我有两个数组,
a = array([
[ 0.93825418, 0.60731973, 0.44218921, 0.90888805, 0.97695114],
[ 0.27422807, 0.75870153, 0.12154102, 0.89137678, 0.04257262],
[ 0.32855867, 0.17215507, 0.00302302, 0.95395069, 0.02596567],
[ 0.18385244, 0.09108341, 0.27925367, 0.0177183 , 0.41035188],
[ 0.87229432, 0.73573982, 0.98554476, 0.72321398, 0.98316711],
[ 0.16474265, 0.5308054 , 0.27913615, 0.59107689, 0.6480463 ],
[ 0.88356436, 0.22343885, 0.74900285, 0.43895017, 0.74993129],
[ 0.08097611, 0.48984607, 0.33991052, 0.06431022, 0.10753135],
[ 0.67351561, 0.13165046, 0.41327765, 0.21768539, 0.7337069 ],
[ 0.65609999, 0.06241059, 0.3400624 , 0.13234171, 0.23679716]
])
b = array([
[False, True, True, False, False],
[ True, …Run Code Online (Sandbox Code Playgroud) 我正在使用正交距离回归方法(scipy.odr)来拟合我的数据,拟合后,我在计算95%置信区间时遇到困难,请帮助我不知道如何计算它〜
这里是代码:
#define fit function
def def f(p,x):
return p*x
linear = scipy.odr.Model(f)
mydata = scipy.odr.RealData(x, y, sx=dx, sy=dy)
myodr = scipy.odr.ODR(mydata, linear, beta0=[1.])
myoutput = myodr.run()
myoutput.pprint()
[solpe] = myoutput.beta
Run Code Online (Sandbox Code Playgroud)
现在我得到了 solpe,得到拟合函数:y=p*x,但是我应该如何得到 95% 置信区间?
我试图用windrose的4个子图制作一个图,但是我意识到windrose仅具有这样的轴:ax = WindroseAxes.from_ax()那么,我如何用windrose绘制一个子图?