On Thu, 24 Apr 2008, Robert Kern wrote:
> No, it's not! scale is the standard deviation of the Normal distribution,
> not half the FWHM!
OK. Thanks for clearing up my mis-understanding.
Rich
--
Richard B. Shepard, Ph.D. | IntegrityCredibility
Applied Ecosystem Ser
On Thu, Apr 24, 2008 at 1:39 PM, Rich Shepard <[EMAIL PROTECTED]> wrote:
> On Thu, 24 Apr 2008, Joris De Ridder wrote:
>
> > Scale is half the width between the inflection points, mind the factor of
> > 2.
>
> Joris,
>
>So, half of the full width at half the maximum height. Thank you.
No, i
On Thu, 24 Apr 2008, Zachary Pincus wrote:
> I assume you 'from numpy import *'? This is why it works -- because that
> import causes the python built-in exp() to be replaced (in the current
> namespace) by numpy.exp().
The equivalent (I think): import numpy as nx.
Rich
--
Richard B. Shepar
On 24/04/2008, Keith Goodman <[EMAIL PROTECTED]> wrote:
> On Thu, Apr 24, 2008 at 10:01 AM, Rich Shepard <[EMAIL PROTECTED]> wrote:
> >In the application's function I now have:
> >
> >from numpy import *
> >
> >x = nx.arange(0, 100, 0.1)
> >y = nx.normal(center,width) # passe
On Thu, 24 Apr 2008, Zachary Pincus wrote:
> Python's built in pow() and exp() functions can't handle numpy arrays, and
> thus try (and fail) to convert arrays to scalar values. You want to use
> numpy.exp and numpy.power (or just the ** operator), to do these
> operations to numpy arrays elementw
On Thu, 24 Apr 2008, Joris De Ridder wrote:
> Scale is half the width between the inflection points, mind the factor of
> 2.
Joris,
So, half of the full width at half the maximum height. Thank you.
Rich
--
Richard B. Shepard, Ph.D. | IntegrityCredibility
Applied
On 24 Apr 2008, at 19:26, Rich Shepard wrote:
>> norm = 1 / (scale * sqrt(2 * pi))
>> y = norm * exp(-power((x - loc), 2) / (2 * scale**2))
>
> Can do. So, scale would equate to width and loc to center, yes?
Scale is half the width between the inflection points, mind the factor
of 2.
J.
D
> It works for me:
>
>>> x = arange(0,10)
>>> scale=1
>>> loc=1
>>> norm = 1 / (scale * sqrt(2 * pi))
>>> y = norm * exp(-power((x - loc), 2) / (2 * scale**2))
>>> y
>
> array([ 1.46762663e-01, 3.98942280e-01, 1.46762663e-01,
> 5.39909665e-02, 2.68805194e-03, 1.33830226e-04,
>
On Thu, Apr 24, 2008 at 10:51 AM, Rich Shepard <[EMAIL PROTECTED]> wrote:
> On Thu, 24 Apr 2008, Keith Goodman wrote:
>
> > norm = 1 / (scale * sqrt(2 * pi))
> > y = norm * exp(-power((x - loc), 2) / (2 * scale**2))
>
>Hmm-m-m. I don't understand the source of the error:
>
>y = norm * exp
>> norm = 1 / (scale * sqrt(2 * pi))
>> y = norm * exp(-power((x - loc), 2) / (2 * scale**2))
>
> Hmm-m-m. I don't understand the source of the error:
>
> y = norm * exp(-pow((x - loc), 2) / (2 * scale**2))
> TypeError: only length-1 arrays can be converted to Python scalars
Python's built in
On Thu, 24 Apr 2008, Keith Goodman wrote:
> norm = 1 / (scale * sqrt(2 * pi))
> y = norm * exp(-power((x - loc), 2) / (2 * scale**2))
Hmm-m-m. I don't understand the source of the error:
y = norm * exp(-pow((x - loc), 2) / (2 * scale**2))
TypeError: only length-1 arrays can be converted to
On Thu, 24 Apr 2008, Zachary Pincus wrote:
> 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:
>
> >
> 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
On Thu, 24 Apr 2008, Keith Goodman wrote:
> It's random.normal(loc, scale). But that will give you random x values
> drawn from the normal distribution, not y values at your x. So you'll have
> to code your own normal, which will probably look something like
Keith,
I overlooked that, thanks.
On Thu, Apr 24, 2008 at 10:01 AM, Rich Shepard <[EMAIL PROTECTED]> wrote:
>In the application's function I now have:
>
>from numpy import *
>
>x = nx.arange(0, 100, 0.1)
>y = nx.normal(center,width) # passed to the function when called
>
> and I then pass x,y to PyX for plotting.
Rereading "Guide to NumPy" once again, I saw what I had missed all the
previous times: the normal() distribution function (Chapter 10, page 173).
I have several questions on using it in my application.
The syntax is normal(loc=0.0, scale=1.0, size=None), but I've not seen
what those represen
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