Get only the code out of Jupyter Notebook

Question

Is there a solution to pull out all the code of the notebook? For example, if I wanted to generate a source file of my notebook "source.py" that contained all the code in the code cells of the notebook, is that possible?

Thanks!

Answer 1

nbconvert

You can use the command line tool nbconvert to convert the ipynb file to various other formats. The easiest way to convert it to a .py file is:

jupyter nbconvert --no-prompt --to script notebook_name.ipynb

It outputs only the code and comments without the markdown, input and output prompts. There is also --stdout option.

• nbconvert documentation

jq

But you can also just parse the JSON of the notebook using jq:

jq -j '
  .cells
  | map( select(.cell_type == "code") | .source + ["\n\n"] )
  | .[][]
  ' \
  notebook.ipynb > source.py

• jq homepage
• Jupyter Notebook format

Answer 2

Since the notebook format is JSON it's relatively easy to extract just the text content of only the code cells. The task is made even easier when you use the Python API for working with notebook files.

The following will get you the code on standard output. You can handle it in other ways similarly easily. Bear in mind code source may not have a terminating newline.

from nbformat import read, NO_CONVERT

with open("Some Notebook.ipynb") as fp:
    notebook = read(fp, NO_CONVERT)
cells = notebook['cells']
code_cells = [c for c in cells if c['cell_type'] == 'code']
for cell in code_cells:
    print(cell['source'])

Notebook nodes are a little more flexible than dictionaries, though, and allow attribute (.name) access to fields as well as subscripting (['name']). As a typing-challenged person I find it preferable to write

cells = notebook.cells
code_cells = [c for c in cells if c.cell_type == 'code']

for cell in code_cells:
    print(cell.source)

In answering this question I became aware that the nbformat library has been unbundled, and can therefore be installed with pip without the rest of Jupyter.