pyplot/matplotlib条形图，其填充颜色取决于值 [英] pyplot/matplotlib Bar chart with fill color depending on value

问题描述

我想用 matplotlib/pyplot 制作 python

I want to produce in python with matplotlib/pyplot

• 根据值填充的条形图.
• 图例颜色栏

同时将模块依赖保持在最低限度.

while keeping module dependencies at a minimum.

有没有比这更简单的东西了?

Is there something simpler than:

import matplotlib.pyplot as plt

def color_gradient ( val, beg_rgb, end_rgb, val_min = 0, val_max = 1):
    val_scale = (1.0 * val - val_min) / (val_max - val_min)
    return ( beg_rgb[0] + val_scale * (end_rgb[0] - beg_rgb[0]),
             beg_rgb[1] + val_scale * (end_rgb[1] - beg_rgb[1]),
             beg_rgb[2] + val_scale * (end_rgb[2] - beg_rgb[2]))

# -----------------------------------------------
x_lbls = [ "09:00", "09:15", "10:10"]
y_vals = [       7,       9,       5]

plt_idx = np.arange( len( x_lbls))
bar_wd  = 0.35

grad_beg, grad_end = ( 0.5, 0.5, 0.5), (1, 1, 0)
col_list = [ color_gradient( val,
                             grad_beg,
                             grad_end,
                             min( y_vals),
                             max( y_vals)) for val in y_vals]

plt.bar( plt_idx, y_vals, color = col_list)
plt.xticks( plt_idx + bar_wd, x_lbls)
plt.show()

这仍然缺少图例颜色栏

我在 R 中使用 ggplot 的解决方案是:

my solution in R with ggplot would be:

library(ggplot2)
df = data.frame( time = 1:10, vals = abs(rnorm( n = 10)))
ggplot( df, aes( x = time, y = vals, fill = vals)) + 
  geom_bar(stat = "identity") + 
  scale_fill_gradient(low="#888888",high="#FFFF00")

并产生所需的输出:

推荐答案

我不知道如何在不绘制其他内容然后清除它的情况下让颜色栏工作，因此这不是最优雅的解决方案.

I couldn't figure out how to get the colorbar to work without plotting something else and then clearing it, so it's not the most elegant solution.

import matplotlib.pyplot as plt
from matplotlib import cm
import numpy as np

y = np.array([1, 4, 3, 2, 7, 11])
colors = cm.hsv(y / float(max(y)))
plot = plt.scatter(y, y, c = y, cmap = 'hsv')
plt.clf()
plt.colorbar(plot)
plt.bar(range(len(y)), y, color = colors)
plt.show()

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