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S3FS is a PyFilesystem interface to Amazon S3 cloud storage.

As a PyFilesystem concrete class, S3FS allows you to work with S3 in the same way as any other supported filesystem.


You can install S3FS from pip as follows:

pip install fs-s3fs

Opening a S3FS

Open an S3FS by explicitly using the constructor:

from fs_s3fs import S3FS
s3fs = S3FS('mybucket')

Or with a FS URL:

  from fs import open_fs
  s3fs = open_fs('s3://mybucket')

Downloading Files

To download files from an S3 bucket, open a file on the S3 filesystem for reading, then write the data to a file on the local filesystem. Here's an example that copies a file from S3 to your HD:

from import copy_file_data
with'', 'rb') as remote_file:
    with open('', 'wb') as local_file:
        copy_file_data(remote_file, local_file)

Although it is preferable to use the higher-level functionality in the fs.copy module. Here's an example:

from fs.copy import copy_file
copy_file(s3fs, '', './', '')

Uploading Files

You can upload files in the same way. Simply copy a file from a source filesystem to the S3 filesystem. See Moving and Copying for more information.


S3 objects have additional properties, beyond a traditional filesystem. These options can be set using the upload_args and download_args properties. which are handed to upload and download methods, as appropriate, for the lifetime of the filesystem instance.

For example, to set the cache-control header of all objects uploaded to a bucket:

import fs, fs.mirror
s3fs = S3FS('example', upload_args={"CacheControl": "max-age=2592000", "ACL": "public-read"})
fs.mirror.mirror('/path/to/mirror', s3fs)

see the Boto3 docs for more information.

acl and cache_control are exposed explicitly for convenience, and can be used in URLs. It is important to URL-Escape the cache_control value in a URL, as it may contain special characters.

import fs, fs.mirror
with open fs.open_fs('s3://example?acl=public-read&cache_control=max-age%3D2592000%2Cpublic') as s3fs
    fs.mirror.mirror('/path/to/mirror', s3fs)


You can get a public URL to a file on a S3 bucket as follows:

movie_url = s3fs.geturl('')