Find slope of function given empirical data.
in [Python]
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From: Thomas on 29 Jun 2010 14:05 Hello all. Trying to find slope of function using numpy. Getting close, but results are a bit off. Hope someone out here can help. import numpy as np def deriv(y): x = list(range(len(y))) x.reverse() # Change from [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] x = np.array(x) # to [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] y = np.array(y) # x.reverse() is used to put point 0 at end of list. z = np.polyfit(x, y, 2) print np.poly1d(z) # Returns: # 2 # 3.142 x - 18.85 x + 35.13 # 2 # Should be closer to 3.142 x - 6.283 + 10 ???????????????????? return [z[0] * 2, z[1]] if __name__=='__main__': # range(-6,5,1) # y = 3.141592 * x ** 2 - 6.283184 * x + 10 for x in range(-6, 5, 1) # 160.796416, 119.95572, 85.398208, 57.12388, 35.132736, 19.424776, 10.0, 6.858408, 10.0, 19.424776, 35.132736 # # y' = 6.283184 * x - 6.283184 for x in range(-6, 5, 1) # -43.982288, -37.699104, -31.41592, -25.132736, -18.849552, -12.566368, -6.283184, 0.0, 6.283184, 12.566368, 18.849552 # z = deriv([160.796416, 119.95572, 85.398208, 57.12388, 35.132736, 19.424776, 10.0, 6.858408, 10.0, 19.424776, 35.132736]) for x in range(-6,5,1): print str(w(x)) + ',' , # Returns: # -56.548656, -50.265472, -43.982288, -37.699104, -31.41592, -25.132736, -18.849552, -12.566368, -6.283184, -1.06581410364e-14, 6.283184 # Should be: # -43.982288, -37.699104, -31.41592, -25.132736, -18.849552, -12.566368, -6.283184, 0.0, 6.283184, 12.566368, 18.849552 # Note that the range is offset by 2 positions
From: duncan smith on 29 Jun 2010 21:33 Thomas wrote: > Hello all. > > Trying to find slope of function using numpy. > Getting close, but results are a bit off. Hope someone out here can > help. > [snip] Why are you generating y-coordinates from the x-coordinates [-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4]? If you're going to use the x-coordinates [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] in your function, then use the same to generate the y-coordinates. Surely, if you have empirical data (which, for some reason, you know are well fitted by a quadratic function?) you'd pass both the x and y coordinates to the function? Maybe (untested), def deriv(x, y): z = np.polyfit(x, y, 2) p = np.poly1d(z) return p.deriv() Duncan
From: Peter Otten on 30 Jun 2010 03:28 Thomas wrote: > Trying to find slope of function using numpy. > Getting close, but results are a bit off. Hope someone out here can > help. You don't make it easy to understand your post. In the future please try to rely more on plain english than on lots of numbers and code that doesn't run. > import numpy as np > > def deriv(y): > x = list(range(len(y))) > x.reverse() # Change from [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] > x = np.array(x) # to [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] > y = np.array(y) # x.reverse() is used to put point 0 at end of > list. > z = np.polyfit(x, y, 2) > print np.poly1d(z) > # Returns: > # 2 > # 3.142 x - 18.85 x + 35.13 > # 2 > # Should be closer to 3.142 x - 6.283 + > 10 ???????????????????? To add one more question mark: how did you find that alternative? Anyway, we can put both polynomials to a test: >>> import numpy as np >>> y = np.array([160.796416, 119.95572, 85.398208, 57.12388, 35.132736,19.424776, 10.0, 6.858408, 10.0, 19.424776, 35.132736]) >>> x = np.arange(len(y), dtype=float)[::-1] >>> p1 = np.poly1d(np.polyfit(x, y, 2)) >>> print p1 2 3.142 x - 18.85 x + 35.13 >>> p2 = np.poly1d([3.142, -6.283, 10.0]) >>> print p2 2 3.142 x - 6.283 x + 10 Now calculate the sum of the squares: >>> np.sum((p1(x)-y)**2) 5.0683524299544787e-26 >>> np.sum((p2(x)-y)**2) 33028.342907811333 Conclusion: numpy's result is much better than what you suggest. Peter
From: Thomas on 30 Jun 2010 09:12 On Jun 30, 3:28 am, Peter Otten <__pete...(a)web.de> wrote: > Thomas wrote: > > Trying to find slope of function using numpy. > > Getting close, but results are a bit off. Hope someone out here can > > help. > > You don't make it easy to understand your post. In the future please try to > rely more on plain english than on lots of numbers and code that doesn't > run. > > > > > > > import numpy as np > > > def deriv(y): > > x = list(range(len(y))) > > x.reverse() # Change from [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] > > x = np.array(x) # to [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] > > y = np.array(y) # x.reverse() is used to put point 0 at end of > > list. > > z = np.polyfit(x, y, 2) > > print np.poly1d(z) > > # Returns: > > # 2 > > # 3.142 x - 18.85 x + 35.13 > > # 2 > > # Should be closer to 3.142 x - 6.283 + > > 10 ???????????????????? > > To add one more question mark: how did you find that alternative? > > Anyway, we can put both polynomials to a test: > > >>> import numpy as np > >>> y = np.array([160.796416, 119.95572, 85.398208, 57.12388, > > 35.132736,19.424776, 10.0, 6.858408, 10.0, 19.424776, 35.132736])>>> x = np.arange(len(y), dtype=float)[::-1] > >>> p1 = np.poly1d(np.polyfit(x, y, 2)) > >>> print p1 > > 2 > 3.142 x - 18.85 x + 35.13>>> p2 = np.poly1d([3.142, -6.283, 10.0]) > >>> print p2 > > 2 > 3.142 x - 6.283 x + 10 > > Now calculate the sum of the squares: > > >>> np.sum((p1(x)-y)**2) > > 5.0683524299544787e-26>>> np.sum((p2(x)-y)**2) > > 33028.342907811333 > > Conclusion: numpy's result is much better than what you suggest. > > Peter As usual, thanks to all for putting me on the right track. Kind regards.
From: Grant Edwards on 30 Jun 2010 10:23
On 2010-06-29, Thomas <thom1948(a)gmail.com> wrote: > Trying to find slope of function using numpy. Getting close, but > results are a bit off. Hope someone out here can help. > > import numpy as np > > def deriv(y): > x = list(range(len(y))) > x.reverse() # Change from [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] > x = np.array(x) # to [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] > y = np.array(y) # x.reverse() is used to put point 0 at end of > list. > z = np.polyfit(x, y, 2) > print np.poly1d(z) > # Returns: > # 2 > # 3.142 x - 18.85 x + 35.13 > # 2 > # Should be closer to 3.142 x - 6.283 + > 10 ???????????????????? Numpy's answer is correct. I've no idea where you got your answer. Perhaps the commented-out stuff was meant to convey that. If so, it went past me. Here's the least-squares fit done by gnuplot, the resulting fuction f(x) plotted against the data, as well as your "should be" function plotted against the data. As you can see, your "should be" function is way off. For prettier results, you can change the gnuplot script to use a different output format -- I just used the dumb terminal so I could post it here. In any case, the values returned by numpy and gnuplot hit the data points pretty much exactly -- far better than your "should be" results. NB: You might want to grab a copy of gnuplot (or a similar) tool to use when tinkering with data analysis and visualisation. It's very handy. ------------------------------foo.gp------------------------------ #!/usr/bin/gnuplot f(x) = a*x**2 + b*x + c fit f(x) "foo.dat" via a,b,c set xra[-1:11] set term dumb 120 40 plot f(x), "foo.dat" plot 3.142*x**2 + 6.283*x + 10, "foo.dat" ------------------------------foo.gp------------------------------ ------------------------------foo.dat------------------------------ 10 160.796416 9 119.95572 8 85.398208 7 57.12388 6 35.132736 5 19.424776 4 10.0 3 6.858408 2 10.0 1 19.424776 0 35.132736 ------------------------------foo.dat------------------------------ --------------------results of running foo.gp-------------------- Iteration 0 [...] Iteration 1 [...] Iteration 2 [...] Iteration 3 [...] Iteration 4 [...] Iteration 5 [...] Iteration 6 [...] Iteration 7 [...] ************************** After 8 iterations the fit converged. final sum of squares of residuals : 7.70717e-28 rel. change during last iteration : 0 degrees of freedom (FIT_NDF) : 8 rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 9.81528e-15 variance of residuals (reduced chisquare) = WSSR/ndf : 9.63396e-29 Final set of parameters Asymptotic Standard Error ======================= ========================== a = 3.14159 +/- 3.351e-16 (1.067e-14%) b = -18.8496 +/- 3.479e-15 (1.846e-14%) c = 35.1327 +/- 7.478e-15 (2.128e-14%) [...] 250 ++-------+-----------------+-----------------+------------------+-----------------+-----------------+-------++ | + + + + + f(x) ****** | | "foo.dat" A | | | | | | | | * 200 ++ *+ | * | | *** | | * | | * | | * | | *A | 150 ++ * ++ | * | | ** | | ** | | ** | | *A | | ** | | * | 100 ++ ** ++ | *** | | **A | | ** | | ** | | ** | ** **A | 50 ++**** *** ++ | ** ** | | *A** **A* | | *** **** | | ***A* *A** | | ******** ******** | | + A********A********A + + + | 0 ++-------+-----------------+-----------------+------------------+-----------------+-----------------+-------++ 0 2 4 6 8 10 500 ++-------+-----------------+-----------------+------------------+-----------------+-----------------+-------++ | + + + + 3.142*x**2 + 6.283*x + 10 ****** | | "foo.dat" A | | * 450 ++ **+ | * | | *** | 400 ++ ** ++ | * | | ** | | ** | 350 ++ *** ++ | * | | ** | 300 ++ ** ++ | ** | | ** | | *** | 250 ++ ** ++ | ** | | ** | | **** | 200 ++ ** ++ | ** | | *** A | 150 ++ **** ++ | ** | | *** A | | **** | 100 ++ **** ++ | **** A | | ***** | 50 ++ **** A ++ | A ******* A | | ******* | | ****************A* A A A + + + | 0 **-------+-----------------+--------A--------+------------------+-----------------+-----------------+-------++ 0 2 4 6 8 10 --------------------results of running foo.gp-------------------- -- Grant Edwards grant.b.edwards Yow! Mary Tyler Moore's at SEVENTH HUSBAND is wearing gmail.com my DACRON TANK TOP in a cheap hotel in HONOLULU! |