[SciPy-User] [SciPy-user] ValueError: The truth value of an array with more than one element is ambiguous.
Fri Apr 6 16:00:21 CDT 2012
Would I be able to use the np.any and np.all
functions to count the number of true occurrences?
You can count the number of true occurences with np.sum():
>>> a = np.array((True, False, False, True, True))
On Fri, Apr 6, 2012 at 3:52 PM, surfcast23 <email@example.com> wrote:
> Hi Tony,
> Thanks for the help. Would I be able to use the np.any and np.all
> functions to count the number of true occurrences?
> Tony Yu-3 wrote:
> > On Fri, Apr 6, 2012 at 12:54 AM, Tony Yu <firstname.lastname@example.org> wrote:
> >> On Thu, Apr 5, 2012 at 6:46 PM, surfcast23 <email@example.com>
> >>> Hi, I have an if statement and what I want it to do is go through
> >>> and find the common elements in all three arrays. When I try the code
> >>> below
> >>> I get this error * ValueError: The truth value of an array with more
> >>> than one element is ambiguous. Use a.any() or a.all()* Can some one
> >>> explain the error to me and how I might be able to fix it. Thanks in
> >>> advance. *if min <= Xa <=max & min <= Ya <=max & min <= Za <=max:
> >>> print("in range") else: print("Not in range")*
> >> This explanation may or may not be clear, but your question is answered
> >> in this
> >> communication<
> >> .
> >> Roughly:
> >> 1) Python's default behavior for chained comparisons don't work as you'd
> >> expect for numpy arrays.
> >> 2) Python doesn't allow numpy to change this default behavior (at least
> >> currently, and maybe
> >> never<
> >> ).
> >> Nevertheless, you can get around this by separating the comparisons
> >> >>> if (min <= Xa) & (Xa <= max):
> >> Note the use of `&` instead of `and`, which is at the heart of the
> >> issue<http://www.python.org/dev/peps/pep-0335/>
> >> .
> >> Hope that helps,
> >> -Tony
> > Oops, I think I got myself mixed up in the explanation. Separating the
> > comparisons fixes one error; For example, the following:
> >>>> (min <= Xa) & (Xa <= max)
> > will return an array of bools instead of raising an error (as you would
> > get
> > with `min <= Xa <= max`). This is what I meant to explain above.
> > But, throwing an `if` in front of that comparison still doesn't work
> > because it's ambiguous: Should `np.array([True False])` be true or false?
> > Instead you should check `np.all(np.array([True False]))`, which
> > as False since not-all elements are True, or `np.any(np.array([True
> > False]))`, which evaluates as True since one element is True.
> > -Tony
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