计算Numpy(或Scipy)中的坡度 [英] Calculating Slopes in Numpy (or Scipy)

问题描述

我正在尝试找到使用Numpy和Scipy计算坡度的最快，最有效的方法.我有一个包含三个Y变量和一个X变量的数据集，我需要计算它们各自的斜率.例如，如下所示，我可以轻松地一次完成这一行，但是我希望有一种更有效的方法.我也不认为linregress是最好的方法，因为我的结果中不需要任何辅助变量，例如拦截，标准错误等.任何帮助是极大的赞赏.

I am trying to find the fastest and most efficient way to calculate slopes using Numpy and Scipy. I have a data set of three Y variables and one X variable and I need to calculate their individual slopes. For example, I can easily do this one row at a time, as shown below, but I was hoping there was a more efficient way of doing this. I also don't think linregress is the best way to go because I don't need any of the auxiliary variables like intercept, standard error, etc in my results. Any help is greatly appreciated.

    import numpy as np
    from scipy import stats

    Y = [[  2.62710000e+11   3.14454000e+11   3.63609000e+11   4.03196000e+11
        4.21725000e+11   2.86698000e+11   3.32909000e+11   4.01480000e+11
        4.21215000e+11   4.81202000e+11]
        [  3.11612352e+03   3.65968334e+03   4.15442691e+03   4.52470938e+03
        4.65011423e+03   3.10707392e+03   3.54692896e+03   4.20656404e+03
        4.34233412e+03   4.88462501e+03]
        [  2.21536396e+01   2.59098311e+01   2.97401268e+01   3.04784552e+01
        3.13667639e+01   2.76377113e+01   3.27846013e+01   3.73223417e+01
        3.51249997e+01   4.42563658e+01]]
    X = [ 1990.  1991.  1992.  1993.  1994.  1995.  1996.  1997.  1998.  1999.] 
    slope_0, intercept, r_value, p_value, std_err = stats.linregress(X, Y[0,:])
    slope_1, intercept, r_value, p_value, std_err = stats.linregress(X, Y[1,:])
    slope_2, intercept, r_value, p_value, std_err = stats.linregress(X, Y[2,:])
    slope_0 = slope/Y[0,:][0]
    slope_1 = slope/Y[1,:][0]
    slope_2 = slope/Y[2,:][0]
    b, a = polyfit(X, Y[1,:], 1)
    slope_1_a = b/Y[1,:][0]

推荐答案

在一个维度上，线性回归计算是

The linear regression calculation is, in one dimension, a vector calculation. This means we can combine the multiplications on the entire Y matrix, and then vectorize the fits using the axis parameter in numpy. In your case that works out to the following

((X*Y).mean(axis=1) - X.mean()*Y.mean(axis=1)) / ((X**2).mean() - (X.mean())**2)

您对拟合质量参数不感兴趣，但是大多数参数都可以类似的方式获得.

You're not interested in fit quality parameters but most of them can be obtained in a similar manner.

这篇关于计算Numpy(或Scipy)中的坡度的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！