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