S3 is not a file server, nor does it offer operating system file services, such as data manipulation.

If there is many "HUGE" files, your best bet is

1. start a simple EC2 instance
2. Download all those files to EC2 instance, compress them, reupload it back to S3 bucket with a new object name

Yes, you can use AWS lambda to do the same thing, but lambda is bounds to 900 seconds (15 mins) execution timeout (Thus it is recommended to allocate more RAM to boost lambda execution performance)

Traffics from S3 to local region EC2 instance and etc services is FREE.

If your main purpose is just to read those file within same AWS region using EC2/etc services, then you don't need this extra step. Just access the file directly.

(Update) : As mentioned by @Robert Reiz, now you can also use AWS Fargate to do the job.

Note :

It is recommended to access and share file using AWS API. If you intend to share the file publicly, you must look into security issue seriously and impose download restriction. AWS traffics out to internet is never cheap.

Hi I recently have to do that for my application -- serve a bundle of files in zip format through a url link that the users can download.

In a nutshell, first create an object using BytesIO method, then use the ZipFile method to write into this object by iterating all the s3 objects, then use put method on this zip object and create a presiged url for it.

The code I used looks like this:

First, call this function to get the zip object, ObjectKeys are the s3 objects that you need to put into the zip file.

def zipResults(bucketName, ObjectKeys):
    buffer = BytesIO()
    with zipfile.ZipFile(buffer, 'w', compression=zipfile.ZIP_DEFLATED) as zip_file:
        for ObjectKey in ObjectKeys:
            objectContent = S3Helper().readFromS3(bucketName, ObjectKey)
            fileName = os.path.basename(ObjectKey)
            zip_file.writestr(fileName, objectContent)

    buffer.seek(0)
    return buffer

Then call this function, key is the key you give to your zip object:

def uploadObject(bucketName, body, key):
    s3client = AwsHelper().getClient("s3")
    try:
        response = s3client.put_object(
            Bucket=bucketName,
            Body=body,
            Key=key
        )
    except ClientError as e:
        logging.error(e)
        return None

    return response

Of course, you would need io, zipfile and boto3 modules.

If you need individual files (objects) in S3 compressed, then it is possible to do so in a round-about way. You can define a CloudFront endpoint pointing to the S3 bucket, then let CloudFront compress the content on the way out: https://docs.aws.amazon.com/AmazonCloudFront/latest/DeveloperGuide/ServingCompressedFiles.html