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TA貢獻1829條經驗 獲得超7個贊
不確定,為什么你認為這是 CDF 圖。如果在 X 軸上放置概率,在 Y 軸上繪制x,它看起來像分位數圖。
我沒有在 Python/Matplotlib 中看到內置的此類比例,logit看起來像是最佳近似值。
或者我誤解了你的問題?
代碼,Python 3.8 x64 Win10
import numpy as np
import matplotlib.pyplot as plt
# exponential distribution
def PDF(x, λ):
return λ*np.exp(-λ*x)
def CDF(x, λ):
return 1.0 - np.exp(-λ*x)
def Q(p, λ): # quantile
x = -np.log(1.0 - p)/λ
return x
# plots
λ = 0.04
p = np.linspace(0.01, 0.99, 101)
x = Q(p, λ)
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)
ax.set_xscale('logit')
ax.plot(p, x, 'r-')
plt.show()
你會得到類似的東西
更新
如果您需要概率圖,這里是
代碼
from scipy.stats import expon
rve = expon(loc=0.0, scale=1.0/λ)
p = np.linspace(0.01, 0.99, 101)
x = rve.ppf(p) # Q(p, λ)
q = rve.rvs(size = 10000)
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)
res = stats.probplot(q, dist=rve, plot=ax)
plt.show()
和圖表
好吧,當你發送消息時我已經寫好了代碼。無論如何將其放置在這里,它看起來與您詢問的圖表完全相同 - 與自定義標簽呈線性。
import matplotlib.ticker as ticker
def linear(x0, xn, y0, yn, x):
q = (x-x0)/(xn-x0)
return q * yn + (1.0 - q) * y0
pmin = 0.009
pmax = 0.991
xmin = Q(pmin, λ)
xmax = Q(pmax, λ)
x = np.linspace(xmin, xmax, 2)
p = np.linspace(pmin, pmax, 2)
tick_values = [0.01, 0.50, 0.80, 0.90, 0.96, 0.99] # probabilies
tick_names = [str(v) for v in tick_values] # ticks to place on graph
tick_xvals = [Q(p, λ) for p in tick_values] # x values for each p
tick_places = [linear(xmin, xmax, pmin, pmax, x) for x in tick_xvals] # x from linear function
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)
ax.axes.xaxis.set_major_locator(ticker.FixedLocator((tick_places)))
ax.xaxis.set_major_formatter(ticker.FixedFormatter((tick_names)))
ax.plot(p, x, 'r-')
plt.show()
和圖表本身
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