Python modules

You can create a module using #[pymodule]:

#![allow(unused)]
fn main() {
use pyo3::prelude::*;

#[pyfunction]
fn double(x: usize) -> usize {
    x * 2
}

/// This module is implemented in Rust.
#[pymodule]
fn my_extension(py: Python<'_>, m: &PyModule) -> PyResult<()> {
    m.add_function(wrap_pyfunction!(double, m)?)?;
    Ok(())
}
}

The #[pymodule] procedural macro takes care of exporting the initialization function of your module to Python.

The module's name defaults to the name of the Rust function. You can override the module name by using #[pyo3(name = "custom_name")]:

#![allow(unused)]
fn main() {
use pyo3::prelude::*;

#[pyfunction]
fn double(x: usize) -> usize {
    x * 2
}

#[pymodule]
#[pyo3(name = "custom_name")]
fn my_extension(py: Python<'_>, m: &PyModule) -> PyResult<()> {
    m.add_function(wrap_pyfunction!(double, m)?)?;
    Ok(())
}
}

The name of the module must match the name of the .so or .pyd file. Otherwise, you will get an import error in Python with the following message: ImportError: dynamic module does not define module export function (PyInit_name_of_your_module)

To import the module, either:

• copy the shared library as described in Manual builds, or
• use a tool, e.g. maturin develop with maturin or python setup.py develop with setuptools-rust.

Documentation

The Rust doc comments of the module initialization function will be applied automatically as the Python docstring of your module.

For example, building off of the above code, this will print This module is implemented in Rust.:

import my_extension

print(my_extension.__doc__)

Python submodules

You can create a module hierarchy within a single extension module by using PyModule.add_submodule(). For example, you could define the modules parent_module and parent_module.child_module.

#![allow(unused)]
fn main() {
use pyo3::prelude::*;

#[pymodule]
fn parent_module(py: Python<'_>, m: &PyModule) -> PyResult<()> {
    register_child_module(py, m)?;
    Ok(())
}

fn register_child_module(py: Python<'_>, parent_module: &PyModule) -> PyResult<()> {
    let child_module = PyModule::new(py, "child_module")?;
    child_module.add_function(wrap_pyfunction!(func, child_module)?)?;
    parent_module.add_submodule(child_module)?;
    Ok(())
}

#[pyfunction]
fn func() -> String {
    "func".to_string()
}

Python::with_gil(|py| {
   use pyo3::wrap_pymodule;
   use pyo3::types::IntoPyDict;
   let parent_module = wrap_pymodule!(parent_module)(py);
   let ctx = [("parent_module", parent_module)].into_py_dict(py);

   py.run("assert parent_module.child_module.func() == 'func'", None, Some(&ctx)).unwrap();
})
}

Note that this does not define a package, so this won’t allow Python code to directly import submodules by using from parent_module import child_module. For more information, see #759 and #1517.

It is not necessary to add #[pymodule] on nested modules, which is only required on the top-level module.