[SciPy-User] Heaviside function in vector form

[Michael Sarahan]

I think I'd do this with indexing:

def Heaviside(x):
    mult = -100.0; diff = mult*(x - Theta_syn);
    output = np.ones_like(x)
    output[x<Theta_syn] = 0
    output[abs(diff < 50)] = 1.0/(1 + np.exp(diff[abs(diff < 50)]))
    return output


On Sat, Jul 5, 2014 at 8:26 PM, Michael Sarahan <msarahan at gmail.com> wrote:

> PS:
>
>
> if abs(diff) < 50: #if x is close to Theta_syn
>
> will also fail with this error if x is a vector.
>
> The error message suggests any() or all(), which will return a single
> value representing if any or all of the values are true (respectively) -
> you might also consider min() or max() as sort of threshold settings, or a
> sum or mean of the diff vector (it will be a vector if x is a vector).
>
>
> On Sat, Jul 5, 2014 at 8:15 PM, Michael Sarahan <msarahan at gmail.com>
> wrote:
>
>> when you say
>>
>> if x < Theta_syn:
>>
>> you are trying to figure out the truth value of a vector.  That's what
>> Numpy is telling you.  It doesn't make sense - only the truth value of a
>> single value makes sense for an if statement.  Instead, you probably want a
>> sum or something similar.
>>
>> HTH.
>> Mike
>>
>>
>> On Sat, Jul 5, 2014 at 8:02 PM, Barrett B <barrett.n.b at gmail.com> wrote:
>>
>>> I am trying to code the Heaviside function in vector form. There are
>>> several different versions of this, but essentially, f(x) = 1 whenever x >=
>>> 0, and f(x) = 0 whenever x < 0. Here is what I have so far (keep in mind, x
>>> is a vector here):
>>>
>>> =========
>>>
>>> def Heaviside(x):
>>>     mult = -100.0; diff = mult*(x - Theta_syn);
>>> #    print x
>>>     if abs(diff) < 50: #if x is close to Theta_syn
>>>         return 1.0/(1 + np.exp(diff))
>>>     if x < Theta_syn:
>>>         return 0
>>>     return 1 #otherwise
>>> #    return 1.0/(1 + np.exp(diff))
>>>
>>> =========
>>>
>>> which produces the following error:
>>>
>>> ValueError: The truth value of an array with more than one element is
>>> ambiguous. Use a.any() or a.all()
>>>
>>> BTW, if I were to comment out every line except the first and last, it
>>> runs fine.
>>>
>>> See what I'm trying to do? Basically, I want to check whether current
>>> value of diff is not very large. But I don't know how to do this without
>>> rewriting the code to include for loops, which is what I've been trying to
>>> get away from all along?
>>>
>>> _______________________________________________
>>> SciPy-User mailing list
>>> SciPy-User at scipy.org
>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>>>
>>>
>>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.scipy.org/pipermail/scipy-user/attachments/20140705/58f0eb98/attachment.html>