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POCalculating Slopes in Numpy (or Scipy)
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copied!<p>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. </p> <pre><code> 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] </code></pre>

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