On 2010-06-29, Thomas <thom1948 at 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--------------------
--
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