I am having following code in AWS lambda to zip S3 files

import zipfile
import boto3
from io import BytesIO

def lambda_handler(event, context):
    in_bucket = 'in_bucket'
    in_key = 'test/run_id=123/'
    out_bucket = 'out_bucket'
    out_key = 'test/zip_output/model.zip'
    
    createZipFileStream(in_bucket, in_key, out_bucket, out_key)

def create_zip(in_bucket, in_key, out_bucket, out_key):
    s3 = boto3.resource('s3')
    bucket = s3.Bucket(in_bucket)
    files_collection = bucket.objects.filter(Prefix=in_key).all()
    
    archive = BytesIO()
    with zipfile.ZipFile(archive, 'w', zipfile.ZIP_DEFLATED) as zip_archive:
        for f in files_collection:
            with zip_archive.open(f.key, 'w') as file1:
                file1.write(f.get()['Body'].read())  

    archive.seek(0)
    s3.Object(out_bucket, out_key).upload_fileobj(archive)
    archive.close()

Above code though work successfully the zip contains entire S3 path. I would like to have only the file objects in the zip and not the path location. For example, if the input path s3://bucket/key/file.csv the zip has folder key and then file.csv and I am looking for a solution to have only the file object.

Thank you

>Solution :

It appears that this line:

with zip_archive.open(f.key, 'w') as file1:

is telling the Zip what to name the file.

Therefore, you could change it like this:

with zip_archive.open(f.key.split('/')[-1], 'w') as file1:

That will give it only the characters found after the last slash.