[SciPy-User] [SciPy-user] ValueError: The truth value of an array with more than one element is ambiguous.
Fri Apr 6 16:11:58 CDT 2012
> 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 <email@example.com> wrote:
>>> On Thu, Apr 5, 2012 at 6:46 PM, surfcast23 <firstname.lastname@example.org> 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
>>>> 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
>>> 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
>>> 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
>>> Hope that helps,
>> 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
>> 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.
>> SciPy-User mailing list
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