從 Plotly 色標訪問顏色

Question

Plotly 中是否有辦法訪問其范圍內任何值的顏色圖顏色？

我知道我可以從

plotly.colors.PLOTLY_SCALES["Viridis"]

但我無法找到如何訪問中間/插值。

此問題中顯示了 Matplotlib 中的等效項。 還有另一個問題解決了colorlover庫中的類似問題，但都沒有提供很好的解決方案。

Answer 1

Plotly 好像沒有這種方法，所以寫了一個：

import plotly.colors

def get_continuous_color(colorscale, intermed):
    """
    Plotly continuous colorscales assign colors to the range [0, 1]. This function computes the intermediate
    color for any value in that range.

    Plotly doesn't make the colorscales directly accessible in a common format.
    Some are ready to use:
    
        colorscale = plotly.colors.PLOTLY_SCALES["Greens"]

    Others are just swatches that need to be constructed into a colorscale:

        viridis_colors, scale = plotly.colors.convert_colors_to_same_type(plotly.colors.sequential.Viridis)
        colorscale = plotly.colors.make_colorscale(viridis_colors, scale=scale)

    :param colorscale: A plotly continuous colorscale defined with RGB string colors.
    :param intermed: value in the range [0, 1]
    :return: color in rgb string format
    :rtype: str
    """
    if len(colorscale) < 1:
        raise ValueError("colorscale must have at least one color")

    if intermed <= 0 or len(colorscale) == 1:
        return colorscale[0][1]
    if intermed >= 1:
        return colorscale[-1][1]

    for cutoff, color in colorscale:
        if intermed > cutoff:
            low_cutoff, low_color = cutoff, color
        else:
            high_cutoff, high_color = cutoff, color
            break

    # noinspection PyUnboundLocalVariable
    return plotly.colors.find_intermediate_color(
        lowcolor=low_color, highcolor=high_color,
        intermed=((intermed - low_cutoff) / (high_cutoff - low_cutoff)),
        colortype="rgb")

挑戰在於內置的 Plotly 色標並未始終如一地暴露。 有些已經被定義為色標，其他的只是必須首先轉換為色標的色板列表。

Viridis 色階是用十六進制值定義的，Plotly 顏色操作方法不喜歡這樣的值，所以最容易從這樣的色板構造它：

viridis_colors, _ = plotly.colors.convert_colors_to_same_type(plotly.colors.sequential.Viridis)
colorscale = plotly.colors.make_colorscale(viridis_colors)

get_continuous_color(colorscale, intermed=0.25)
# rgb(58.75, 80.75, 138.25)

Answer 2

這個答案擴展了亞當提供的已經很好的答案。 特別是，它處理 Plotly 色標的不一致問題。

在 Plotly 中，您可以通過編寫colorscale="name_of_the_colorscale"來指定內置色標。 這表明 Plotly 已經有一個內置工具，可以以某種方式將色標轉換為適當的值，並且能夠處理這些不一致。 通過搜索 Plotly 的源代碼，我們找到了有用的ColorscaleValidator class。 讓我們看看如何使用它：

def get_color(colorscale_name, loc):
    from _plotly_utils.basevalidators import ColorscaleValidator
    # first parameter: Name of the property being validated
    # second parameter: a string, doesn't really matter in our use case
    cv = ColorscaleValidator("colorscale", "")
    # colorscale will be a list of lists: [[loc1, "rgb1"], [loc2, "rgb2"], ...] 
    colorscale = cv.validate_coerce(colorscale_name)
    
    if hasattr(loc, "__iter__"):
        return [get_continuous_color(colorscale, x) for x in loc]
    return get_continuous_color(colorscale, loc)
        

# Identical to Adam's answer
import plotly.colors
from PIL import ImageColor

def get_continuous_color(colorscale, intermed):
    """
    Plotly continuous colorscales assign colors to the range [0, 1]. This function computes the intermediate
    color for any value in that range.

    Plotly doesn't make the colorscales directly accessible in a common format.
    Some are ready to use:
    
        colorscale = plotly.colors.PLOTLY_SCALES["Greens"]

    Others are just swatches that need to be constructed into a colorscale:

        viridis_colors, scale = plotly.colors.convert_colors_to_same_type(plotly.colors.sequential.Viridis)
        colorscale = plotly.colors.make_colorscale(viridis_colors, scale=scale)

    :param colorscale: A plotly continuous colorscale defined with RGB string colors.
    :param intermed: value in the range [0, 1]
    :return: color in rgb string format
    :rtype: str
    """
    if len(colorscale) < 1:
        raise ValueError("colorscale must have at least one color")

    hex_to_rgb = lambda c: "rgb" + str(ImageColor.getcolor(c, "RGB"))

    if intermed <= 0 or len(colorscale) == 1:
        c = colorscale[0][1]
        return c if c[0] != "#" else hex_to_rgb(c)
    if intermed >= 1:
        c = colorscale[-1][1]
        return c if c[0] != "#" else hex_to_rgb(c)

    for cutoff, color in colorscale:
        if intermed > cutoff:
            low_cutoff, low_color = cutoff, color
        else:
            high_cutoff, high_color = cutoff, color
            break

    if (low_color[0] == "#") or (high_color[0] == "#"):
        # some color scale names (such as cividis) returns:
        # [[loc1, "hex1"], [loc2, "hex2"], ...]
        low_color = hex_to_rgb(low_color)
        high_color = hex_to_rgb(high_color)

    return plotly.colors.find_intermediate_color(
        lowcolor=low_color,
        highcolor=high_color,
        intermed=((intermed - low_cutoff) / (high_cutoff - low_cutoff)),
        colortype="rgb",
    )

此時，您所要做的就是：

get_color("phase", 0.5)
# 'rgb(123.99999999999999, 112.00000000000001, 236.0)'

import numpy as np
get_color("phase", np.linspace(0, 1, 256))
# ['rgb(167, 119, 12)',
#  'rgb(168.2941176470588, 118.0078431372549, 13.68235294117647)',
#  ...

編輯：處理特殊情況的改進。

Answer 3

從plotly.express.colors到sample_colorscale有一個內置方法可以提供顏色樣本：

from plotly.express.colors import sample_colorscale

import plotly.graph_objects as go
import numpy as np

x = np.linspace(0, 1, 25)
c = sample_colorscale('jet', list(x))

fig = go.FigureWidget()
fig.add_trace(
    go.Bar(x=x, y=y, marker_color=c)
)
fig.show()

參見 output 圖 -> sampled_colors

Answer 4

官方參考解釋。 這里

import plotly.express as px

print(px.colors.sequential.Viridis)
['#440154', '#482878', '#3e4989', '#31688e', '#26828e', '#1f9e89', '#35b779', '#6ece58', '#b5de2b', '#fde725']

print(px.colors.sequential.Viridis[0])
#440154