[SciPy-User] [SciPy-user] ValueError: The truth value of an array with more than one element is ambiguous.
surfcast23
surfcast23@gmail....
Fri Apr 6 15:52:30 CDT 2012
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 <tsyu80@gmail.com> wrote:
>
>>
>>
>> On Thu, Apr 5, 2012 at 6:46 PM, surfcast23 <surfcast23@gmail.com> wrote:
>>
>>> Hi, I have an if statement and what I want it to do is go through arrays
>>> 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<http://mail.python.org/pipermail/python-ideas/2011-October/012278.html>
>> .
>>
>> 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<http://mail.python.org/pipermail/python-dev/2012-March/117510.html>
>> ).
>>
>> 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 evaluates
> 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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