
                contents = [line for line in reader]
            with open('output.csv', 'wb') as fout:
                writer = csv.writer(fout, quotechar = '', quoting = csv.QUOTE_NONE, escapechar = '\\')
                writer.writerows.content

IOError: [Errno 2] 没有这样的文件或目录：'my_prefix/54c91e35-4dd0-4da6-a7b7-283dff0f4483-000000'

该文件存在，这是我要检索的正确文件夹和文件。

Answer 1

就像@roganjosh 所说，在您测试名称匹配后，您似乎还没有downloaded the file。我在下面添加了 cmets 来向您展示如何在 python 2 中处理内存中的文件：

    from io import StringIO # alternatively use BytesIO
    import contextlib

    bucket = conn.get_bucket(bucket_name)
    # use re.compile outside of the for loop
    # it has slightly better performance characteristics
    matcher = re.compile(read_key_pattern)

    for sub in bucket.list(prefix = 'some_prefix'):
        # bucket.list returns an iterator over s3.Key objects
        # so we can use `sub` directly as the Key object
        matched = matcher.search(sub.name)
        if matched:
            # download the file to an in-memory buffer
            with contextlib.closing(StringIO()) as fp:
                sub.get_contents_to_file(fp)
                fp.seek(0)
                # read straight from the memory buffer
                reader = csv.reader(fp, delimiter = '\x01')
                contents = [line for line in reader]
            with open('output.csv', 'wb') as fout:
                writer = csv.writer(fout, quotechar = '', quoting = csv.QUOTE_NONE, escapechar = '\\')
                writer.writerows.content    

对于 python 3，您需要将 cmets 中讨论的 with 语句更改为答案 for this question。