latex2markdown 0.2.1

Demonstration

Check out tullo.ch/projects/LaTeX2Markdown for a live demonstration of the converter.

Getting Started

pip install latex2markdown

easy_install latex2markdown

python -m latex2markdown path/to/latex/file path/to/output/markdown/file

python -m latex2markdown sample.tex sample.md

import latex2markdown
with open("latex_file.tex", "r") as f:
    latex_string = f.read()

l2m = latex2markdown.LaTeX2Markdown(latex_string)

markdown_string = l2m.to_markdown()

with open("markdown_file.md", "w") as f:
    f.write(markdown_string)

<script type="text/x-mathjax-config">
    MathJax.Hub.Config({
        extensions: ["tex2jax.js", "AMSmath.js"],
        jax: ["input/TeX", "output/HTML-CSS"],
        tex2jax: {
            inlineMath: [ ['$','$'], ["\\(","\\)"] ],
            displayMath: [ ['$$','$$'], ["\[","\]"] ],
            processEscapes: true
        },
    });
</script>
<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>

Why not use Pandoc?

Pandoc is an excellent document converter for less complex LaTeX documents. Indeed, I’ve used it to convert this README document to a reST version for use on PyPI.

Unfortunately, it is not designed to deal with documents that use the AMSTeX extensions - which include the theorem, lemma, proof, and exercise environments that are heavily used for typesetting papers, lecture notes, and other documents.

As neither Pandoc nor MathJaX can deal with these documents, I hacked together a set of regular expressions that can convert a subset of LaTeX to Markdown, and used a few more to convert the sMarkdown to MathJaX-convertible Markdown.

Example

As an example, the following LaTeX code:

\section{Example Section}
\begin{thm}[Euclid]
    There are infinitely many primes.
\end{thm}

\begin{proof}
    Suppose that $p_1 < p_2 < \dots < p_n$ are all of the primes.
    Let $P = 1 + \prod_{i=1}^n p_i$ and let $p$ be a prime dividing $P$.

    Then $p$ can not be any of $p_i$, for otherwise $p$ would divide the
    difference $P - \left(\prod_{i=1}^n p_i \right) - 1$, which is impossible.
    So this prime $p$ is still another prime, and $p_1, p_2, \dots p_n$
    cannot be all of the primes.
\end{proof}

is converted into the following Markdown:

###  Example Section
#### Theorem 1 (Euclid)

> There are infinitely many primes.

#### Proof

Suppose that $p_1 < p_2 < \dots < p_n$ are all of the primes.
Let $P = 1 + \prod_{i=1}^n p_i$ and let $p$ be a prime dividing $P$.

Then $p$ can not be any of $p_i$, for otherwise $p$ would divide the difference
$P - \left(\prod_{i=1}^n p_i \right) - 1$, which is impossible. So this prime
$p$ is still another prime, and $p_1, p_2, \dots p_n$ cannot be all of the primes.

Supported LaTeX/AMSTeX Environments

• emph, textbf, texttt

• thm

• prop

• lem

• exer

• proof

• chapter

• section

• subsection

• itemize

• enumerate

along with everything supported by MathJax - list available online.