[SciPy-User] [SciPy-user] ValueError: The truth value of an array with more than one element is ambiguous.

surfcast23 surfcast23@gmail....
Fri Apr 6 21:16:33 CDT 2012


Thanks Zachary

Zachary Pincus-2 wrote:
> 
>> Thanks for the help. Would I be able to use the np.any and np.all
>> functions to count the number of true occurrences? 
> 
> I typically use np.sum() (or arr.sum() where arr is a numpy array) to
> count the number of True values (which count as 1 in boolean arrays, where
> Falses are 0.)
> 
>> 
>> 
>> 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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>>> 
>>> 
>> 
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