如何将可迭代对象转换为流? [英] How to convert an iterable to a stream?

问题描述

如果我有一个包含字符串的可迭代对象，是否有一种简单的方法可以将其转换为流?我想做这样的事情:

If I've got an iterable containing strings, is there a simple way to turn it into a stream? I want to do something like this:

def make_file():
    yield "hello\n"
    yield "world\n"

output = tarfile.TarFile(…)
stream = iterable_to_stream(make_file())
output.addfile(…, stream)

推荐答案

这是我的流式迭代器，它是 urllib3 支持通过可迭代的流式分块请求:

Here's my streaming iterator an experimental branch of urllib3 supporting streaming chunked request via iterables:

class IterStreamer(object):
    """
    File-like streaming iterator.
    """
    def __init__(self, generator):
        self.generator = generator
        self.iterator = iter(generator)
        self.leftover = ''

    def __len__(self):
        return self.generator.__len__()

    def __iter__(self):
        return self.iterator

    def next(self):
        return self.iterator.next()

    def read(self, size):
        data = self.leftover
        count = len(self.leftover)

        if count < size:
            try:
                while count < size:
                    chunk = self.next()
                    data += chunk
                    count += len(chunk)
            except StopIteration:
                pass

        self.leftover = data[size:]

        return data[:size]

带有上下文的来源: https://github.com/shazow/urllib3/blob/filepost-stream/urllib3/filepost.py#L23

Source with context: https://github.com/shazow/urllib3/blob/filepost-stream/urllib3/filepost.py#L23

相关单元测试: https://github.com/shazow/urllib3/blob/filepost-stream/test/test_filepost.py#L9

可惜的是，此代码尚未将其放入稳定分支中，因为对无大小的分块请求的支持较差，但它应该为您要执行的操作奠定良好的基础.有关显示如何使用它的示例，请参见源链接.

Alas this code hasn't made it into the stable branch yet as sizeless chunked requests are poorly supported, but it should be a good foundation for what you're trying to do. See the source link for examples showing how it can be used.

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