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Joshua Tanke
Joshua Tanke

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Deploy Python APIs to AWS Lambda In Minutes

This short guide will show you how to deploy a Python API to AWS Lambda using LaunchFlow, an open source deployment tool for AWS & GCP.

Here’s what the final code will look like:

import launchflow as lf
import numpy as np

def handler(event, context):
    result = np.random.randint(0, 100)
    return {
        "statusCode": 200,
        "body": f"Result: {result}",
    }

api = lf.aws.LambdaService(
    name="my-lambda-api",
    handler=handler,
    runtime=lf.aws.lambda_service.PythonRuntime(
        requirements_txt_path="requirements.txt"
    ),
)
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You'll also be able to add other AWS services (i.e. RDS) to your API with little to no effort:

import launchflow as lf

rds = lf.aws.RDS(
    name="my-rds-cluster",
    publicly_accessible=False,
    engine_version=lf.aws.rds.RDSEngineVersion.POSTGRES16,
)

def handler(event, context):
    result = rds.query("SELECT 1")  # TODO: Add your SQL query here
    return {
        "statusCode": 200,
        "body": f"Query result: {result}",
    }

api = lf.aws.LambdaService(
    name="my-lambda-api",
    handler=handler,
    runtime=lf.aws.lambda_service.PythonRuntime(
        requirements_txt_path="requirements.txt"
    ),
)
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Let's get started!

Step 1: Install the LaunchFlow SDK + CLI

You can install LaunchFlow using pip:

pip install launchflow
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Step 2: Initialize your project directory

Create / navigate to a new project directory, the run:

lf init
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This will prompt you for things like your project name, where you want to store your deployment state, etc. I would recommend choosing the options shown below to get started quickly:

LaunchFlow CLI init command

Step 3: Create your API

NOTE: You will need local AWS credentials for this step

If you choose the same prompts as above, you should have an infra.py file that looks like this:

import launchflow as lf

api = lf.aws.LambdaService("my-lambda-api", handler="TODO") 
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You can now run lf create to spin up an empty Lambda API in a dedicated environment in your AWS account.

lf create
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This will first prompt you to create a new Environment in your AWS account (this is essentially just a VPC + role + s3 bucket):

LaunchFlow CLI create command

Once your environment is created, you'll see the AWS Lambda infrastructure that will be deployed to the environment:
LaunchFlow CLI create plan

Approve the plan and watch your API get created in a couple of minutes:

LaunchFlow CLI create approve

Step 4: Deploy your custom code

Now that we have everything set up on AWS, let's deploy a custom handler function that depends on the Numpy Python package.

Create a main.py file for the handler code:

import numpy

def handler(event, context):
    result = np.random.randint(0, 100)
    return {
        "statusCode": 200,
        "body": f"Random Number: {result}",
    }
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Add a requirements.txt with numpy added:

numpy
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Then update the LambdaService handler + runtime options in the infra.py file:

import launchflow as lf

api = lf.aws.LambdaService(
    name="my-lambda-api",
    handler="main.handler",
    runtime=lf.aws.lambda_service.PythonRuntime(
        requirements_txt_path="requirements.txt"
    ),
)
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Now simply run lf deploy to launch it to your Lambda function!

lf deploy
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If all goes well, you should see the url to access your API:
LaunchFlow CLI deploy command

That's it! You now have a production-grade API deployed to AWS Lambda.

Bonus: Add AWS services to your Lambda

LaunchFlow includes a high-level Python SDK for interacting with RDS, S3, SQS, and other AWS services at runtime. For example, here's how you would connect to a private Postgres databases hosted on RDS inside your lambda function:

import launchflow as lf

rds = lf.aws.RDS(
    name="my-rds-cluster",
    publicly_accessible=False,
    engine_version=lf.aws.rds.RDSEngineVersion.POSTGRES16,
)

def handler(event, context):
    result = rds.query("SELECT 1")  # TODO: Add your SQL query here
    return {
        "statusCode": 200,
        "body": f"Query result: {result}",
    }

api = lf.aws.LambdaService(
    name="my-lambda-api",
    handler=handler,
    runtime=lf.aws.lambda_service.PythonRuntime(
        requirements_txt_path="requirements.txt"
    ),
)
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Simply add launchflow[aws] to your requirements.txt, then run lf deploy to create the RDS database + deploy the Lambda API changes!

Notice that none of these code examples include networking, roles, or any other low-level configuration. The LaunchFlow client handles that for us, while still exposing the options we need to customize our deployment environment.

NOTE: You can plug in your own Terraform modules to customize the low-level environment configuration if needed.

I hope this helps save you from getting lost in the AWS sauce!

Happy to connect over email (josh@launchflow.com) or LinkedIn if you need any help trying out the code examples.

Happy Launching! πŸš€

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