从seaborn swarmplot获取绘制点的跨度

Question

我有以下数据：

import pandas as pd
import numpy as np

# Generate dummy data.
a = np.random.random(75)
b = np.random.random(75) - 0.6
c = np.random.random(75) + 0.75

# Collate into a DataFrame
df = pd.DataFrame({'a': a, 'b': b, 'c': c}) 
df.columns = [list(['WT', 'MUT', 'WTxMUT']), list(['Parent', 'Parent', 'Offspring'])]
df.columns.names = ['Genotype', 'Status']
df_melt = pd.melt(df)

我使用以下代码将其绘制在seaborn中：

import seaborn as sb
sb.swarmplot(data = df_melt, x = "Status", y = "value", hue = "Genotype")

如何获得每个组的X跨度？例如，父组的swarmplot的水平跨度的范围是多少？

Answer 1

您可以从中collections创建的信息swarmplot。

swarmplot实际返回matplotlibAxes实例，然后从那里可以找到PathCollections它创建的实例。要获取职位，我们可以使用.get_offsets()。

这是您的示例，进行了修改以查找并打印群限制，然后使用它们在群周围绘制方框。

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sb
from matplotlib.patches import Rectangle

# Generate dummy data.
a = np.random.random(75)
b = np.random.random(75) - 0.6
c = np.random.random(75) + 0.75

# Collate into a DataFrame
df = pd.DataFrame({'a': a, 'b': b, 'c': c}) 
df.columns = [list(['WT', 'MUT', 'WTxMUT']), list(['Parent', 'Parent', 'Offspring'])]
df.columns.names = ['Genotype', 'Status']
df_melt = pd.melt(df)

ax = sb.swarmplot(data = df_melt, x = "Status", y = "value", hue = "Genotype")

def getdatalim(coll):
    x,y = np.array(coll.get_offsets()).T
    try:
        print 'xmin={}, xmax={}, ymin={}, ymax={}'.format(
                x.min(), x.max(), y.min(), y.max())
        rect = Rectangle((x.min(),y.min()),x.ptp(),y.ptp(),edgecolor='k',facecolor='None',lw=3)
        ax.add_patch(rect)
    except ValueError:
        pass

getdatalim(ax.collections[0]) # "Parent"
getdatalim(ax.collections[1]) # "Offspring"

plt.show()

打印：

xmin=-0.107313729132, xmax=0.10661092707, ymin=-0.598534246847, ymax=0.980441247759
xmin=0.942829146473, xmax=1.06105941656, ymin=0.761277608688, ymax=1.74729717464

这是图：