更改matplotlib.pyplot点的颜色

Question

这是我写的剧情代码：

import matplotlib.pyplot as plt

Y = [ 1 , 2 , 3 ]
X = [ 1 , 2 , 4 ]
vocabulary = [1 , 2 , 3]

plt.scatter(X , Y)
for label, x, y in zip(vocabulary, X, Y):
    if(label == 1):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='red' , textcoords='offset points')
    elif(label == 1):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='green' , textcoords='offset points')
    elif(label == 1):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='blue' , textcoords='offset points')
    else :
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='black' , textcoords='offset points')
plt.show()

我正在尝试根据数组vocabulary中的值更改颜色 如果1然后将数据点着色为红色，如果2则为绿色，如果为3则为蓝色，否则将该点着色为黑色。但是对于所有点，每个点的颜色都设置为蓝色。如何根据vocabulary的当前值

为数据点着色

以上代码产生：

Answer 1

您可以制作颜色词典并在散点图中查找，如下所示

%matplotlib inline
import matplotlib.pyplot as plt

Y = [ 1 , 2 , 3 ,6]
X = [ 1 , 2 , 4 ,5]
vocabulary = [1 , 2 , 3, 0]
my_colors = {1:'red',2:'green',3:'blue'}

for i,j in enumerate(X):
    # look for the color based on vocabulary, if not found in vocubulary, then black is returned.
    plt.scatter(X[i] , Y[i], color = my_colors.get(vocabulary[i], 'black'))

plt.show()

结果

Answer 2

你刚做了一个副本＆amp;粘贴错误。 只是评论你的风格：你可以在使用颜色列表时避免这么多ifs，所以：

colors=[red,green,blue,black]

然后：

plt.annotate('', xy=(x, y), xytext=(0, 0), color=colors[max(3,label)] , textcoords='offset points')

你的代码必须如此，你总是写elif label=1，这完全没有意义：

import matplotlib.pyplot as plt

Y = [ 1 , 2 , 3 ]
X = [ 1 , 2 , 4 ]
vocabulary = [1 , 2 , 3]

plt.scatter(X , Y)
for label, x, y in zip(vocabulary, X, Y):
    if(label == 1):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='red' , textcoords='offset points')
    elif(label == 2):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='green' , textcoords='offset points')
    elif(label == 3):
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='blue' , textcoords='offset points')
    else :
        plt.annotate('', xy=(x, y), xytext=(0, 0), color='black' , textcoords='offset points')
plt.show()