Python 在范围内绘制矩形和颜色

Question

import brewer2mpl
import numpy as np

a = np.random.rand(3) # a[0] represents value of rect1, a[1] of rect[2]....

def get_colors():    
    """
    Get colorbrewer colors, which are nicer
    """            
    bmap   = brewer2mpl.get_map('Accent','qualitative',6)
    return bmap.mpl_colors

rect1 = matplotlib.patches.Rectangle((2,2), 1, 1, color='yellow'))
ax.add_patch(rect1)
rect2 = matplotlib.patches.Rectangle((3,3), 1, 1, color='green'))
ax.add_patch(rect2)
rect3 = matplotlib.patches.Rectangle((5,5), 1, 1, color='red'))
ax.add_patch(rect3)

我希望矩形的颜色根据向量的值而变化 'a'。不是纯 yellow/green/red 颜色，而是 select 范围内的颜色，最好是 brewer2mpl 颜色

Answer 1

据我所知，mpl_color 是一个颜色列表。颜色是三元组，范围在 (0,1) 代表 rgb 量。你可以把一个颜色设置成这样的三元组。

import pylab as py

py.plot([0,1],[2,3], color = (1,0,0.))
py.savefig('tmp.png')

因此，您所要做的就是获取向量 a 中的条目（我将在下面将其称为 avalue，使用 0<avalue<1）并将其映射到适当的整数.例如

colorlist = get_colors()
colorlength = len(colorlist)
py.plot([0,1],[2,3], color = colorlist[int(avalue*colorlength)])

Answer 2

您不一定需要创建一个 np 数组来保存您要使用的颜色的随机索引。

为了给每个矩形分配一种颜色，您可以使用 itertools 轻松循环颜色：

import brewer2mpl
import numpy as np
import itertools

color_map = brewer2mpl.get_map('Accent', 'qualitative', 6)
colors = itertools.cycle(color_map.mpl_colors)

rect1 = matplotlib.patches.Rectangle((2,2), 1, 1, color=next(colors))

这将按照原始顺序分配颜色。如果你想随机化颜色，你可以很容易地做到：

import random
import itertools

color_map = brewer2mpl.get_map('Accent', 'qualitative', 6)
colors = [c for c in color_map.mpl_colors]
random.shuffle(colors)
rnd_colors = itertools.cycle(colors)

rect1 = matplotlib.patches.Rectangle((2,2), 1, 1, color=next(rnd_colors))

如果你确定你的矩形有足够的颜色，你真的不需要循环，可以随机弹出一些：

color_map = brewer2mpl.get_map('Accent', 'qualitative', 6)
colors = [c for c in color_map.mpl_colors]
random.shuffle(colors)

rect1 = matplotlib.patches.Rectangle((2,2), 1, 1, color=colors.pop())

Answer 3

不确定是 colorbrewer 还是使用 a 更重要，而且它们似乎不适合同时使用。几种方法：

import brewer2mpl
import numpy as np
import matplotlib


bigness = 8
a = np.random.rand(bigness) # a[0] represents value of rect1, a[1] of rect[2]....

def get_colors(numpatches):
    """
    Get colorbrewer colors, which are nicer
    """
    bmap   = brewer2mpl.get_map('Accent','qualitative', numpatches)
    return bmap.mpl_colors


fig, axs = matplotlib.pyplot.subplots(ncols=3, nrows=1)
colors = get_colors(bigness)

for i in (0,2,3,5,bigness-1):
    rect = matplotlib.patches.Rectangle((i,i), 1, 1, color=colors[i])
    axs[0].add_patch(rect)
axs[0].set_title('Color evenly spaced\nfrom colorbrewer')

patches = []
for i in (0,2,3,5,bigness-1):
    rect = matplotlib.patches.Rectangle((i,i), 1, 1)
    patches.append(rect)
collection = matplotlib.collections.PatchCollection(patches,
                                                    cmap=matplotlib.cm.hot)
collection.set_array(a)
collection.set_edgecolor('none')
axs[1].add_collection(collection) # and why not?
axs[1].set_title('Color from\nrandom array a')


patches = []
for i in a:
    loc = i*bigness
    rect = matplotlib.patches.Rectangle((loc,loc), 1, 1)
    patches.append(rect)
collection = matplotlib.collections.PatchCollection(patches,
                                                    cmap=matplotlib.cm.hot)
collection.set_array(a)
collection.set_edgecolor('none')
axs[2].add_collection(collection) # and why not?


axs[2].set_title('Color matches location\nboth from a')

for ax in axs:
    ax.set_xlim((0,bigness))
    ax.set_ylim((0,bigness))

from os.path import realpath, basename
s = basename(realpath(__file__))
fig = matplotlib.pyplot.gcf()
fig.savefig(s.split('.')[0])

matplotlib.pyplot.show()

Python 在范围内绘制矩形和颜色

Python draw rectangle and color in range

python

matplotlib