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Let x be a NumPy array. The following:
(x > 1) and (x < 3)
Gives the error message:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
How do I fix this?
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Use a. any() or a. all() if you try to get the truth value of a numpy array with more than one element. To solve this error, you can use the any() function if you want at least one element in the array to return True or the all() function if you want all the elements in the array to return True.
The Python any() and all() functions evaluate the items in a list to see which are true. The any() method returns true if any of the list items are true, and the all() function returns true if all the list items are true.
Compare Two Arrays in Python Using the numpy. array_equiv() Method. The numpy. array_equiv(a1, a2) method takes array a1 and a2 as input and returns True if both arrays' shape and elements are the same; otherwise, returns False .
If a and b are Boolean NumPy arrays, the & operation returns the elementwise-and of them:
a & b
That returns a Boolean array. To reduce this to a single Boolean value, use either
(a & b).any()
or
(a & b).all()
Note: if a and b are non-Boolean arrays, consider (a - b).any() or (a - b).all() instead.
The NumPy developers felt there was no one commonly understood way to evaluate an array in Boolean context: it could mean True if any element is True, or it could mean True if all elements are True, or True if the array has non-zero length, just to name three possibilities.
Since different users might have different needs and different assumptions, the
NumPy developers refused to guess and instead decided to raise a ValueError whenever one tries to evaluate an array in Boolean context. Applying and to two numpy arrays causes the two arrays to be evaluated in Boolean context (by calling __bool__ in Python3 or __nonzero__ in Python2).
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