在matplotlib上的散点图中为每个系列设置不同的颜色

Question

假设我有三个数据集:

X = [1,2,3,4]
Y1 = [4,8,12,16]
Y2 = [1,4,9,16]

我可以分散这个情节:

from matplotlib import pyplot as plt
plt.scatter(X,Y1,color='red')
plt.scatter(X,Y2,color='blue')
plt.show()

我怎么能用10套呢？

我搜索了这个,可以找到我所问的任何参考.

编辑:澄清(希望)我的问题

如果我多次调用散点图,我只能在每个散点图上设置相同的颜色.另外,我知道我可以手动设置颜色数组,但我确信有更好的方法可以做到这一点.我的问题是,"我怎样才能自动散布我的几个数据集,每个数据集都有不同的颜色.

如果这有帮助,我可以轻松地为每个数据集分配一个唯一的编号.

Answer 1

我不知道'手动'是什么意思.您可以选择颜色图并轻松制作颜色数组:

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

x = np.arange(10)
ys = [i+x+(i*x)**2 for i in range(10)]

colors = cm.rainbow(np.linspace(0, 1, len(ys)))
for y, c in zip(ys, colors):
    plt.scatter(x, y, color=c)

或者您可以使用itertools.cycle并指定要循环的颜色来制作自己的颜色循环仪,next以获得所需的颜色.例如,有3种颜色:

import itertools

colors = itertools.cycle(["r", "b", "g"])
for y in ys:
    plt.scatter(x, y, color=next(colors))

想想看,也许更清洁,不要zip与第一个一起使用:

colors = iter(cm.rainbow(np.linspace(0, 1, len(ys))))
for y in ys:
    plt.scatter(x, y, color=next(colors))

Answer 2

在matplotlib中绘制具有不同颜色的点的绘图的常规方法是将颜色列表作为参数传递.

例如:

import matplotlib.pyplot
matplotlib.pyplot.scatter([1,2,3],[4,5,6],color=['red','green','blue'])

如果您有列表列表,并且希望每个列表都有颜色.我认为最优雅的方式是由@DSM建议的,只需做一个循环,多次调用分散.

但是如果由于某种原因你想只用一个电话就可以做到这一点,你可以制作一个大的颜色列表,列表理解和一些地板划分:

import matplotlib
import numpy as np

X = [1,2,3,4]
Ys = np.array([[4,8,12,16],
      [1,4,9,16],
      [17, 10, 13, 18],
      [9, 10, 18, 11],
      [4, 15, 17, 6],
      [7, 10, 8, 7],
      [9, 0, 10, 11],
      [14, 1, 15, 5],
      [8, 15, 9, 14],
       [20, 7, 1, 5]])
nCols = len(X)  
nRows = Ys.shape[0]

colors = matplotlib.cm.rainbow(np.linspace(0, 1, len(Ys)))

cs = [colors[i//len(X)] for i in range(len(Ys)*len(X))] #could be done with numpy's repmat
Xs=X*nRows #use list multiplication for repetition
matplotlib.pyplot.scatter(Xs,Ys.flatten(),color=cs)

cs = [array([ 0.5,  0. ,  1. ,  1. ]),
 array([ 0.5,  0. ,  1. ,  1. ]),
 array([ 0.5,  0. ,  1. ,  1. ]),
 array([ 0.5,  0. ,  1. ,  1. ]),
 array([ 0.28039216,  0.33815827,  0.98516223,  1.        ]),
 array([ 0.28039216,  0.33815827,  0.98516223,  1.        ]),
 array([ 0.28039216,  0.33815827,  0.98516223,  1.        ]),
 array([ 0.28039216,  0.33815827,  0.98516223,  1.        ]),
 ...
 array([  1.00000000e+00,   1.22464680e-16,   6.12323400e-17,
          1.00000000e+00]),
 array([  1.00000000e+00,   1.22464680e-16,   6.12323400e-17,
          1.00000000e+00]),
 array([  1.00000000e+00,   1.22464680e-16,   6.12323400e-17,
          1.00000000e+00]),
 array([  1.00000000e+00,   1.22464680e-16,   6.12323400e-17,
          1.00000000e+00])]

Answer 3

一个简单的修复

如果您只有一种类型的集合(例如散射而没有误差条),您也可以在绘制它们之后更改颜色,这有时更容易执行.

import matplotlib.pyplot as plt
from random import randint
import numpy as np

#Let's generate some random X, Y data X = [ [frst group],[second group] ...]
X = [ [randint(0,50) for i in range(0,5)] for i in range(0,24)]
Y = [ [randint(0,50) for i in range(0,5)] for i in range(0,24)]
labels = range(1,len(X)+1)

fig = plt.figure()
ax = fig.add_subplot(111)
for x,y,lab in zip(X,Y,labels):
        ax.scatter(x,y,label=lab)

您需要的唯一代码:

#Now this is actually the code that you need, an easy fix your colors just cut and paste not you need ax.
colormap = plt.cm.gist_ncar #nipy_spectral, Set1,Paired  
colorst = [colormap(i) for i in np.linspace(0, 0.9,len(ax.collections))]       
for t,j1 in enumerate(ax.collections):
    j1.set_color(colorst[t])


ax.legend(fontsize='small')

即使在同一子图中有许多不同的散点图,输出也会为您提供不同的颜色.

Answer 4

你可以随时使用这样的plot()功能:

import matplotlib.pyplot as plt

import numpy as np

x = np.arange(10)
ys = [i+x+(i*x)**2 for i in range(10)]
plt.figure()
for y in ys:
    plt.plot(x, y, 'o')
plt.show()

Answer 5

这个问题在2013年1月和matplotlib 1.3.1(2013年8月)之前有点棘手,这是你可以在matpplotlib网站上找到的最古老的稳定版本.但在那之后它是相当微不足道的.

因为matplotlib.pylab.scatter支持的当前版本分配:颜色名称字符串数组,带颜色映射的浮点数数组,RGB或RGBA数组.

这个答案专注于@ Oxinabox对于在2015年纠正2013版本的无尽热情.

您可以选择在单个调用中使用具有多种颜色的scatter命令.

1. 作为pylab.scatter命令支持使用RGBA数组来做你想要的任何颜色;

2. 早在2013年初,就没有办法这样做,因为该命令仅支持整个散点集合的单色.当我在进行10000线项目时,我找到了绕过它的一般解决方案.所以它非常俗气,但我可以做任何形状,颜色,大小和透明.这个技巧也可以应用于绘制路径集合,行集合....

代码也受到源代码的启发pyplot.scatter,我只是复制了散布所做的事情而没有触发它来绘制.

该命令pyplot.scatter返回一个PatchCollection对象,文件"matplotlib/collections.py"私有变量在_facecolors中Collection类和方法set_facecolors.

所以每当你有一个分散点来绘制你可以这样做:

# rgbaArr is a N*4 array of float numbers you know what I mean
# X is a N*2 array of coordinates
# axx is the axes object that current draw, you get it from
# axx = fig.gca()

# also import these, to recreate the within env of scatter command 
import matplotlib.markers as mmarkers
import matplotlib.transforms as mtransforms
from matplotlib.collections import PatchCollection
import matplotlib.markers as mmarkers
import matplotlib.patches as mpatches


# define this function
# m is a string of scatter marker, it could be 'o', 's' etc..
# s is the size of the point, use 1.0
# dpi, get it from axx.figure.dpi
def addPatch_point(m, s, dpi):
    marker_obj = mmarkers.MarkerStyle(m)
    path = marker_obj.get_path()
    trans = mtransforms.Affine2D().scale(np.sqrt(s*5)*dpi/72.0)
    ptch = mpatches.PathPatch(path, fill = True, transform = trans)
    return ptch

patches = []
# markerArr is an array of maker string, ['o', 's'. 'o'...]
# sizeArr is an array of size float, [1.0, 1.0. 0.5...]

for m, s in zip(markerArr, sizeArr):
    patches.append(addPatch_point(m, s, axx.figure.dpi))

pclt = PatchCollection(
                patches,
                offsets = zip(X[:,0], X[:,1]),
                transOffset = axx.transData)

pclt.set_transform(mtransforms.IdentityTransform())
pclt.set_edgecolors('none') # it's up to you
pclt._facecolors = rgbaArr

# in the end, when you decide to draw
axx.add_collection(pclt)
# and call axx's parent to draw_idle()