I have a python app running on a Jupiter-notebook on AWS. I loaded a C-library into my python code which expects a path to a file. I would like to access this file from the S3 bucket. I tried to use s3fs: s3 = s3fs.S3FileSystem(anon=False) using s3.ls('..') lists all my bucket files... this is ok so far. But, the library I am using should actually use the s3 variable inside where I have no access. I can only pass the path to the c library. Is there a way to mount the s3 bucket in a way, where I don't have to call s3.open(), and can just call open(/path/to/s3) were somewhere hidden the s3 bucket is really mounted as a local filesystem? I think it should work like this without using s3. Because I can't change the library I am using internally to use the s3 variable... with s3.open("path/to/s3/file",'w') as f: df.to_csv(f) with open("path/to/s3/file",'w') as f: df.to_csv(f) Or am I doing it completely wrong? The c library iam using is loaded as DLL in python and i call a function : lib.OpenFile(path/to/s3/file) I have to pass the path to s3 into the library OpenFile function. Continue reading...