How to read content from the s3 bucket as url

Question

I have s3 bucket url is below

s3_filename is s3://xx/xx/y/z/ion.csv

if its is bucket i can read like below code

def read_s3(bucket,         


        

Answer 1

Since you appear to be using Pandas, please note that it actually uses s3fs under the cover. So, if your install is relatively recent and standard, you may directly do:

df = pd.read_csv(s3_path)

If you have some specific config for your bucket, for example special credentials, KMS encryption, etc., you may use an explicitly configured s3fs filesystem, for example:

fs = s3fs.S3FileSystem(
    key=my_aws_access_key_id,
    secret=my_aws_secret_access_key,
    s3_additional_kwargs={
            'ServerSideEncryption': 'aws:kms',
            'SSEKMSKeyId': my_kms_key,
    },
)
# note: KMS encryption only used when writing; when reading, it is automatic if you have access

with fs.open(s3_path, 'r') as f:
    df = pd.read_csv(f)

# here we write the same df at a different location, making sure
# it is using my_kms_key:
with fs.open(out_s3_path, 'w') as f:
    df.to_csv(f)

That said, if you are really interested to deal yourself with getting the object, and the question is just about how to remove a potential s3:// prefix and then split bucket/key, you could simply use:

bucket, key = re.sub(r'^s3://', '', s3_path).split('/', 1)

But that may miss more general cases and conventions handled by systems such as awscli or the very s3fs referenced above.

For more generality, you can take a look at how they do this in awscli. In general, doing so often provides a good indication of whether or not some functionality may already be built in boto3 or botocore. In this case however, it would appear not (looking at a local clone of release-1.18.126). They simply do this from first principles: see awscli.customizations.s3.utils.split_s3_bucket_key as it is implemented here.

From the regex that is eventually used in that code, you can infer that the kind of cases awscli allows for s3_path is quite diverse indeed:

_S3_ACCESSPOINT_TO_BUCKET_KEY_REGEX = re.compile(
    r'^(?P<bucket>arn:(aws).*:s3:[a-z\-0-9]+:[0-9]{12}:accesspoint[:/][^/]+)/?'
    r'(?P<key>.*)$'
)