问题描述
我正在使用 Seaborn 绘制相对频率直方图。因为我还没有找到一种方法来保存与最高峰相关的值，所以我使用 stats.relfreq 来做到这一点。然而相对频率似乎并不匹配。
我在 Jupyter Notebook 中使用 Python。
我的数据：
my_data = [0.9995,0.9995,-0.0803,-0.7736,0.9418,0.3612,0.5023,0.9686,0.5574,0.8629,0.5226,0.9947,-0.8391,-0.4767,0.4215,0.8176,0.5106,-0.0772,0.0865,-0.6739,-0.5574,-0.6776,0.4588,-0.2263,0.8224,0.3804,-0.0516,-0.3818,0.0325,0.6341,0.0516,-0.5859,-0.5106,-0.0258,0.128,0.8126,-0.4201,-0.2449,-0.4215,-0.3506,-0.872,0.7506,-0.5719,0.7003,-0.235,0.1747,0.5994,0.5423,-0.25,0.8834,0.1761,-0.7691,0.6249,0.7819,-0.34700000000000003,-0.6486,0.2955,0.6486,0.1734,-0.2732,-0.6049,-0.8622,0.4404,0.25,0.5519,0.5583,-0.1027,0.4939,-0.2144,0.2247,0.9079,-0.7273,-0.4329,0.2263,-0.5423,-0.7362,0.34,-0.6115,-0.5994,-0.6697,0.9201,0.1027,0.5922,0.3822,0.5667,0.8316,0.9679,0.29600000000000004,0.3169,-0.9413,0.6478,0.29600000000000004]
我的代码：
from scipy import stats
import seaborn as sns
# Calculate relative frequency of values,using 10 bins.
res = stats.relfreq(points,numbins = 10)
relative_frequency = res.frequency
print(relative_frequency)
#find highest value and corresponding index
highest_val = np.max(relative_frequency)
highest_index = np.where(relative_frequency == highest_val)
highest_index = int(highest_index[0])
print(highest_index)
# Ordered list with possible scores associated to each frequency bin
possible_scores = [-0.9,-0.7,-0.5,-0.3,-0.1,0.1,0.3,0.5,0.7,0.9]
averaged_relative_frequency_score = possible_scores[highest_index]
print(averaged_relative_frequency_score)
# Plot histogram with Seaborn
ax = sns.histplot(data = date_result['score'],stat = 'probability',bins = 10,binwidth = 0.2,binrange = [-1,1])
plt.xlim(-1,1)
plt.show()
以下是我得到的不同输出。
print(relative_frequency)
#relative_frequency [0.0610687 0.06870229 0.09923664 0.07633588 0.04580153 0.08396947
0.16793893 0.17557252 0.07633588 0.14503817]
print(highest_index)
# highest index = 7
print(averaged_relative_frequency_score)
# averaged_relative_frequency_score = 0.5
还有 Seaborn 情节：
如您所知，如果一切正常，Seaborn 图中的相应指数在使用 stats 模块计算的频率中将为 9。与 Seaborn 相比，stats.relfreq 中 bin 的大小是否不同？
我是否误解了任何明显的内容？我似乎无法理解为什么这两种方法会得到不同的峰值。
呸！
解决方法
刚写完这篇我就知道哪里出了问题。
stats.relfreq 中的 bin 默认有点 oversized。
要获得相同的结果，您必须使用 defaultreallimits 参数指定直方图的范围。
在代码中实现：
res = stats.relfreq(points,numbins = 10,defaultreallimits = [-1,1])