A decision-tree based conditional independence test
Project description
.. image:: https://img.shields.io/badge/License-MIT-yellow.svg
:target: https://opensource.org/licenses/MIT
:alt: License
*A Decision Tree (Conditional) Independence Test (DTIT).*
Introduction
-----------
Let *x, y, z* be random variables. Then deciding whether *P(y | x, z) = P(y | z)*
can be difficult, especially if the variables are continuous. This package
implements a simple yet efficient and effective conditional independence test,
described in [link to arXiv when we write it up!]. Important features that differentiate
this test from competition:
* It is fast. Worst-case speed scales as O(n_data * log(n_data) * dim), where dim is max(x_dim + z_dim, y_dim). However, amortized speed is O(n_data * log(n_data) * log(dim)).
* It applies to cases where some of x, y, z are continuous and some are discrete, or categorical (one-hot-encoded).
* It is very simple to understand and modify.
We have applied this test to tens of thousands of samples of thousand-dimensional datapoints in seconds. For smaller dimensionalities and sample sizes, it takes a fraction of a second. The algorithm is described in [arXiv link coming], where we also provide detailed experimental results and comparison with other methods. However for now, you should be able to just look through the code to understand what's going on -- it's only 90 lines of Python, including detailed comments!
Usage
-----
Basic usage is simple:
.. code:: python
import numpy as np
import dtit
# Generate some data such that x is indpendent of y given z.
n_samples = 300
z = np.random.dirichlet(alpha=np.ones(2), size=n_samples)
x = np.vstack([np.random.multinomial(20, p) for p in z])
y = np.vstack([np.random.multinomial(20, p) for p in z])
# Run the conditional independence test.
pval = dtit.test(x, y, z)
Here, we created discrete variables *x* and *y*, d-separated by a "common cause"
*z*. The null hypothesis is that *x* is independent of *y* given *z*. Since in this
case the variables are independent given *z*, pval should be distributed uniformly on [0, 1].
Requirements
------------
To use the nn methods:
* numpy >= 1.12
* scikit-learn >= 0.18.1
* scipy >= 0.16.1
.. _pip: http://www.pip-installer.org/en/latest/
:target: https://opensource.org/licenses/MIT
:alt: License
*A Decision Tree (Conditional) Independence Test (DTIT).*
Introduction
-----------
Let *x, y, z* be random variables. Then deciding whether *P(y | x, z) = P(y | z)*
can be difficult, especially if the variables are continuous. This package
implements a simple yet efficient and effective conditional independence test,
described in [link to arXiv when we write it up!]. Important features that differentiate
this test from competition:
* It is fast. Worst-case speed scales as O(n_data * log(n_data) * dim), where dim is max(x_dim + z_dim, y_dim). However, amortized speed is O(n_data * log(n_data) * log(dim)).
* It applies to cases where some of x, y, z are continuous and some are discrete, or categorical (one-hot-encoded).
* It is very simple to understand and modify.
We have applied this test to tens of thousands of samples of thousand-dimensional datapoints in seconds. For smaller dimensionalities and sample sizes, it takes a fraction of a second. The algorithm is described in [arXiv link coming], where we also provide detailed experimental results and comparison with other methods. However for now, you should be able to just look through the code to understand what's going on -- it's only 90 lines of Python, including detailed comments!
Usage
-----
Basic usage is simple:
.. code:: python
import numpy as np
import dtit
# Generate some data such that x is indpendent of y given z.
n_samples = 300
z = np.random.dirichlet(alpha=np.ones(2), size=n_samples)
x = np.vstack([np.random.multinomial(20, p) for p in z])
y = np.vstack([np.random.multinomial(20, p) for p in z])
# Run the conditional independence test.
pval = dtit.test(x, y, z)
Here, we created discrete variables *x* and *y*, d-separated by a "common cause"
*z*. The null hypothesis is that *x* is independent of *y* given *z*. Since in this
case the variables are independent given *z*, pval should be distributed uniformly on [0, 1].
Requirements
------------
To use the nn methods:
* numpy >= 1.12
* scikit-learn >= 0.18.1
* scipy >= 0.16.1
.. _pip: http://www.pip-installer.org/en/latest/
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