Colorscales in templates are not valid list of lists, which breaks `plotly.colors.sample_colorscale`

plotly.colors.validate_colorscale と sample_colorscale から始め、plotly.io.templates が ColorscaleValidator を通じて layout.colorscale の値をどのように公開しているかを追跡します。組み込みの list-of-lists 結果とテンプレートの tuple-of-tuples 結果、および既存の強制変換経路を比較します。テンプレートの colorscale を報告された検証エラーなしで sample_colorscale に渡せるようになれば完了です。

Hello! I have something which is maybe a bug. It's certainly how I expected plotly.py to work, although maybe it's just a known limitation of how layout templates work.

Plotly comes with various colorscales built in, which work great:

import plotly
rainbow = plotly.colors.get_colorscale("rainbow")
# gives
[[0.0, 'rgb(150,0,90)'],
 [0.125, 'rgb(0,0,200)'],
 [0.25, 'rgb(0,25,255)'],
 [0.375, 'rgb(0,152,255)'],
 [0.5, 'rgb(44,255,150)'],
 [0.625, 'rgb(151,255,0)'],
 [0.75, 'rgb(255,234,0)'],
 [0.875, 'rgb(255,111,0)'],
 [1.0, 'rgb(255,0,0)']]

This is a list of lists and works as expected: plotly.colors.validate_colorscale(rainbow) passes.

Now I'd like to extract the colorscale from a template. This where things don't work as expected:

diverging  = plotly.io.templates["plotly"].layout.colorscale.diverging
# gives
((0, '#8e0152'),
 (0.1, '#c51b7d'),
 (0.2, '#de77ae'),
 (0.3, '#f1b6da'),
 (0.4, '#fde0ef'),
 (0.5, '#f7f7f7'),
 (0.6, '#e6f5d0'),
 (0.7, '#b8e186'),
 (0.8, '#7fbc41'),
 (0.9, '#4d9221'),
 (1, '#276419'))

This is a tuple of tuples and does not pass plotly.colors.validate_colorscale(diverging), which raises:

_plotly_utils.exceptions.PlotlyError: A valid colorscale must be a list.

It's easy to convert this into a valid colorscale as diverging = [list(x) for x in diverging] but it seems strange that this is necessary.

Should the colorscale coming from a template be a list of lists already? Or should validate_colorscale accept tuple of tuples as valid? The real reason this matters is because plotly.colors.sample_colorscale runs validate_colorscale under the hood, which raises an error and means that sample_colorscale doesn't work.

Side note

AFAICT there's no way register a new colorscale in plotly other than through a template. plotly.colors.named_colorscales comes from _plotly_utils.basevalidators.ColorscaleValidator.named_colorscales, but that's private and has no method for adding a new scale.

ColorscaleValidator("", "").validate_coerce(diverging) works to do the tuple of tuples to list of lists conversion, but again it's not public, and it seems weird that it would be necessary to run any coercion on a colorscale coming from a template anyway.