convert io.StringIO to io.BytesIO

Question:

original question: i got a StringIO object, how can i convert it into BytesIO?

update: The more general question is, how to convert a binary (encoded) file-like object into decoded file-like object in python3?

the naive approach i got is:

import io
sio = io.StringIO('wello horld')
bio = io.BytesIO(sio.read().encode('utf8'))
print(bio.read())  # prints b'wello horld'

is there more efficient and elegant way of doing this? the above code just reads everything into memory, encodes it instead of streaming the data in chunks.

for example, for the reverse question (BytesIO -> StringIO) there exist a class – io.TextIOWrapper which does exactly that (see this answer)

Asked By: ShmulikA

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Answers:

It could be a generally useful tool to convert a character stream into a byte stream, so here goes:

import io

class EncodeIO(io.BufferedIOBase):
  def __init__(self,s,e='utf-8'):
    self.stream=s               # not raw, since it isn't
    self.encoding=e
    self.buf=b""                # encoded but not yet returned
  def _read(self,s): return self.stream.read(s).encode(self.encoding)
  def read(self,size=-1):
    b=self.buf
    self.buf=b""
    if size is None or size<0: return b+self._read(None)
    ret=[]
    while True:
      n=len(b)
      if size<n:
        b,self.buf=b[:size],b[size:]
        n=size
      ret.append(b)
      size-=n
      if not size: break
      b=self._read(min((size+1024)//2,size))
      if not b: break
    return b"".join(ret)
  read1=read

Obviously write could be defined symmetrically to decode input and send it to the underlying stream, although then you have to deal with having enough bytes for only part of a character.

Answered By: Davis Herring

bio from your example is _io.BytesIO class object.
You have used 2 times the read() function.

I came up with bytes conversion and one read() method:

sio = io.StringIO('wello horld')
b = bytes(sio.read(), encoding='utf-8')
print(b)

But the second variant should be even faster:

sio = io.StringIO('wello horld')
b = sio.read().encode()
print(b)
Answered By: prosti

As some pointed out, you need to do the encoding/decoding yourself.

However, you can achieve this in an elegant way – implementing your own TextIOWrapper for string => bytes.

Here is such a sample:

class BytesIOWrapper:
    def __init__(self, string_buffer, encoding='utf-8'):
        self.string_buffer = string_buffer
        self.encoding = encoding

    def __getattr__(self, attr):
        return getattr(self.string_buffer, attr)

    def read(self, size=-1):
        content = self.string_buffer.read(size)
        return content.encode(self.encoding)

    def write(self, b):
        content = b.decode(self.encoding)
        return self.string_buffer.write(content)

Which produces an output like this:

In [36]: bw = BytesIOWrapper(StringIO("some lengt˙˚hyÔstring in here"))

In [37]: bw.read(15)
Out[37]: b'some lengtxcbx99xcbx9ahyxc3x94'

In [38]: bw.tell()
Out[38]: 15

In [39]: bw.write(b'ME')
Out[39]: 2

In [40]: bw.seek(15)
Out[40]: 15

In [41]: bw.read()
Out[41]: b'MEring in here'

Hope it clears your thoughts!

Answered By: foobarna

@foobarna answer can be improved by inheriting some io base-class

import io
sio = io.StringIO('wello horld')


class BytesIOWrapper(io.BufferedReader):
    """Wrap a buffered bytes stream over TextIOBase string stream."""

    def __init__(self, text_io_buffer, encoding=None, errors=None, **kwargs):
        super(BytesIOWrapper, self).__init__(text_io_buffer, **kwargs)
        self.encoding = encoding or text_io_buffer.encoding or 'utf-8'
        self.errors = errors or text_io_buffer.errors or 'strict'

    def _encoding_call(self, method_name, *args, **kwargs):
        raw_method = getattr(self.raw, method_name)
        val = raw_method(*args, **kwargs)
        return val.encode(self.encoding, errors=self.errors)

    def read(self, size=-1):
        return self._encoding_call('read', size)

    def read1(self, size=-1):
        return self._encoding_call('read1', size)

    def peek(self, size=-1):
        return self._encoding_call('peek', size)


bio = BytesIOWrapper(sio)
print(bio.read())  # b'wello horld'
Answered By: imposeren

It’s interesting that though the question might seem reasonable, it’s not that easy to figure out a practical reason why I would need to convert a StringIO into a BytesIO. Both are basically buffers and you usually need only one of them to make some additional manipulations either with the bytes or with the text.

I may be wrong, but I think your question is actually how to use a BytesIO instance when some code to which you want to pass it expects a text file.

In which case, it is a common question and the solution is codecs module.

The two usual cases of using it are the following:

Compose a File Object to Read

In [16]: import codecs, io

In [17]: bio = io.BytesIO(b'qwenasdn')

In [18]: StreamReader = codecs.getreader('utf-8')  # here you pass the encoding

In [19]: wrapper_file = StreamReader(bio)

In [20]: print(repr(wrapper_file.readline()))
'qwen'

In [21]: print(repr(wrapper_file.read()))
'asdn'

In [26]: bio.seek(0)
Out[26]: 0

In [27]: for line in wrapper_file:
    ...:     print(repr(line))
    ...:
'qwen'
'asdn'

Compose a File Object to Write To

In [28]: bio = io.BytesIO()

In [29]: StreamWriter = codecs.getwriter('utf-8')  # here you pass the encoding

In [30]: wrapper_file = StreamWriter(bio)

In [31]: print('жаба', 'цап', file=wrapper_file)

In [32]: bio.getvalue()
Out[32]: b'xd0xb6xd0xb0xd0xb1xd0xb0 xd1x86xd0xb0xd0xbfn'

In [33]: repr(bio.getvalue().decode('utf-8'))
Out[33]: "'жаба цап\n'"
Answered By: newtover

I had the exact same need, so I created an EncodedStreamReader class in the nr.utils.io package. It also solves the issue with actually reading the number of bytes requested instead of the number of characters from the wrapped stream.

$ pip install 'nr.utils.io>=0.1.0,<1.0.0'

Example usage:

import io
from nr.utils.io.readers import EncodedStreamReader
fp = EncodedStreamReader(io.StringIO('ä'), 'utf-8')
assert fp.read(1) == b'xc3'
assert fp.read(1) == b'xa4'
Answered By: Niklas R