How to convert an iterable to a stream?

Question

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)

Answer 1

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]

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

Related unit tests: https://github.com/shazow/urllib3/blob/filepost-stream/test/test_filepost.py#L9

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.

Answer 2

Python 3 has a new I/O stream API (library docs), replacing the old file-like object protocol. (The new API is also available in Python 2 in the io module, and it's backwards-compatible with the file-like object protocol.)

Here's an implementation for the new API, in Python 2 and 3:

import io

def iterable_to_stream(iterable, buffer_size=io.DEFAULT_BUFFER_SIZE):
    """
    Lets you use an iterable (e.g. a generator) that yields bytestrings as a read-only
    input stream.

    The stream implements Python 3's newer I/O API (available in Python 2's io module).
    For efficiency, the stream is buffered.
    """
    class IterStream(io.RawIOBase):
        def __init__(self):
            self.leftover = None
        def readable(self):
            return True
        def readinto(self, b):
            try:
                l = len(b)  # We're supposed to return at most this much
                chunk = self.leftover or next(iterable)
                output, self.leftover = chunk[:l], chunk[l:]
                b[:len(output)] = output
                return len(output)
            except StopIteration:
                return 0    # indicate EOF
    return io.BufferedReader(IterStream(), buffer_size=buffer_size)

Example usage:

with iterable_to_stream(str(x**2).encode('utf8') for x in range(11)) as s:
    print(s.read())

Answer 3

Since it doesn't look like there is a "standard" way of doing it, I've banged together a simple implementation:

class iter_to_stream(object):
    def __init__(self, iterable):
        self.buffered = ""
        self.iter = iter(iterable)

    def read(self, size):
        result = ""
        while size > 0:
            data = self.buffered or next(self.iter, None)
            self.buffered = ""
            if data is None:
                break
            size -= len(data)
            if size < 0:
                data, self.buffered = data[:size], data[size:]
            result += data
        return result

Answer 4

A starting point:

class iterable_to_stream:
    def __init__(self, iterable):
        self.iter = iter(iterable)

    def read(self):
        try:
            return self.iter.next()
        except StopIteration:
            return ""