Download multiple files from S3 bucket using boto3 - python

I have a csv file containing numerous uuids
I'd like to write a python script using boto3 which:
Connects to an AWS S3 bucket
Uses each uuid contained in the CSV to copy the file contained
Files are all contained in a filepath like this: BUCKET/ORG/FOLDER1/UUID/DATA/FILE.PNG
However, the file contained in DATA/ can be different file types.
Put the copied file in a new S3 bucket
So far, I have successfully connected to the s3 bucket and checked its contents in python using boto3, but need help implementing the rest
import boto3
#Create Session
session = boto3.Session(
aws_access_key_id='ACCESS_KEY_ID',
aws_secret_access_key='SECRET_ACCESS_KEY',
)
#Initiate S3 Resource
s3 = session.resource('s3')
your_bucket = s3.Bucket('BUCKET-NAME')
for s3_file in your_bucket.objects.all():
print(s3_file.key) # prints the contents of bucket


To read the CSV file you can use csv library (see: https://docs.python.org/fr/3.6/library/csv.html)
Example:
import csv
with open('file.csv', 'r') as file:
reader = csv.reader(file)
for row in reader:
print(row)
To push files to the new bucket, you can use the copy method (see: https://boto3.amazonaws.com/v1/documentation/api/latest/reference/services/s3.html#S3.Client.copy)
Example:
import boto3
s3 = boto3.resource('s3')
source = {
'Bucket': 'BUCKET-NAME',
'Key': 'mykey'
}
bucket = s3.Bucket('SECOND_BUCKET-NAME')
bucket.copy(source, 'SECOND_BUCKET-NAME')

Related

File conversion XML to JSON in S3 through AWS Glue

I have my bucket structure like below and i have xml files landing in this s3 bucket folder.
S3:/Fin-app-ops/data-ops/raw-d
Need to convert those xml files to JSON files and put back to s3 in same bucket but different folder:
S3:/Fin-app-ops/data-ops/con-d
I tried by this way but did not work:
import os
import json
import boto3
import xmltodict
s3 = boto3.resource('s3')
s3_bucket = s3.bucket('Fin-app-ops')
file_in_path = 'data-ops/raw-d/'
file_dest_path = 'data-ops/con-d/'
Datafiles = [f.key for f in s3_bucket.objects.filter(prefix = file_in_path)]
for datafile in datafiles:
if "xml" in obj.key:
datafile = obj.get()['Body']
data_dict = xmltodict.parse(datafile .read())
datafile.close()
json_data = json.dumps(data_dict)
s3.Object(bucket_name, file_dest_path `enter code here`+'.json').put(Body=json.dumps(data_dict))
is there any other way I can achieve this, help please i'm new to Python and Glue

How to write pyarrow table as csv to s3 directly? [duplicate]

In boto 2, you can write to an S3 object using these methods:
Key.set_contents_from_string()
Key.set_contents_from_file()
Key.set_contents_from_filename()
Key.set_contents_from_stream()
Is there a boto 3 equivalent? What is the boto3 method for saving data to an object stored on S3?

In boto 3, the 'Key.set_contents_from_' methods were replaced by
Object.put()
Client.put_object()
For example:
import boto3
some_binary_data = b'Here we have some data'
more_binary_data = b'Here we have some more data'
# Method 1: Object.put()
s3 = boto3.resource('s3')
object = s3.Object('my_bucket_name', 'my/key/including/filename.txt')
object.put(Body=some_binary_data)
# Method 2: Client.put_object()
client = boto3.client('s3')
client.put_object(Body=more_binary_data, Bucket='my_bucket_name', Key='my/key/including/anotherfilename.txt')
Alternatively, the binary data can come from reading a file, as described in the official docs comparing boto 2 and boto 3:
Storing Data
Storing data from a file, stream, or string is easy:
# Boto 2.x
from boto.s3.key import Key
key = Key('hello.txt')
key.set_contents_from_file('/tmp/hello.txt')
# Boto 3
s3.Object('mybucket', 'hello.txt').put(Body=open('/tmp/hello.txt', 'rb'))

boto3 also has a method for uploading a file directly:
s3 = boto3.resource('s3')
s3.Bucket('bucketname').upload_file('/local/file/here.txt','folder/sub/path/to/s3key')
http://boto3.readthedocs.io/en/latest/reference/services/s3.html#S3.Bucket.upload_file

You no longer have to convert the contents to binary before writing to the file in S3. The following example creates a new text file (called newfile.txt) in an S3 bucket with string contents:
import boto3
s3 = boto3.resource(
's3',
region_name='us-east-1',
aws_access_key_id=KEY_ID,
aws_secret_access_key=ACCESS_KEY
)
content="String content to write to a new S3 file"
s3.Object('my-bucket-name', 'newfile.txt').put(Body=content)

Here's a nice trick to read JSON from s3:
import json, boto3
s3 = boto3.resource("s3").Bucket("bucket")
json.load_s3 = lambda f: json.load(s3.Object(key=f).get()["Body"])
json.dump_s3 = lambda obj, f: s3.Object(key=f).put(Body=json.dumps(obj))
Now you can use json.load_s3 and json.dump_s3 with the same API as load and dump
data = {"test":0}
json.dump_s3(data, "key") # saves json to s3://bucket/key
data = json.load_s3("key") # read json from s3://bucket/key

A cleaner and concise version which I use to upload files on the fly to a given S3 bucket and sub-folder-
import boto3
BUCKET_NAME = 'sample_bucket_name'
PREFIX = 'sub-folder/'
s3 = boto3.resource('s3')
# Creating an empty file called "_DONE" and putting it in the S3 bucket
s3.Object(BUCKET_NAME, PREFIX + '_DONE').put(Body="")
Note: You should ALWAYS put your AWS credentials (aws_access_key_id and aws_secret_access_key) in a separate file, for example- ~/.aws/credentials

After some research, I found this. It can be achieved using a simple csv writer. It is to write a dictionary to CSV directly to S3 bucket.
eg: data_dict = [{"Key1": "value1", "Key2": "value2"}, {"Key1": "value4", "Key2": "value3"}]
assuming that the keys in all the dictionary are uniform.
import csv
import boto3
# Sample input dictionary
data_dict = [{"Key1": "value1", "Key2": "value2"}, {"Key1": "value4", "Key2": "value3"}]
data_dict_keys = data_dict[0].keys()
# creating a file buffer
file_buff = StringIO()
# writing csv data to file buffer
writer = csv.DictWriter(file_buff, fieldnames=data_dict_keys)
writer.writeheader()
for data in data_dict:
writer.writerow(data)
# creating s3 client connection
client = boto3.client('s3')
# placing file to S3, file_buff.getvalue() is the CSV body for the file
client.put_object(Body=file_buff.getvalue(), Bucket='my_bucket_name', Key='my/key/including/anotherfilename.txt')

it is worth mentioning smart-open that uses boto3 as a back-end.
smart-open is a drop-in replacement for python's open that can open files from s3, as well as ftp, http and many other protocols.
for example
from smart_open import open
import json
with open("s3://your_bucket/your_key.json", 'r') as f:
data = json.load(f)
The aws credentials are loaded via boto3 credentials, usually a file in the ~/.aws/ dir or an environment variable.

You may use the below code to write, for example an image to S3 in 2019. To be able to connect to S3 you will have to install AWS CLI using command pip install awscli, then enter few credentials using command aws configure:
import urllib3
import uuid
from pathlib import Path
from io import BytesIO
from errors import custom_exceptions as cex
BUCKET_NAME = "xxx.yyy.zzz"
POSTERS_BASE_PATH = "assets/wallcontent"
CLOUDFRONT_BASE_URL = "https://xxx.cloudfront.net/"
class S3(object):
def __init__(self):
self.client = boto3.client('s3')
self.bucket_name = BUCKET_NAME
self.posters_base_path = POSTERS_BASE_PATH
def __download_image(self, url):
manager = urllib3.PoolManager()
try:
res = manager.request('GET', url)
except Exception:
print("Could not download the image from URL: ", url)
raise cex.ImageDownloadFailed
return BytesIO(res.data) # any file-like object that implements read()
def upload_image(self, url):
try:
image_file = self.__download_image(url)
except cex.ImageDownloadFailed:
raise cex.ImageUploadFailed
extension = Path(url).suffix
id = uuid.uuid1().hex + extension
final_path = self.posters_base_path + "/" + id
try:
self.client.upload_fileobj(image_file,
self.bucket_name,
final_path
)
except Exception:
print("Image Upload Error for URL: ", url)
raise cex.ImageUploadFailed
return CLOUDFRONT_BASE_URL + id

Extracting gzip file in an S3 bucket to another S3 Bucket

im trying to copy the gzip file from one S3 bucket and extract its content to another S3 bucket using gzip library.
im getting an error
Seek from end not supported
import boto3, json
from io import BytesIO
import gzip
def lambda_handler():
try:
s3 = boto3.resource('s3')
copy_source = {
'Bucket': 'srcbucket',
'Key': 'samp.gz'
}
bucket = s3.Bucket('destbucket')
bucketSrc = s3.Bucket('srcbucket')
s3Client = boto3.client('s3', use_ssl=False)
s3Client.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(s3Client.get_object(Bucket='srcbucket', Key='samp.gz')['Body'].read())),
Bucket='destbucket', # target bucket, writing to
Key="") # target key, writing to
except Exception as e:
print(e)

You can't unzip the ZIP file and upload its constituent files the way you're trying to.
You could unzip the entire ZIP file to Lambda local disk in /tmp (note this has a limit of 512MB diskspace) then upload file by file. Or, if it will not fit on disk or you prefer not to persist to desk, then you can stream the contents of the ZIP file into memory, file by file, and then upload each stream to S3). In both solutions, you will need to supply an appropriate key for each and every upload.

how do I write a list of data to S3 in ORC format

I need to write a file in ORC format directly to an S3 bucket. the file will be a result of a query to a db.
I know how to write a CSV file directly to S3 but couldn't find a way to write directly in ORC.. any recommendations?

save ORC content to file
using default values as per the linked documentation as there is no code sample to work with
df = spark.read.load("examples/src/main/resources/users.parquet")
df.select("name", "favorite_color").write.save("namesAndFavColors.parquet")
upload file
import boto3
# Create an S3 client
s3 = boto3.client('s3')
filename = 'file.txt'
bucket_name = 'my-bucket'
# Uploads the given file using a managed uploader, which will split up large
# files automatically and upload parts in parallel.
s3.upload_file(filename, bucket_name, filename)

Writing json to file in s3 bucket

This code writes json to a file in s3,
what i wanted to achieve is instead of opening data.json file and writing to s3 (sample.json) file,
how do i pass the json directly and write to a file in s3 ?
import boto3
s3 = boto3.resource('s3', aws_access_key_id='aws_key', aws_secret_access_key='aws_sec_key')
s3.Object('mybucket', 'sample.json').put(Body=open('data.json', 'rb'))

I'm not sure, if I get the question right. You just want to write JSON data to a file using Boto3? The following code writes a python dictionary to a JSON file.
import json
import boto3
s3 = boto3.resource('s3')
s3object = s3.Object('your-bucket-name', 'your_file.json')
s3object.put(
Body=(bytes(json.dumps(json_data).encode('UTF-8')))
)

I don't know if anyone is still attempting to use this thread, but I was attempting to upload a JSON to s3 trying to use the method above, which didnt quite work for me. Boto and s3 might have changed since 2018, but this achieved the results for me:
import json
import boto3
s3 = boto3.client('s3')
json_object = 'your_json_object here'
s3.put_object(
Body=json.dumps(json_object),
Bucket='your_bucket_name',
Key='your_key_here'
)

Amazon S3 is an object store (File store in reality). The primary operations are PUT and GET. You can not add data into an existing object in S3. You can only replace the entire object itself.
For a list of available operations you can perform on s3 see this link
http://docs.aws.amazon.com/AmazonS3/latest/API/RESTObjectOps.html

An alternative to Joseph McCombs answer can be achieved using s3fs.
from s3fs import S3FileSystem
json_object = {'test': 3.14}
path_to_s3_object = 's3://your-s3-bucket/your_json_filename.json'
s3 = S3FileSystem()
with s3.open(path_to_s3_object, 'w') as file:
json.dump(json_object, file)

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