I am trying to upload a file in S3 using boto3.I tried below code.
import boto3
s3 = boto3.resource('s3')
buck_name = s3.create_bucket(Bucket='trubuckboto')
s3.Object('trubuckboto','tlearn.txt').upload_file(
Filename='G:\tlearn.txt')
My bucket creation is successfull but i am not able to upload file from location G:\tlearn.txt inside that bucket.Below is the error i am getting
return os.stat(filename).st_size
OSError: [WinError 123] The filename, directory name, or volume label syntax is incorrect: 'G:\tlearn.txt'
Can someone suggest what i am missing here ?
In Python strings, the backslash "\" is a special character, also called the "escape" character. If you want a literal backslash then you need to escape the escape character, for example G:\\tlearn.txt:
import boto3
s3 = boto3.resource('s3')
# buck_name = s3.create_bucket(Bucket='trubuckboto')
s3.Object('trubuckboto', 'tlearn.txt').upload_file(
Filename='G:\\tlearn.txt')
Related
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 · 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 new to this, and trying to download a snappy.parquet file from Amazon s3 I can later convert to CSV file.
I tried working with the following example I've found online, and I get an empty folder. can anyone please help me?
import boto
import sys, os
from boto.s3.key import Key
from boto.exception import S3ResponseError
DOWNLOAD_LOCATION_PATH =""
BUCKET_NAME = ""
AWS_ACCESS_KEY_ID= ""
AWS_ACCESS_SECRET_KEY = ""
conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_ACCESS_SECRET_KEY)
bucket = conn.get_bucket(BUCKET_NAME)
#goto through the list of files
bucket_list = bucket.list()
for l in bucket_list:
key_string = str(l.key)
s3_path = DOWNLOAD_LOCATION_PATH + key_string
try:
print ("Current File is ", s3_path)
l.get_contents_to_filename(s3_path)
except (OSError, S3ResponseError) as e:
pass
# check if the file has been downloaded locally
if not os.path.exists(s3_path):
try:
os.makedirs(s3_path)
except OSError as exc:
# let guard againts race conditions
import errno
if exc.errno != errno.EEXIST:
raise
The script you are using appears to recursively download the contents of the specified S3 bucket (BUCKET_NAME) to the specified local directory (DOWNLOAD_LOCATION_PATH). FWIW, I notice this script looks like it comes from here.
The "Current File is ..." output line should show you the progress of these files being written. One problem you might be having is due to this line:
s3_path = DOWNLOAD_LOCATION_PATH + key_string
If you had specified DOWNLOAD_LOCATION_PATH at the top as a directory without a trailing '/' character, e.g. like this:
DOWNLOAD_LOCATION_PATH = '/tmp/my_dir'
then the files being downloaded would be written not underneath the /tmp/my_dir directory, but directly in /tmp/ with a my_dir prefix on each filename! You can fix this by changing this line to:
s3_path = os.path.join(DOWNLOAD_LOCATION_PATH, key_string)
Other than that, the script appears to work alright. You may want to add this line at the very top:
from __future__ import print_function
if you are still using Python 2.x, otherwise the print output will look a bit odd (print will think you are printing a 2-Tuple).
Your question also makes it sound like you really only want/need to download a single file from the bucket -- if so, this isn't really a great script to be using, since it's downloading everything.
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.
I’m doing simple operation to of downloading the gzip files from S3 bucket to the local directory. I’m extracting those into another local directory and then uploading them back to S3 bucket again into archive folder path. While doing this operation I want to make sure I am processing same set of files that I initially download from S3 bucket which is (f_name) in below code. Now, below code is not uploading those back to S3 , that’s where I’m stuck. But able to download from S3 and extract it into local directory. Can you please help me understand what is wrong with the _uploadFile function?
from boto.s3.connection import S3Connection
from boto.s3.key import *
import os
import os.path
aws_bucket= "event-logs-dev” ## S3 Bucket name
local_download_directory= "/Users/TargetData/Download/test_queue1/“ ## local directory to download the gzip files from S3.
Target_directory_to_extract = "/Users/TargetData/unzip” ##local directory to gunzip the downloaded files.
Target_s3_path_to_upload= "event-logs-dev/data/clean/xact/logs/archive/“ ## S3 bucket path to upload the files.
def decompressAllFilesFromNetfiler(self,aws_bucket,local_download_directory,Target_d irectory_to_extract,Target_s3_path_to_upload):
zipFiles = [f for f in os.listdir(local_download_directory) if re.match(r'.*\.tar\.gz', f)]
for f_name in zipFiles:
if os.path.exists(Target_directory_to_extract+"/"+f_name[:-len('.tar.gz')]) and os.access(Target_directory_to_extract+"/"+f_name[:-len('.tar.gz')], os.R_OK):
print ('File {} already exists!'.format(f_name))
else:
f_name_with_path = os.path.join(local_download_directory, f_name)
os.system('mkdir -p {} && tar vxzf {} -C {}'.format(Target_directory_to_extract, f_name_with_path, Target_directory_to_extract))
print ('Extracted file {}'.format(f_name))
self._uploadFile(aws_bucket,f_name,Target_s3_path_to_upload,Target_directory_to_extract)
def _uploadFile(self, aws_bucket, f_name,Target_s3_path_to_upload,Target_directory_to_extract):
full_key_name = os.path.expanduser(os.path.join(Target_s3_path_to_upload, f_name))
path = os.path.expanduser(os.path.join(Target_directory_to_extract, f_name))
try:
print "Uploaded extracted file to: %s" % (full_key_name)
key = aws_bucket.new_key(full_key_name)
key.set_contents_from_filename(path)
except:
if full_key_name is None:
print "Error uploading”
Currently, the output prints that Uploaded extracted file to: event-logs-dev/data/clean/xact/logs/archive/1442235602129200000.tar.gz, but nothing is uploaded to S3 bucket. Your help is greatly appreciated!! Thank you in advance!
It appears that you have cut and pasted parts of your code - and maybe formatting was lost as your code above will not work as pasted. I've taken the liberty to make it PEP8 (mostly) however there is still some missing code to create the S3 objects. Since your import the modules, I presume that you have that section of code and just didn't paste it.
here is a cleaned up version of your code formatted correctly. I also added a Exception code to your try: block to print out the error you get. You should update the Exception to be more specific to the Exceptions thrown for make_key or set_contents_... but the general Exception will get you started. If nothing more this is more readable, but you should include your S3 connection code too - and remove anything that is specific to your domain (e.g. keys, trade secrets, etc).
#!/usr/bin/env python
"""
do some download
some extract
and some upload
"""
from boto.s3.connection import S3Connection
from boto.s3.key import *
import os
import os.path
aws_bucket = 'event-logs-dev'
local_download_directory = '/Users/TargetData/Download/test_queue1/'
Target_directory_to_extract = '/Users/TargetData/unzip'
Target_s3_path_to_upload = 'event-logs-dev/data/clean/xact/logs/archive/'
'''
MUST BE SOME MAGIC HERE TO GET AN S3 CONNECTION ???
aws_bucket IS NOT A BUCKET OBJECT ...
'''
def decompressAllFilesFromNetfiler(self,
aws_bucket,
local_download_directory,
Target_directory_to_extract,
Target_s3_path_to_upload):
'''
decompress stuff
'''
zipFiles = [f for f in os.listdir(
local_download_directory) if re.match(r'.*\.tar\.gz', f)]
for f_name in zipFiles:
if os.path.exists(
"{}/{}".format(Target_directory_to_extract,
f_name[:len('.tar.gz')])) and os.access(
"{}/{}".format(Target_directory_to_extract,
f_name[:len('.tar.gz')])) and os.R_OK:
print ('File {} already exists!'.format(f_name))
else:
f_name_with_path = os.path.join(local_download_directory, f_name)
os.system('mkdir -p {} && tar vxzf {} -C {}'.format(
Target_directory_to_extract,
f_name_with_path,
Target_directory_to_extract))
print ('Extracted file {}'.format(f_name))
self._uploadFile(aws_bucket,
f_name,
Target_s3_path_to_upload,
Target_directory_to_extract)
def _uploadFile(self,
aws_bucket,
f_name,
Target_s3_path_to_upload,
Target_directory_to_extract):
full_key_name = os.path.expanduser(os.path.join(Target_s3_path_to_upload,
f_name))
path = os.path.expanduser(os.path.join(Target_directory_to_extract, f_name))
try:
S3CONN = S3Connection()
BUCKET = S3CONN.get_bucket(aws_bucket)
key = BUCKET.new_key(full_key_name)
key.set_contents_from_filename(path)
print "Uploaded extracted file to: {}".format(full_key_name)
except Exception as UploadERR:
if full_key_name is None:
print 'Error uploading'
else:
print "Error : {}".format(UploadERR)