matplotlib中用于SciPy树状图(Python)的更大的颜色调色板 [英] Bigger color-palette in matplotlib for SciPy's dendrogram (Python)
问题描述
我想在 目前, 如何使用更多的颜色从这种类型的图形中的巨大调色板? 大多数matplotlib色彩映射将给你一个介于0和1之间的值。例如, 将打印 所以你不再需要被限制为提供给你的价值。只需选择一个色彩映射,并根据某些分数从该色彩映射中获取颜色。例如,在您的代码中,您可以考虑, 或某事。我不确定你想要如何显示你的颜色,但我相信你可以很容易地修改你的代码来实现任何你想要的颜色组合... 另一件事你可以尝试是 玩一下... I'm trying to expand my Currently, the How can I use more colors from that huge palette in this type of figure? Most matplotlib colormaps will give you a value given a value between 0 and 1. For example, will print So you no longer need to be restricted to values provided to you. Just choose a colormap, and get a color from that colormap depending upon some fraction. For example, in your code, you could consider, or something to that effect. I am unsure how exactly you want to display your colors, but I am sure you can modify your code very easily for achieving any color combinations you want ... Another thing you can try is and pass on that Play around a bit ... 这篇关于matplotlib中用于SciPy树状图(Python)的更大的颜色调色板的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋! matplotlib
或<$ 中尝试扩大 color_palette
c / c> 使用
color_palette
只有几种颜色,因此多个集群被映射为相同的颜色。我知道有1600万 RGB
颜色,所以...
#!/ usr / bin / python
从__future__导入print_function
导入pandas为pd
导入matplotlib.pyplot as plt
导入numpy为np
导入colorsys
从scipy.cluster.hierarchy import dendrogram,linkage,fcluster
从scipy.spatial导入距离
np.random.seed(0)#43984
#Dims
n,m = 10,1000
#DataFrame:rows = Samples,cols = Attributes
attributes = [a+ str(j)for j in range(m)]
DF_data = pd.DataFrame (n.random.randn(n,m),#
columns = attributes)
A_dist = distance.cdist(DF_data.as_matrix()。T,DF_data.as_matrix )
DF_dist = pd.DataFrame(A_dist,index = attributes,columns = attributes)
#Linkage Matrix
Z = linkage(正方形(DF_dist.as_matrix =average)#metric =euclidead因为输入是一个不相似度量?
#Create dendrogram
D_dendro = dendrogram(
Z,
labels = DF_dist.index,
no_plot = True ,
color_threshold = 3.5,
count_sort =ascending,
#link_color_func = lambda k:colors [k]
)
#显示树状图
def plotTree(D_dendro):
fig,ax = plt.subplots(figsize =(25,10))
icoord = np.array(D_dendro ['icoord'])
dcoord = np.array(D_dendro ['dcoord'])
color_list = np.array(D_dendro ['color_list'])
x_min,x_max = icoord.min(),icoord.max b $ b y_min,y_max = dcoord.min(),dcoord.max()
对于xs,ys,zip中的颜色(icoord,dcoord,color_list):
plt.plot xs,ys,color)
plt.xlim(x_min-10,x_max + 0.1 * abs(x_max))
plt.ylim(y_min,y_max + 0.1 * abs(y_max))
plt.title(Dendrogram,fontsize = 30)
plt.xlabel(Clusters,fontsize = 25)
plt.ylabel(Distance,fontsize = 25)
plt。 yticks(fontsize = 20)
plt.show()
return(fig,ax)
fig,ax = plotTree(D_dendro)#wrapper b
$ b #Dims
print(
len(set(D_dendro [color_list])),^来自树形图的颜色数,
len(D_dendro [ ivb],^#of labels,sep =\\\
)
#7
#^树状图中的颜色数量
#1000
#的标签
import matplotlib.pyplot as plt
import numpy as np
print [plt.cm.Greens(i)for i in np.linspace(0,1, 5)]
[(0.9686274528503418,0.98823529481887817,0.96078431606292725,1.0),
(0.77922338878407194,0.91323337695177864,0.75180316742728737,1.0),
(0.45176470875740049,0.76708959481295413,0.46120723030146432,1.0),
(0.13402538141783546,0.54232989970375511,0.226828144368003398,1.0),
(0.0,0.26666668057441711,0.10588235408067703,1.0)]
$ b b
代表xs,ys代表zip(icoord,dcoord):
color = plt.cm.Spectral(ys / 6.0)
plt.plot(xs,ys,color)
b $ b
N = D_dendro [color_list]
colorList = [plt.cm.Spectral (i)/(N-1))for a in range(N)]
colorList
。
color_palette
in either matplotlib
or seaborn
for use in scipy
's dendrogram so it colors each cluster differently.color_palette
only has a few colors so multiple clusters are getting mapped to the same color. I know there's like 16 million RGB
colors, so...#!/usr/bin/python
from __future__ import print_function
import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import colorsys
from scipy.cluster.hierarchy import dendrogram,linkage,fcluster
from scipy.spatial import distance
np.random.seed(0) #43984
#Dims
n,m = 10,1000
#DataFrame: rows = Samples, cols = Attributes
attributes = ["a" + str(j) for j in range(m)]
DF_data = pd.DataFrame(np.random.randn(n, m),#
columns = attributes)
A_dist = distance.cdist(DF_data.as_matrix().T, DF_data.as_matrix().T)
DF_dist = pd.DataFrame(A_dist, index = attributes, columns = attributes)
#Linkage Matrix
Z = linkage(squareform(DF_dist.as_matrix()),method="average") #metric="euclidead" necessary since the input is a dissimilarity measure?
#Create dendrogram
D_dendro = dendrogram(
Z,
labels=DF_dist.index,
no_plot=True,
color_threshold=3.5,
count_sort = "ascending",
#link_color_func=lambda k: colors[k]
)
#Display dendrogram
def plotTree(D_dendro):
fig,ax = plt.subplots(figsize=(25, 10))
icoord = np.array( D_dendro['icoord'] )
dcoord = np.array( D_dendro['dcoord'] )
color_list = np.array( D_dendro['color_list'] )
x_min, x_max = icoord.min(), icoord.max()
y_min, y_max = dcoord.min(), dcoord.max()
for xs, ys, color in zip(icoord, dcoord, color_list):
plt.plot(xs, ys, color)
plt.xlim( x_min-10, x_max + 0.1*abs(x_max) )
plt.ylim( y_min, y_max + 0.1*abs(y_max) )
plt.title("Dendrogram", fontsize=30)
plt.xlabel("Clusters", fontsize=25)
plt.ylabel("Distance", fontsize=25)
plt.yticks(fontsize = 20)
plt.show()
return(fig,ax)
fig,ax = plotTree(D_dendro) #wrapper I made
#Dims
print(
len(set(D_dendro["color_list"])), "^ # of colors from dendrogram",
len(D_dendro["ivl"]), "^ # of labels",sep="\n")
# 7
# ^ # of colors from dendrogram
# 1000
# ^ # of labels
import matplotlib.pyplot as plt
import numpy as np
print [plt.cm.Greens(i) for i in np.linspace(0, 1, 5)]
[(0.9686274528503418, 0.98823529481887817, 0.96078431606292725, 1.0),
(0.77922338878407194, 0.91323337695177864, 0.75180316742728737, 1.0),
(0.45176470875740049, 0.76708959481295413, 0.46120723030146432, 1.0),
(0.13402538141783546, 0.54232989970375511, 0.26828144368003398, 1.0),
(0.0, 0.26666668057441711, 0.10588235408067703, 1.0)]
for xs, ys in zip(icoord, dcoord):
color = plt.cm.Spectral( ys/6.0 )
plt.plot(xs, ys, color)
N = D_dendro["color_list"]
colorList = [ plt.cm.Spectral( float(i)/(N-1) ) for i in range(N)]
colorList
.