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

從 plotly.colors.validate_colorscale 和 sample_colorscale 開始，接著追蹤 plotly.io.templates 如何透過 ColorscaleValidator 暴露 layout.colorscale 的值。比較內建的 list-of-lists 結果與範本的 tuple-of-tuples 結果，以及現有的強制轉換路徑；當範本 colorscale 可以傳遞給 sample_colorscale 且不會出現回報的驗證錯誤時，即表示完成。

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.