I need to know if all the elements of an array of numpy are equal to a number
It would be like:
numbers = np.zeros(5) # array[0,0,0,0,0]
print numbers.allEqual(0) # return True because all elements are 0
I can make an algorithm but, there is some method implemented in numpy library?
You can break that down into np.all(), which takes a boolean array and checks it's all True, and an equality comparison:
np.all(numbers == 0)
# or equivalently
(numbers == 0).all()
all(numbers==0)
If you want to compare floats, use np.isclose instead:
np.all(np.isclose(numbers, numbers[0]))
The following version checks the equality of all entries of the array without requiring the repeating number.
numbers_0 = np.zeros(5) # array[0,0,0,0,0]
numbers.ptp() == 0.0 # True
# checking an array having some random repeating entry
numbers_r = np.ones((10, 10))*np.random.rand()
numbers_r.ptp() == 0.0 # True
np.array_equal() also works (for Python 3).
tmp0 = np.array([0]*5)
tmp1 = np.array([0]*5)
np.array_equal(tmp0, tmp1)
returns True
[1, 0] it should be False since 1 != 0, but [1, 0] == [1, 0]. To be precise, this is a roundabout way of evaluating a == a, which is true iff a does not contain a nanAre numpy methods necessary? If all the elements are equal to a number, then all the elements are the same. You could do the following, which takes advantage of short circuiting.
numbers[0] == 0 and len(set(numbers)) == 1
This way is faster than using np.all()
timeit or plots of the scaling of your solution compared to numpy might be in order. len(set(numbers)) doesn't short circuit at all, it needs to hash and compare each value to the existing set.