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Why is FastAPI's UploadFile Handling Slower Than Flask's?

Patricia Arquette
Patricia ArquetteOriginal
2024-11-08 11:18:021037browse

Why is FastAPI's UploadFile Handling Slower Than Flask's?

FastAPI UploadFile Performance Issues Compared to Flask

FastAPI's UploadFile handling can be slower than Flask's due to differences in file handling. Flask uses synchronous file writing, while FastAPI's UploadFile methods are asynchronous and use a buffer with a default size of 1 MB.

Improved Performance Solution

To improve performance, implement file writing asynchronously with the aiofiles library:

<code class="python">from fastapi import File, UploadFile
import aiofiles

@app.post("/upload")
async def upload_async(file: UploadFile = File(...)):
    try:
        contents = await file.read()
        async with aiofiles.open(file.filename, 'wb') as f:
            await f.write(contents)
    except Exception:
        return {"message": "There was an error uploading the file"}
    finally:
        await file.close()

    return {"message": f"Successfully uploaded {file.filename}"}</code>

Additional Notes

  • This approach keeps the entire file in memory, which may not be ideal for large files.
  • For chunked file upload, consider using the await file.read() method with a smaller buffer size.
  • Alternately, you can use shutil.copyfileobj() with run_in_threadpool() to perform blocking operations in a separate thread, ensuring that the main thread remains responsive.

Streaming Solution

For even better performance, consider accessing the request body as a stream without storing the entire body in memory or a temporary directory:

<code class="python">from fastapi import Request
import aiofiles

@app.post("/upload")
async def upload_stream(request: Request):
    try:
        filename = request.headers['filename']
        async with aiofiles.open(filename, 'wb') as f:
            async for chunk in request.stream():
                await f.write(chunk)
    except Exception:
        return {"message": "There was an error uploading the file"}

    return {"message": f"Successfully uploaded {filename}"}</code>

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