I am uploading a relatively small(<1 MiB) .jsonl file on Google CLoud storage using the python API. The function I used is from the gcp documentation:
def upload_blob(key_path,bucket_name, source_file_name, destination_blob_name):
"""Uploads a file to the bucket."""
# The ID of your GCS bucket
# bucket_name = "your-bucket-name"
# The path to your file to upload
# source_file_name = "local/path/to/file"
# The ID of your GCS object
# destination_blob_name = "storage-object-name"
storage_client = storage.Client.from_service_account_json(key_path)
bucket = storage_client.bucket(bucket_name)
blob = bucket.blob(destination_blob_name)
blob.upload_from_filename(source_file_name)
print(
"File {} uploaded to {}.".format(
source_file_name, destination_blob_name
)
)
The issue I am having is that the .jsonl file is getting truncated at 9500 lines after the upload. In fact, the 9500th line is not complete. I am not sure what the issue is and don't think there would be any limit for this small file. Any help is appreciated.
I had a similar problem some time ago. In my case the upload to bucket was called inside a with python clause right after the line where I recorded contents to source_file_name, so I just needed to move the upload line outside the with in order to properly recorded and close local file to be uploaded.
Related
I am developing a webapplication with Flask as the backend and Nuxt JS as the frontend. I receive an image file from the frontend and can save it to my Flask directory structure locally. The file is ok and the images is being shown if I open it. Now i want to upload this image to AWS S3 instead of saving it to my disk. I use the boto3 SDK, here is my code:
Here is my save_picture method, that opens the image file and resizes it. I had the save method, but commented it out to avoid saving the file to disk as I want it only on S3.
def save_picture(object_id, form_picture, path):
if form_picture is None:
return None
random_hex = token_hex(8)
filename = form_picture.filename
if '.' not in filename:
return None
extension = filename.rsplit('.', 1)[1].lower()
if not allowed_file(extension, form_picture):
return None
picture_fn = f'{object_id}_{random_hex}.{extension}'
picture_path = current_app.config['UPLOAD_FOLDER'] / path / picture_fn
# resizing image and saving the small version
output_size = (1280, 720)
i = Image.open(form_picture)
i.thumbnail(output_size)
# i.save(picture_path)
return picture_fn
image_name = save_picture(object_id=new_object.id, form_picture=file, path=f'{object_type}_images')
s3 = boto3.client(
's3',
aws_access_key_id=current_app.config['AWS_ACCESS_KEY'],
aws_secret_access_key=current_app.config['AWS_SECRET_ACCESS_KEY']
)
print(file) # this prints <FileStorage: 'Capture.JPG' ('image/jpeg')>, so the file is ok
try:
s3.upload_fileobj(
file,
current_app.config['AWS_BUCKET_NAME'],
image_name,
ExtraArgs={
'ContentType': file.content_type
}
)
except Exception as e:
print(e)
return make_response({'msg': 'Something went wrong.'}, 500)
I can see the uploaded file in my S3, but it shows 0 B in size and if I download it, it says that it cannot be viewed.
I have tried different access policies in S3, as well as many tutorials online, nothing seems to help. Changing the version of S3 to v3 when creating the client breaks the whole system and the file is not being uploaded at all with an access error.
What could be the reason for this upload failure? I it the config of AWS or something else?
Thank you!
Thanks to #jarmod I tried to avoid the image processing and it worked. I am now resizing the image, saving it to disk, opening the saved image, not the initial file, and sending it to S3. I then delete the image on disk as I don't need it.
Problem: I am trying to combine large amounts of small-sized text files into 1 large-sized file in S3 bucket. Using python:
The code I tested to try this locally is below. It works perfectly. (obtained from another post):
with open(outfilename, 'wb') as outfile:
for filename in glob.glob('UBXEvents*'):
if filename == outfilename: # don't want to copy the output into the output
continue
with open(filename, 'rb') as readfile:
shutil.copyfileobj(readfile, outfile)
Now, since my files are located in an S3 bucket, I have trouble referencing the S3 bucket. I wanted to run this code for all files (using wild card *) in an S3 but I am having a hard time connecting the two.
Below is the s3 object I created:
object = client.get_object(
Bucket= 'my_bucket_name',
Key='bucket_path/prefix_of_file_name*'
)
Question: How would I reference the S3 bucket/path in my combining code above?
Obtaining a list of files
You can obtain a list of files in the bucket like this:
import boto3
s3_client = boto3.client('s3')
response = s3_client.list_objects_v2(Bucket='my-bucket', Prefix = 'folder1/')
for object in response['Contents']:
# Do stuff here
print(object['Key'])
Reading & Writing to Amazon S3
Normally, you would need to download each file from Amazon S3 to the local disk (using download_file() and then read the contents). However, you might instead want to use smart-open ยท PyPI, which is a library that allows files to be opened on S3 using similar syntax to the normal Python open() command.
Here's a program that uses smart-open to read files from S3 and combine them into an output file in S3:
import boto3
from smart_open import open
BUCKET = 'my-bucket'
PREFIX = 'folder1/' # Optional
s3_client = boto3.client('s3')
# Open output file with smart-open
with open(f's3://{BUCKET}/out.txt', 'w') as out_file:
response = s3_client.list_objects_v2(Bucket=BUCKET, Prefix = PREFIX)
for object in response['Contents']:
print(f"Copying {object['Key']}")
# Open input file with smart-open
with open(f"s3://{BUCKET}/{object['Key']}", 'r') as in_file:
# Read content from input file
for line in in_file:
# Write content to output file
out_file.write(line)
I'm trying to find a way to extract .gz files in S3 on the fly, that is no need to download it to locally, extract and then push it back to S3.
With boto3 + lambda, how can i achieve my goal?
I didn't see any extract part in boto3 document.
You can use BytesIO to stream the file from S3, run it through gzip, then pipe it back up to S3 using upload_fileobj to write the BytesIO.
# python imports
import boto3
from io import BytesIO
import gzip
# setup constants
bucket = '<bucket_name>'
gzipped_key = '<key_name.gz>'
uncompressed_key = '<key_name>'
# initialize s3 client, this is dependent upon your aws config being done
s3 = boto3.client('s3', use_ssl=False) # optional
s3.upload_fileobj( # upload a new obj to s3
Fileobj=gzip.GzipFile( # read in the output of gzip -d
None, # just return output as BytesIO
'rb', # read binary
fileobj=BytesIO(s3.get_object(Bucket=bucket, Key=gzipped_key)['Body'].read())),
Bucket=bucket, # target bucket, writing to
Key=uncompressed_key) # target key, writing to
Ensure that your key is reading in correctly:
# read the body of the s3 key object into a string to ensure download
s = s3.get_object(Bucket=bucket, Key=gzip_key)['Body'].read()
print(len(s)) # check to ensure some data was returned
The above answers are for gzip files, for zip files, you may try
import boto3
import zipfile
from io import BytesIO
bucket = 'bucket1'
s3 = boto3.client('s3', use_ssl=False)
Key_unzip = 'result_files/'
prefix = "folder_name/"
zipped_keys = s3.list_objects_v2(Bucket=bucket, Prefix=prefix, Delimiter = "/")
file_list = []
for key in zipped_keys['Contents']:
file_list.append(key['Key'])
#This will give you list of files in the folder you mentioned as prefix
s3_resource = boto3.resource('s3')
#Now create zip object one by one, this below is for 1st file in file_list
zip_obj = s3_resource.Object(bucket_name=bucket, key=file_list[0])
print (zip_obj)
buffer = BytesIO(zip_obj.get()["Body"].read())
z = zipfile.ZipFile(buffer)
for filename in z.namelist():
file_info = z.getinfo(filename)
s3_resource.meta.client.upload_fileobj(
z.open(filename),
Bucket=bucket,
Key='result_files/' + f'{filename}')
This will work for your zip file and your result unzipped data will be in result_files folder. Make sure to increase memory and time on AWS Lambda to maximum since some files are pretty large and needs time to write.
Amazon S3 is a storage service. There is no in-built capability to manipulate the content of files.
However, you could use an AWS Lambda function to retrieve an object from S3, decompress it, then upload content back up again. However, please note that there is default limit of 500MB in temporary disk space for Lambda, so avoid decompressing too much data at the same time.
You could configure the S3 bucket to trigger the Lambda function when a new file is created in the bucket. The Lambda function would then:
Use boto3 to download the new file
Use the gzip Python library to extract files
Use boto3 to upload the resulting file(s)
Sample code:
import gzip
import io
import boto3
bucket = '<bucket_name>'
key = '<key_name>'
s3 = boto3.client('s3', use_ssl=False)
compressed_file = io.BytesIO(
s3.get_object(Bucket=bucket, Key=key)['Body'].read())
uncompressed_file = gzip.GzipFile(None, 'rb', fileobj=compressed_file)
s3.upload_fileobj(Fileobj=uncompressed_file, Bucket=bucket, Key=key[:-3])
I'm trying to download a large amount of small files from an S3 bucket - I'm doing this by using the following:
s3 = boto3.client('s3')
kwargs = {'Bucket': bucket}
with open('/Users/hr/Desktop/s3_backup/files.csv','w') as file:
while True:
# The S3 API response is a large blob of metadata.
# 'Contents' contains information about the listed objects.
resp = s3.list_objects_v2(**kwargs)
try:
contents = resp['Contents']
except KeyError:
return
for obj in contents:
key = obj['Key']
file.write(key)
file.write('\n')
# The S3 API is paginated, returning up to 1000 keys at a time.
# Pass the continuation token into the next response, until we
# reach the final page (when this field is missing).
try:
kwargs['ContinuationToken'] = resp['NextContinuationToken']
except KeyError:
break
However, after a certain amount of time I received this error message 'EndpointConnectionError: Could not connect to the endpoint URL'.
I know that there is still considerably more files on the s3 bucket. I have three questions:
Why is this error occurring when I haven't downloaded all files in the bucket?
Is there a way to start my code from the last file I downloaded from the S3 bucket (I don't want to have to re-download the file names I've already downloaded)
Is there a default ordering of the S3 bucket, is it alphabetical?
The GCP python docs have a script with the following function:
def upload_pyspark_file(project_id, bucket_name, filename, file):
"""Uploads the PySpark file in this directory to the configured
input bucket."""
print('Uploading pyspark file to GCS')
client = storage.Client(project=project_id)
bucket = client.get_bucket(bucket_name)
blob = bucket.blob(filename)
blob.upload_from_file(file)
I've created an argument parsing function in my script that takes in multiple arguments (file names) to upload to a GCS bucket. I'm trying to adapt the above function to parse those multiple args and upload those files, but am unsure how to proceed. My confusion is with the 'filename' and 'file' variables above. How can I adapt the function for my specific purpose?
I don't suppose you're still looking for something like this?
from google.cloud import storage
import os
files = os.listdir('data-files')
client = storage.Client.from_service_account_json('cred.json')
bucket = client.get_bucket('xxxxxx')
def upload_pyspark_file(filename, file):
# """Uploads the PySpark file in this directory to the configured
# input bucket."""
# print('Uploading pyspark file to GCS')
# client = storage.Client(project=project_id)
# bucket = client.get_bucket(bucket_name)
print('Uploading from ', file, 'to', filename)
blob = bucket.blob(filename)
blob.upload_from_file(file)
for f in files:
upload_pyspark_file(f, "data-files\\{0}".format(f))
The difference between file and filename is as you may have guessed, file is the source file and filename is the destination file.