2016-09-27 282 views
4

我有散点图,我想根据另一个值(在本例中天真地分配给np.random.random())对它进行着色。如何使用连续值[`seaborn`调色板?]颜色`matplotlib` scatterplot

是否有一种方法可以使用seaborn将每个点的连续值(与绘制的数据不直接关联)映射为seaborn中连续梯度的值?

这里是我的代码来生成数据:

import numpy as np 
import pandas as pd 
import matplotlib.pyplot as plt 
from sklearn.datasets import load_iris 
from sklearn.preprocessing import StandardScaler 
from sklearn import decomposition 
import seaborn as sns; sns.set_style("whitegrid", {'axes.grid' : False}) 

%matplotlib inline 
np.random.seed(0) 

# Iris dataset 
DF_data = pd.DataFrame(load_iris().data, 
         index = ["iris_%d" % i for i in range(load_iris().data.shape[0])], 
         columns = load_iris().feature_names) 

Se_targets = pd.Series(load_iris().target, 
         index = ["iris_%d" % i for i in range(load_iris().data.shape[0])], 
         name = "Species") 

# Scaling mean = 0, var = 1 
DF_standard = pd.DataFrame(StandardScaler().fit_transform(DF_data), 
          index = DF_data.index, 
          columns = DF_data.columns) 

# Sklearn for Principal Componenet Analysis 
# Dims 
m = DF_standard.shape[1] 
K = 2 

# PCA (How I tend to set it up) 
Mod_PCA = decomposition.PCA(n_components=m) 
DF_PCA = pd.DataFrame(Mod_PCA.fit_transform(DF_standard), 
         columns=["PC%d" % k for k in range(1,m + 1)]).iloc[:,:K] 
# Plot 
fig, ax = plt.subplots() 
ax.scatter(x=DF_PCA["PC1"], y=DF_PCA["PC2"], color="k") 
ax.set_title("No Coloring") 

enter image description here

理想情况下,我想要做这样的事情:

# Color classes 
cmap = {obsv_id:np.random.random() for obsv_id in DF_PCA.index} 

# Plot 



fig, ax = plt.subplots() 
ax.scatter(x=DF_PCA["PC1"], y=DF_PCA["PC2"], color=[cmap[obsv_id] for obsv_id in DF_PCA.index]) 
ax.set_title("With Coloring") 

# ValueError: to_rgba: Invalid rgba arg "0.2965562650640299" 
# to_rgb: Invalid rgb arg "0.2965562650640299" 
# cannot convert argument to rgb sequence 

,但它并没有像连续值。

我想用一个调色板一样:

​​

enter image description here

我也试着做一些像下面,但它是没有意义的B/C不知道哪个值我曾经在我的字典cmap上面:

ax.scatter(x=DF_PCA["PC1"], y=DF_PCA["PC2"],cmap=sns.cubehelix_palette(as_cmap=True) 

回答

12
import numpy as np 
import seaborn as sns 
import matplotlib.pyplot as plt 

x, y, z = np.random.rand(3, 100) 
cmap = sns.cubehelix_palette(as_cmap=True) 

f, ax = plt.subplots() 
points = ax.scatter(x, y, c=z, s=50, cmap=cmap) 
f.colorbar(points) 

enter image description here

+0

也有颜色条!是。那正是我所期待的。非常感谢ヾ(⌐■_■)ノ♪@mwaskom –

+1

这很好,谢谢。有没有一种明智的方式通过seaborn的'hue'参数来做到这一点?我尝试过,但由于色调参数似乎将连续变量的每个值都视为分类级别,所以产生的传说毫无意义。 – user2428107

0
from matplotlib.cm import ScalarMappable 
from matplotlib.colors import Normalize 


cmap = {obsv_id:np.random.random() for obsv_id in DF_PCA.index} 
sm = ScalarMappable(norm=Normalize(vmin=min(list(cmap.values())), vmax=max(list(cmap.values()))), cmap=sns.cubehelix_palette(as_cmap=True)) 

# Plot 
fig, ax = plt.subplots() 
ax.scatter(x=DF_PCA["PC1"], y=DF_PCA["PC2"], color=[sm.to_rgba(cmap[obsv_id]) for obsv_id in DF_PCA.index]) 
ax.set_title("With Coloring") 

enter image description here

+0

如果任何人有,你没有导入'ScalarMappable'和'Normalize'那么我一定会选择是正确的更简单的方法。 –