[Numpy-discussion] Using normal()
Zachary Pincus
zachary.pincus@yale....
Thu Apr 24 12:28:38 CDT 2008
> The syntax is normal(loc=0.0, scale=1.0, size=None), but I've not
> seen
> what those represent, nor how to properly invoke this function. A
> clue will
> be much appreciated.
>
> I want to call normal() passing at least the width of the curve(at
> the end
> points where y=0.0), and the center (where y=1.0). Being able to
> specify the
> y value of the inflection point (by default 0.5) would also be
> helpful.
The 'normal dstribution' is the Gaussian function:
N(x) = 1/(std*sqrt(2*pi))*exp(-(x-mean)**2/(2*std**2)), where N(x) is
the probability density at position x, given a normal distribution
characterized by 'mean' and 'std' (standard deviation).
http://en.wikipedia.org/wiki/Normal_distribution
Now, numpy.random.normal gives random samples distributed according to
the above probability density function. The only remaining mystery is
how 'loc' and 'scale' -- the parameters of numpy.random.normal -- map
to 'mean' and 'standard deviation', which is how a normal distribution
is usually parameterized. Fortunately, the documentation reveals this:
>>> print numpy.random.normal.__doc__
Normal distribution (mean=loc, stdev=scale).
normal(loc=0.0, scale=1.0, size=None) -> random values
If you need an alternate parameterization of the normal (e.g. in terms
of the y value of the inflection point), just solve that out
analytically from the definition of the normal in terms of mean and std.
However, it looks like you're trying to plot the normal function, not
get random samples. Just evaluate the function (as above) at the x
positions:
mean, std = (0, 1)
x = numpy.linspace(-10, 10, 200) # 200 evenly-spaced points from -10
to 10
y = 1/(std*numpy.sqrt(2*numpy.pi))*numpy.exp(-(x-mean)**2/(2*std**2))
Zach
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