Python-lambda is a toolset for developing and deploying serverless Python code in AWS Lambda.
AWS Lambda is a service that allows you to write Python, Java, or Node.js code that gets executed in response to events like http requests or files uploaded to S3.
Working with Lambda is relatively easy, but the process of bundling and deploying your code is not as simple as it could be.
The Python-Lambda library takes away the guess work of developing your Python-Lambda services by providing you a toolset to streamline the annoying parts.
- Python 2.7 (At the time of writing this, AWS Lambda only supports Python 2.7).
- Pip (~8.1.1)
- Virtualenv (~15.0.0)
- Virtualenvwrapper (~4.7.1)
Begin by creating a new virtualenv and project folder.
$ mkvirtualenv pylambda
(pylambda) $ mkdir pylambdaNext, download Python-Lambda using pip via pypi.
(pylambda) $ pip install python-lambdaFrom your pylambda directory, run the following to bootstrap your project.
(pylambda) $ lambda initThis will create the following files: event.json, __init__.py, service.py, and config.yaml.
Let's begin by opening config.yaml in the text editor of your choice. For the purpose of this tutorial, the only required information is aws_access_key_id and aws_secret_access_key. You can find these by logging into the AWS management console.
Next let's open service.py, in here you'll find the following function:
def handler(event, context):
# Your code goes here!
e = event.get('e')
pi = event.get('pi')
return e + piThis is the handler function; this is the function AWS Lambda will invoke in response to an event. You will notice that in the sample code e and pi are values in a dict. AWS Lambda uses the event parameter to pass in event data to the handler.
So if, for example, your function is responding to an http request, event will be the POST JSON data and if your function returns something, the contents will be in your http response payload.
Next let's open the event.json file:
{
"pi": 3.14,
"e": 2.718
}Here you'll find the values of e and pi that are being referenced in the sample code.
If you now try and run:
(pylambda) $ lambda invoke -vYou will get:
# 5.858
# execution time: 0.00000310s
# function execution
6A09
timeout: 15sAs you probably put together, the lambda invoke command grabs the values stored in the event.json file and passes them to your function.
The event.json file should help you develop your Lambda service locally. You can specify an alternate event.json file by passing the --event-file=<filename>.json argument to lambda invoke.
When you're ready to deploy your code to Lambda simply run:
(pylambda) $ lambda deployThe deploy script will evaluate your virtualenv and identify your project dependencies. It will package these up along with your handler function to a zip file that it then uploads to AWS Lambda.
You can now log into the AWS Lambda management console to verify the code deployed successfully.
If you're looking to develop a simple microservice you can easily wire your function up to an http endpoint.
Begin by navigating to your AWS Lambda management console and clicking on your function. Click the API Endpoints tab and click "Add API endpoint".
Under API endpoint type select "API Gateway".
Next change Method to POST and Security to "Open" and click submit (NOTE: you should secure this for use in production, open security is used for demo purposes).
At last you need to change the return value of the function to comply with the standard defined for the API Gateway endpoint, the function should now look like this:
def handler(event, context):
# Your code goes here!
e = event.get('e')
pi = event.get('pi')
return {
"statusCode": 200,
"headers": { "Content-Type": "application/json"},
"body": e + pi
}Now try and run:
$ curl --header "Content-Type:application/json" \
--request POST \
--data '{"pi": 3.14, "e": 2.718}' \
https://<API endpoint URL>
# 5.8580000000000005