如何计算numpy的坡度

Question

如果我有50个元素的数组,我将如何计算3个周期斜率和5个周期斜率？文档不添加太多.....

>>> from scipy import stats
>>> import numpy as np
>>> x = np.random.random(10)
>>> y = np.random.random(10)
>>> slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)

这会有用吗？

def slope(x, n): 
   if i<len(x)-n: 
        slope = stats.linregress(x[i:i+n],y[i:i+n])[0]
        return slope 

但是数组的长度是相同的

@joe :::

xx = [2.0 ,4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30]
x  = np.asarray(xx, np.float)
s  = np.diff(x[::3])/3

window  = [1, 0, 0, 0,  -1]
window2 = [1, 0,  -1]
slope   = np.convolve(x, window, mode='same') / (len(window) - 1)
slope2  = np.convolve(x, window2, mode='same') / (len(window2) - 1)

print x
print s

print slope
print slope2

结果.....

[  2.   4.   6.   8.  10.  12.  14.  16.  18.  20.  22.  24.  26.  28.  30.]
[ 2.  2.  2.  2.]
[ 1.5  2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2.  -6.  -6.5]
[  2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2.   2. -14.]

斜率和斜率2是Im之后的除了-6,-6.5和-14不是我正在寻找的结果.

这工作.......

window    = [1, 0, 0, -1]
slope     = np.convolve(xx, window, mode='valid') / float(len(window) - 1)
padlength = len(window) -1
slope     = np.hstack([np.ones(padlength), slope])
print slope

Answer 1

我假设你的意思是计算每个第3和第5个元素的斜率,这样你就有了一系列(精确的,非最小二乘)斜率？

如果是这样,你只需要做一些事情:

third_period_slope = np.diff(y[::3]) / np.diff(x[::3])
fifth_period_slope = np.diff(y[::5]) / np.diff(x[::5])

不过,我可能完全误解了你的意思.我以前从未领过"3期斜率"一词......

如果你想要更多的"移动窗口"计算(这样输入元素的数量与输出元素相同),只需将其建模为带有[-1, 0, 1]或的窗口的卷积[-1, 0, 0, 0, 1].

例如

window = [-1, 0, 1]
slope = np.convolve(y, window, mode='same') / np.convolve(x, window, mode='same')