I want to change this line
files = os.listdir('/Users/milenko/mario/Json_gzips')
in my code,to read .gz files from my bucket straight into list.
I tried
>>> import boto3
>>> s3 = boto3.resource('s3')
>>> s3
s3.ServiceResource()
>>> my_bucket = s3.Bucket('cw-dushpica-tests')
>>> for object_summary in my_bucket.objects.filter(Prefix='*.gz'):
... print(object_summary)
There is no output,it does print nothing.
for object_summary in my_bucket.objects.filter(Prefix='/'):
... print(object_summary)
The same,got nothing.
How should my prefix look like?
The prefix parameter of the filter method means that
Prefix (string) -- Limits the response to keys that begin with the specified prefix.
So, you can limit the path to the specific folder and then filter by yourself for the file extension.
import boto3
s3 = boto3.resource('s3')
bucket = s3.Bucket('your_bucket')
keys = []
for obj in bucket.objects.filter(Prefix='path/to/files/'):
if obj.key.endswith('gz'):
keys.append(obj.key)
print(keys)
If I have a file in multiple folders in S3, how do I combine them together using boto3 python
Say in a bucket I have
bucket_a
ts
ts_folder
a_date.csv
b_date.csv
c_date.csv
d_date.csv
ts_folder2
a_date.csv
b_date.csv
c_date.csv
d_date.csv
I need to combine these two files into one file, also ignoring header in second file
I am trying to figure out how to achieve using boto3 python or aws
Try something like this. I assume you have your AWS credentials set up properly on your system. My suggestion would be to first add the lines of the CSV to a new variable. For the second CSV you will skip the first line. After finding all the lines you join them as a string so they can be written to an S3 object.
import boto3
# Output will contain the CSV lines
output = []
with open("first.csv", "r") as fh:
output.extend(fh.readlines())
with open("second.csv", "r") as fh:
# Skip header
output.extend(fh.readlines()[1:])
# Combine the lines as string
body = "".join(output)
# Create the S3 client (assuming credentials are setup)
s3_client = boto3.client("s3")
# Write the object
s3_client.put_object(Bucket="my-bucket",
Key="combined.csv",
Body=body)
Update
This should help you with the S3 setup
import boto3
session = boto3.session.Session(profile_name='dev')
s3_client = session.client("s3")
bucket = "my-bucket"
files = []
for item in s3_client.list_objects_v2(Bucket=bucket, Prefix="ts/")['Contents']:
if item['Key'].endswith(".csv"):
files.append(item['Key'])
output = []
for file in files:
body = s3_client.get_object(Bucket=bucket,
Key=file)["Body"].read()
output.append(body)
# Combine the lines as string
outputbody = "".join(output)
# Write the object
s3_client.put_object(Bucket=bucket,
Key="combined.csv",
Body=outputbody)
I am trying to get all the files content sorted by the time the file was uploaded, I have this code:
keys = []
kwargs = {'Bucket': 'my_bucket'}
while True:
resp = s3.list_objects_v2(**kwargs)
for obj in resp['Contents']:
keys.append(obj['Key'])
try:
kwargs['ContinuationToken'] = resp['NextContinuationToken']
except KeyError:
break
but it returns me only the keys...
if I'm doing:
keys.append(obj)
it returns me a dict but I can't figure out how to get the files content in a normal run-time... what am I missing?
usually when you try to go through all the files on a S3 bucket you will encounter performance problems.
I would suggest to add S3 events and inform the "listener" about the new files through an queue (SQS).
Thus as soon as you copy a new file a new message is sent to the SQS (since we have created a S3 event), then the client will consume the SQS queue and do whatever is appropriate with the pending files. This is much faster than traversing the whole S3 bucket over and over.
Here is a link for more information: https://docs.aws.amazon.com/AmazonS3/latest/dev/NotificationHowTo.html
Another solution would be to analyse your S3 bucket with the help of Amazon Athena. https://aws.amazon.com/athena/
I hope that you will find this useful.
You'll need to use LastModified to sort by time. Something like:
import boto3
s3_client = boto3.client('s3', region_name='ap-southeast-2')
all_objects = []
kwargs = {'Bucket': 'my-bucket'}
while True:
# List objects in bucket
response = s3_client.list_objects_v2(**kwargs)
# Create an array of (Modified Date, Key) tuples
for object in response['Contents']:
all_objects.append((object['LastModified'], object['Key']))
try:
# Next page
kwargs['ContinuationToken'] = response['NextContinuationToken']
except KeyError:
break
# Sort by LastModified
sorted_keys = [object[1] for object in sorted(all_objects)]
print (sorted_keys)
You can use a boto3 paginator and iterate through each page and access the obj Key & LastModified
s3_client = boto3.client('s3', region_name=REGION)
pages = s3_client.get_paginator('list_objects_v2')
page_it = pages.paginate(Bucket=BUCKET)
files = {}
for page in page_it:
for content in page['Contents']
files[content['Key']] = content['LastModified']
sorted(files, key=lambda k: k[1])
How can I see what's inside a bucket in S3 with boto3? (i.e. do an "ls")?
Doing the following:
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('some/path/')
returns:
s3.Bucket(name='some/path/')
How do I see its contents?
One way to see the contents would be:
for my_bucket_object in my_bucket.objects.all():
print(my_bucket_object)
This is similar to an 'ls' but it does not take into account the prefix folder convention and will list the objects in the bucket. It's left up to the reader to filter out prefixes which are part of the Key name.
In Python 2:
from boto.s3.connection import S3Connection
conn = S3Connection() # assumes boto.cfg setup
bucket = conn.get_bucket('bucket_name')
for obj in bucket.get_all_keys():
print(obj.key)
In Python 3:
from boto3 import client
conn = client('s3') # again assumes boto.cfg setup, assume AWS S3
for key in conn.list_objects(Bucket='bucket_name')['Contents']:
print(key['Key'])
I'm assuming you have configured authentication separately.
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('bucket_name')
for file in my_bucket.objects.all():
print(file.key)
My s3 keys utility function is essentially an optimized version of #Hephaestus's answer:
import boto3
s3_paginator = boto3.client('s3').get_paginator('list_objects_v2')
def keys(bucket_name, prefix='/', delimiter='/', start_after=''):
prefix = prefix.lstrip(delimiter)
start_after = (start_after or prefix) if prefix.endswith(delimiter) else start_after
for page in s3_paginator.paginate(Bucket=bucket_name, Prefix=prefix, StartAfter=start_after):
for content in page.get('Contents', ()):
yield content['Key']
In my tests (boto3 1.9.84), it's significantly faster than the equivalent (but simpler) code:
import boto3
def keys(bucket_name, prefix='/', delimiter='/'):
prefix = prefix.lstrip(delimiter)
bucket = boto3.resource('s3').Bucket(bucket_name)
return (_.key for _ in bucket.objects.filter(Prefix=prefix))
As S3 guarantees UTF-8 binary sorted results, a start_after optimization has been added to the first function.
In order to handle large key listings (i.e. when the directory list is greater than 1000 items), I used the following code to accumulate key values (i.e. filenames) with multiple listings (thanks to Amelio above for the first lines). Code is for python3:
from boto3 import client
bucket_name = "my_bucket"
prefix = "my_key/sub_key/lots_o_files"
s3_conn = client('s3') # type: BaseClient ## again assumes boto.cfg setup, assume AWS S3
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter = "/")
if 'Contents' not in s3_result:
#print(s3_result)
return []
file_list = []
for key in s3_result['Contents']:
file_list.append(key['Key'])
print(f"List count = {len(file_list)}")
while s3_result['IsTruncated']:
continuation_key = s3_result['NextContinuationToken']
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter="/", ContinuationToken=continuation_key)
for key in s3_result['Contents']:
file_list.append(key['Key'])
print(f"List count = {len(file_list)}")
return file_list
If you want to pass the ACCESS and SECRET keys (which you should not do, because it is not secure):
from boto3.session import Session
ACCESS_KEY='your_access_key'
SECRET_KEY='your_secret_key'
session = Session(aws_access_key_id=ACCESS_KEY,
aws_secret_access_key=SECRET_KEY)
s3 = session.resource('s3')
your_bucket = s3.Bucket('your_bucket')
for s3_file in your_bucket.objects.all():
print(s3_file.key)
#To print all filenames in a bucket
import boto3
s3 = boto3.client('s3')
def get_s3_keys(bucket):
"""Get a list of keys in an S3 bucket."""
resp = s3.list_objects_v2(Bucket=bucket)
for obj in resp['Contents']:
files = obj['Key']
return files
filename = get_s3_keys('your_bucket_name')
print(filename)
#To print all filenames in a certain directory in a bucket
import boto3
s3 = boto3.client('s3')
def get_s3_keys(bucket, prefix):
"""Get a list of keys in an S3 bucket."""
resp = s3.list_objects_v2(Bucket=bucket, Prefix=prefix)
for obj in resp['Contents']:
files = obj['Key']
print(files)
return files
filename = get_s3_keys('your_bucket_name', 'folder_name/sub_folder_name/')
print(filename)
Update:
The most easiest way is to use awswrangler
import awswrangler as wr
wr.s3.list_objects('s3://bucket_name')
import boto3
s3 = boto3.resource('s3')
## Bucket to use
my_bucket = s3.Bucket('city-bucket')
## List objects within a given prefix
for obj in my_bucket.objects.filter(Delimiter='/', Prefix='city/'):
print obj.key
Output:
city/pune.csv
city/goa.csv
A more parsimonious way, rather than iterating through via a for loop you could also just print the original object containing all files inside your S3 bucket:
session = Session(aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key)
s3 = session.resource('s3')
bucket = s3.Bucket('bucket_name')
files_in_s3 = bucket.objects.all()
#you can print this iterable with print(list(files_in_s3))
So you're asking for the equivalent of aws s3 ls in boto3. This would be listing all the top level folders and files. This is the closest I could get; it only lists all the top level folders. Surprising how difficult such a simple operation is.
import boto3
def s3_ls():
s3 = boto3.resource('s3')
bucket = s3.Bucket('example-bucket')
result = bucket.meta.client.list_objects(Bucket=bucket.name,
Delimiter='/')
for o in result.get('CommonPrefixes'):
print(o.get('Prefix'))
ObjectSummary:
There are two identifiers that are attached to the ObjectSummary:
bucket_name
key
boto3 S3: ObjectSummary
More on Object Keys from AWS S3 Documentation:
Object Keys:
When you create an object, you specify the key name, which uniquely identifies the object in the bucket. For example, in the Amazon S3 console (see AWS Management Console), when you highlight a bucket, a list of objects in your bucket appears. These names are the object keys. The name for a key is a sequence of Unicode characters whose UTF-8 encoding is at most 1024 bytes long.
The Amazon S3 data model is a flat structure: you create a bucket, and the bucket stores objects. There is no hierarchy of subbuckets or subfolders; however, you can infer logical hierarchy using key name prefixes and delimiters as the Amazon S3 console does. The Amazon S3 console supports a concept of folders. Suppose that your bucket (admin-created) has four objects with the following object keys:
Development/Projects1.xls
Finance/statement1.pdf
Private/taxdocument.pdf
s3-dg.pdf
Reference:
AWS S3: Object Keys
Here is some example code that demonstrates how to get the bucket name and the object key.
Example:
import boto3
from pprint import pprint
def main():
def enumerate_s3():
s3 = boto3.resource('s3')
for bucket in s3.buckets.all():
print("Name: {}".format(bucket.name))
print("Creation Date: {}".format(bucket.creation_date))
for object in bucket.objects.all():
print("Object: {}".format(object))
print("Object bucket_name: {}".format(object.bucket_name))
print("Object key: {}".format(object.key))
enumerate_s3()
if __name__ == '__main__':
main()
Here is a simple function that returns you the filenames of all files or files with certain types such as 'json', 'jpg'.
def get_file_list_s3(bucket, prefix="", file_extension=None):
"""Return the list of all file paths (prefix + file name) with certain type or all
Parameters
----------
bucket: str
The name of the bucket. For example, if your bucket is "s3://my_bucket" then it should be "my_bucket"
prefix: str
The full path to the the 'folder' of the files (objects). For example, if your files are in
s3://my_bucket/recipes/deserts then it should be "recipes/deserts". Default : ""
file_extension: str
The type of the files. If you want all, just leave it None. If you only want "json" files then it
should be "json". Default: None
Return
------
file_names: list
The list of file names including the prefix
"""
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket(bucket)
file_objs = my_bucket.objects.filter(Prefix=prefix).all()
file_names = [file_obj.key for file_obj in file_objs if file_extension is not None and file_obj.key.split(".")[-1] == file_extension]
return file_names
One way that I used to do this:
import boto3
s3 = boto3.resource('s3')
bucket=s3.Bucket("bucket_name")
contents = [_.key for _ in bucket.objects.all() if "subfolders/ifany/" in _.key]
Here is the solution
import boto3
s3=boto3.resource('s3')
BUCKET_NAME = 'Your S3 Bucket Name'
allFiles = s3.Bucket(BUCKET_NAME).objects.all()
for file in allFiles:
print(file.key)
I just did it like this, including the authentication method:
s3_client = boto3.client(
's3',
aws_access_key_id='access_key',
aws_secret_access_key='access_key_secret',
config=boto3.session.Config(signature_version='s3v4'),
region_name='region'
)
response = s3_client.list_objects(Bucket='bucket_name', Prefix=key)
if ('Contents' in response):
# Object / key exists!
return True
else:
# Object / key DOES NOT exist!
return False
With little modification to #Hephaeastus 's code in one of the above comments, wrote the below method to list down folders and objects (files) in a given path. Works similar to s3 ls command.
from boto3 import session
def s3_ls(profile=None, bucket_name=None, folder_path=None):
folders=[]
files=[]
result=dict()
bucket_name = bucket_name
prefix= folder_path
session = boto3.Session(profile_name=profile)
s3_conn = session.client('s3')
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter = "/", Prefix=prefix)
if 'Contents' not in s3_result and 'CommonPrefixes' not in s3_result:
return []
if s3_result.get('CommonPrefixes'):
for folder in s3_result['CommonPrefixes']:
folders.append(folder.get('Prefix'))
if s3_result.get('Contents'):
for key in s3_result['Contents']:
files.append(key['Key'])
while s3_result['IsTruncated']:
continuation_key = s3_result['NextContinuationToken']
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter="/", ContinuationToken=continuation_key, Prefix=prefix)
if s3_result.get('CommonPrefixes'):
for folder in s3_result['CommonPrefixes']:
folders.append(folder.get('Prefix'))
if s3_result.get('Contents'):
for key in s3_result['Contents']:
files.append(key['Key'])
if folders:
result['folders']=sorted(folders)
if files:
result['files']=sorted(files)
return result
This lists down all objects / folders in a given path. Folder_path can be left as None by default and method will list the immediate contents of the root of the bucket.
It can also be done as follows:
csv_files = s3.list_objects_v2(s3_bucket_path)
for obj in csv_files['Contents']:
key = obj['Key']
Using cloudpathlib
cloudpathlib provides a convenience wrapper so that you can use the simple pathlib API to interact with AWS S3 (and Azure blob storage, GCS, etc.). You can install with pip install "cloudpathlib[s3]".
Like with pathlib you can use glob or iterdir to list the contents of a directory.
Here's an example with a public AWS S3 bucket that you can copy and past to run.
from cloudpathlib import CloudPath
s3_path = CloudPath("s3://ladi/Images/FEMA_CAP/2020/70349")
# list items with glob
list(
s3_path.glob("*")
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#> S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#> S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]
# list items with iterdir
list(
s3_path.iterdir()
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#> S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#> S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]
Created at 2021-05-21 20:38:47 PDT by reprexlite v0.4.2
A good option may also be to run aws cli command from lambda functions
import subprocess
import logging
logger = logging.getLogger()
logger.setLevel(logging.INFO)
def run_command(command):
command_list = command.split(' ')
try:
logger.info("Running shell command: \"{}\"".format(command))
result = subprocess.run(command_list, stdout=subprocess.PIPE);
logger.info("Command output:\n---\n{}\n---".format(result.stdout.decode('UTF-8')))
except Exception as e:
logger.error("Exception: {}".format(e))
return False
return True
def lambda_handler(event, context):
run_command('/opt/aws s3 ls s3://bucket-name')
I was stuck on this for an entire night because I just wanted to get the number of files under a subfolder but it was also returning one extra file in the content that was the subfolder itself,
After researching about it I found that this is how s3 works but I had
a scenario where I unloaded the data from redshift in the following directory
s3://bucket_name/subfolder/<10 number of files>
and when I used
paginator.paginate(Bucket=price_signal_bucket_name,Prefix=new_files_folder_path+"/")
it would only return the 10 files, but when I created the folder on the s3 bucket itself then it would also return the subfolder
Conclusion
If the whole folder is uploaded to s3 then listing the only returns the files under prefix
But if the fodler was created on the s3 bucket itself then listing it using boto3 client will also return the subfolder and the files
First, create an s3 client object:
s3_client = boto3.client('s3')
Next, create a variable to hold the bucket name and folder. Pay attention to the slash "/" ending the folder name:
bucket_name = 'my-bucket'
folder = 'some-folder/'
Next, call s3_client.list_objects_v2 to get the folder's content object's metadata:
response = s3_client.list_objects_v2(
Bucket=bucket_name,
Prefix=folder
)
Finally, with the object's metadata, you can obtain the S3 object by calling the s3_client.get_object function:
for object_metadata in response['Contents']:
object_key = object_metadata['Key']
response = s3_client.get_object(
Bucket=bucket_name,
Key=object_key
)
object_body = response['Body'].read()
print(object_body)
As you can see, the object content in the string format is available by calling response['Body'].read()