python - 从 seaborn swarmplot 获取标绘点的跨度

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) 

然后我使用这段代码在 seaborn 中绘制它:

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

如何获得每个组的 x-span？例如，Parent 组群图的水平跨度范围是多少？

Answer 1

您可以从 swarmplot 创建的 collections 中获取信息。

swarmplot 实际上返回 matplotlib Axes 实例，从那里我们可以找到它创建的 PathCollections。要获取位置，我们可以使用 .get_offsets()。

这是您的示例，经过修改以查找和打印群体限制，然后使用它们在群体周围绘制一个框。

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: