Matplotlib通过分类因素散布颜色

Question

我有一个基本的散点图，其中x和y是 float 的。但是我想根据第三个类别变量来更改标记的颜色。类别变量为字符串形式。这似乎引起了问题。

要使用虹膜数据集，这是我想使用的代码:

#Scatter of Petal
x=df['Petal Length']
y=df['Petal Width']
z=df['Species']
plt.scatter(x, y, c=z, s=15, cmap='hot')
plt.xlabel('Petal Width')
plt.ylabel('Petal Length')
plt.title('Petal Width vs Length')

但我收到一个错误:无法将字符串转换为浮点:iris-setosa

在运行之前是否必须将分类变量更改为数字变量，还是可以对当前格式的数据进行某些处理？

谢谢

更新:整个回溯是:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-47-d67ee3bffc3b> in <module>()
      3 y=df['Petal Width']
      4 z=df['Species']
----> 5 plt.scatter(x, y, c=z, s=15, cmap='hot')
      6 plt.xlabel('Petal Width')
      7 plt.ylabel('Petal Length')

/Users/mpgartland1/anaconda/lib/python2.7/site-packages/matplotlib/pyplot.pyc in scatter(x, y, s, c, marker, cmap, norm, vmin, vmax, alpha, linewidths, verts, hold, **kwargs)
   3198         ret = ax.scatter(x, y, s=s, c=c, marker=marker, cmap=cmap, norm=norm,
   3199                          vmin=vmin, vmax=vmax, alpha=alpha,
-> 3200                          linewidths=linewidths, verts=verts, **kwargs)
   3201         draw_if_interactive()
   3202     finally:

/Users/mpgartland1/anaconda/lib/python2.7/site-packages/matplotlib/axes/_axes.pyc in scatter(self, x, y, s, c, marker, cmap, norm, vmin, vmax, alpha, linewidths, verts, **kwargs)
   3605 
   3606         if c_is_stringy:
-> 3607             colors = mcolors.colorConverter.to_rgba_array(c, alpha)
   3608         else:
   3609             # The inherent ambiguity is resolved in favor of color

/Users/mpgartland1/anaconda/lib/python2.7/site-packages/matplotlib/colors.pyc in to_rgba_array(self, c, alpha)
    420             result = np.zeros((nc, 4), dtype=np.float)
    421             for i, cc in enumerate(c):
--> 422                 result[i] = self.to_rgba(cc, alpha)
    423             return result
    424 

/Users/mpgartland1/anaconda/lib/python2.7/site-packages/matplotlib/colors.pyc in to_rgba(self, arg, alpha)
    374         except (TypeError, ValueError) as exc:
    375             raise ValueError(
--> 376                 'to_rgba: Invalid rgba arg "%s"\n%s' % (str(arg), exc))
    377 
    378     def to_rgba_array(self, c, alpha=None):

ValueError: to_rgba: Invalid rgba arg "Iris-setosa"
to_rgb: Invalid rgb arg "Iris-setosa"
could not convert string to float: iris-setosa

Answer 1

正如您的回溯告诉您的那样，您不能将字符串传递给color参数。您可以传递颜色，也可以传递将其本身解释为颜色的值数组。

看:
http://matplotlib.org/api/pyplot_api.html?highlight=plot#matplotlib.pyplot.plot

也许有一种更优雅的方法，但是以下是一种实现方式(我使用了以下数据集:https://raw.githubusercontent.com/pydata/pandas/master/pandas/tests/data/iris.csv):

import matplotlib.pyplot as plt
import matplotlib.colors as colors
import matplotlib.cm as cmx
from pandas import read_csv

df = read_csv('iris.csv')

#Scatter of Petal
x=df['PetalLength']
y=df['PetalWidth']

# Get unique names of species
uniq = list(set(df['Name']))

# Set the color map to match the number of species
z = range(1,len(uniq))
hot = plt.get_cmap('hot')
cNorm  = colors.Normalize(vmin=0, vmax=len(uniq))
scalarMap = cmx.ScalarMappable(norm=cNorm, cmap=hot)

# Plot each species
for i in range(len(uniq)):
    indx = df['Name'] == uniq[i]
    plt.scatter(x[indx], y[indx], s=15, color=scalarMap.to_rgba(i), label=uniq[i])

plt.xlabel('Petal Width')
plt.ylabel('Petal Length')
plt.title('Petal Width vs Length')
plt.legend(loc='upper left')
plt.show()

给出这样的东西:

编辑:为图例明确添加标签。