从 Plotly 色标访问颜色
Access Color from Plotly Color Scale
Plotly 中有没有一种方法可以访问其范围内任意值的颜色图颜色？
我知道我可以从
访问色标的定义颜色
plotly.colors.PLOTLY_SCALES["Viridis"]
但我找不到如何访问中间值/内插值。
显示了 Matplotlib 中的等效项 in this question. There is also ，它解决了 colorlover 库中的类似问题，但都没有提供很好的解决方案。
官方参考说明。 Here
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
plotly好像没有这样的方法，所以写了一个：
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")
挑战在于 built-in Plotly 色标的曝光不一致。有些已经定义为色标，有些只是色样列表，必须先转换为色标。
Viridis 色标是用十六进制值定义的，Plotly 颜色处理方法不喜欢它，因此最容易从这样的色板构建它：
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)
这个答案扩展了 Adam 提供的已经很好的答案。特别是，它处理了 Plotly 的色标不一致问题。
在 Plotly 中，您可以通过编写 colorscale="name_of_the_colorscale" 来指定内置色标。这表明 Plotly 已经有一个内置工具，可以以某种方式将色标转换为适当的值，并且能够处理这些不一致。通过搜索 Plotly 的源代码，我们发现有用的 ColorscaleValidator class。让我们看看如何使用它：
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",
)
此时，您要做的就是：
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)',
# ...
编辑：处理特殊情况的改进。
Plotly 中有没有一种方法可以访问其范围内任意值的颜色图颜色？
我知道我可以从
访问色标的定义颜色plotly.colors.PLOTLY_SCALES["Viridis"]
但我找不到如何访问中间值/内插值。
显示了 Matplotlib 中的等效项 in this question. There is also colorlover 库中的类似问题，但都没有提供很好的解决方案。
官方参考说明。 Here
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
plotly好像没有这样的方法，所以写了一个：
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")
挑战在于 built-in Plotly 色标的曝光不一致。有些已经定义为色标，有些只是色样列表，必须先转换为色标。
Viridis 色标是用十六进制值定义的，Plotly 颜色处理方法不喜欢它，因此最容易从这样的色板构建它：
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)
这个答案扩展了 Adam 提供的已经很好的答案。特别是，它处理了 Plotly 的色标不一致问题。
在 Plotly 中，您可以通过编写 colorscale="name_of_the_colorscale" 来指定内置色标。这表明 Plotly 已经有一个内置工具，可以以某种方式将色标转换为适当的值，并且能够处理这些不一致。通过搜索 Plotly 的源代码，我们发现有用的 ColorscaleValidator class。让我们看看如何使用它：
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",
)
此时，您要做的就是：
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)',
# ...
编辑：处理特殊情况的改进。