我正在将seaborn.catplot
与kind='point'
一起使用以绘制数据。我想使用与seaborn的方法相同为每个色调变量和每个类别计算平均值(SEM)的标准误差,以确保我的计算值与绘制的误差线完全匹配。用于计算SEM和95%置信区间(CIs)的默认解决方案包含自举算法,其中将平均值自举1000次以计算SEM / CI。在earlier post中,我看到了一种可能为此提供功能的方法(使用诸如seaborn.utils.ci()
和seaborn.algorithms.bootstrap()
之类的源代码),但是我不确定如何实现它。由于自举使用随机抽样,因此也有必要确保产生相同的1000均值阵列以进行绘图和获得SEM。
这里是一个代码示例:
import numpy as np
import pandas as pd
import seaborn as sns
# simulate data
rng = np.random.RandomState(42)
measure_names = np.tile(np.repeat(['Train BAC','Test BAC'],10),2)
model_numbers = np.repeat([0,1],20)
measure_values = np.concatenate((rng.uniform(low=0.6,high=1,size=20),
rng.uniform(low=0.5,high=0.8,size=20)
))
folds=np.tile([1,2,3,4,5,6,7,8,9,10],4)
plot_df = pd.DataFrame({'model_number':model_numbers,
'measure_name':measure_names,
'measure_value':measure_values,
'outer_fold':folds})
# plot data as pointplot
g = sns.catplot(x='model_number',
y='measure_value',
hue='measure_name',
kind='point',
seed=rng,
data=plot_df)
产生:
我想获得两个模型的所有训练和考试分数的SEM。那是:
# obtain SEM for each score in each model using the same method as in sns.catplot
model_0_train_bac = plot_df.loc[((plot_df['model_number'] == 0) & (plot_df['measure_name'] == 'Train BAC')),'measure_value']
model_0_test_bac = plot_df.loc[((plot_df['model_number'] == 0) & (plot_df['measure_name'] == 'Test BAC')),'measure_value']
model_1_train_bac = plot_df.loc[((plot_df['model_number'] == 1) & (plot_df['measure_name'] == 'Train BAC')),'measure_value']
model_1_test_bac = plot_df.loc[((plot_df['model_number'] == 1) & (plot_df['measure_name'] == 'Test BAC')),'measure_value']
我不确定我是否要求您抽取完全相同的样本。根据定义,引导程序是通过随机抽样来进行的,因此从一次运行到下一次运行会有一定的可变性(除非我弄错了)。
您可以按照seaborn的计算方式来计算CI:
# simulate data
rng = np.random.RandomState(42)
measure_names = np.tile(np.repeat(['Train BAC','Test BAC'],10),2)
model_numbers = np.repeat([0,1],20)
measure_values = np.concatenate((rng.uniform(low=0.6,high=1,size=20),
rng.uniform(low=0.5,high=0.8,size=20)
))
folds=np.tile([1,2,3,4,5,6,7,8,9,10],4)
plot_df = pd.DataFrame({'model_number':model_numbers,
'measure_name':measure_names,
'measure_value':measure_values,
'outer_fold':folds})
x_col = 'model_number'
y_col = 'measure_value'
hue_col = 'measure_name'
ci = 95
est = np.mean
n_boot = 1000
for gr,temp_df in plot_df.groupby([hue_col,x_col]):
print(gr,est(temp_df[y_col]), sns.utils.ci(sns.algorithms.bootstrap(temp_df[y_col], func=est,
n_boot=n_boot,
units=None,
seed=rng)))
哪个输出:
('Test BAC', 0) 0.7581071363371585 [0.69217109 0.8316217 ]
('Test BAC', 1) 0.6527812067134964 [0.59523784 0.71539669]
('Train BAC', 0) 0.8080546943810699 [0.73214414 0.88102816]
('Train BAC', 1) 0.6201161718490218 [0.57978654 0.66241543]
注意,如果您第二次运行循环,则会获得相似但不完全相同的配置项。
如果您确实想获得seaborn在绘图中使用的确切值(请注意,如果您再次绘制相同的数据,这些值将略有不同,那么您可以直接从Line2D艺术家过去常常绘制错误栏:
g = sns.catplot(x=x_col,
y=y_col,
hue=hue_col,
kind='point',
ci=ci,
estimator=est,
n_boot=n_boot,
seed=rng,
data=plot_df)
for l in g.ax.lines:
print(l.get_data())
输出:
(array([0., 1.]), array([0.80805469, 0.62011617]))
(array([0., 0.]), array([0.73203808, 0.88129836])) # <<<<
(array([1., 1.]), array([0.57828366, 0.66300033])) # <<<<
(array([0., 1.]), array([0.75810714, 0.65278121]))
(array([0., 0.]), array([0.69124145, 0.83297914])) # <<<<
(array([1., 1.]), array([0.59113739, 0.71572469])) # <<<<