An interpreter package for AIML, the Artificial Intelligence Markup Language

pip install aiml==0.9.2



This is a fork of the PyAIML Python AIML interpreter. It has been refactored to make it install and work in both Python 2.7 and Python 3.

PyAIML is (c) Cort Stratton. aiml uses the same license as PyAIML (2-clause BSD), except for the ALICE AIML files taken from the Free ALICE AIML set, which are licensed with the LGPL license.


Two small scripts are added upon installation:

  • aiml-validate can be used to validate AIML files
  • aiml-bot can be used to start a simple interactive session with a bot, after loading either AIML files or a saved brain file.


The installation includes two AIML datasets:

They can be loaded via the bootstrap method in the Kernel class. See the script for an example.


There are a number of unit tests included (in the test subdirectory); they can be executed by the setup infrastructure as:

python test

or they can also be launched by directly calling:

python test [testname ...]

This last version allows executing only some of the test files by explicitly naming them in the command line; if none is specified all will be executed.

Original README from PyAIML

PyAIML is an interpreter for AIML (the Artificial Intelligence Markup Language), implemented entirely in standard Python. It strives for simple, austere, 100% compliance with the AIML 1.0.1 standard, no less and no more.

This is currently pre-alpha software. Use at your own risk!

For information on what's new in this version, see the CHANGES.txt file.

For information on the state of development, including the current level of AIML 1.0.1 compliance, see the SUPPORTED_TAGS.txt file.

Quick & dirty example (assuming you've downloaded the "standard" AIML set):

import aiml

# The Kernel object is the public interface to
# the AIML interpreter.
k = aiml.Kernel()

# Use the 'learn' method to load the contents
# of an AIML file into the Kernel.

# Use the 'respond' method to compute the response
# to a user's input string.  respond() returns
# the interpreter's response, which in this case
# we ignore.
k.respond("load aiml b")

# Loop forever, reading user input from the command
# line and printing responses.
while True: print k.respond(raw_input("> "))